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Time: 03 Hours
COURSE STRUCTURE
CLASS IX (2025-26)
(Annual Examination)
Unit I: Matter-Nature and Behaviour
Theme: Materials
Matter in Our Surroundings: Definition of matter; Particulate Nature of Matter; States of Matter: solid, liquid and gas and their characteristics; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation.
Is Matter Around Us Pure: Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Physical and chemical changes (excluding separating the components of a mixture); Pure and Impure substances.
Atoms and Molecules: Atoms and molecules, Law of Chemical Combination, Chemical formula of common compounds, Atomic and molecular masses.
Structure of atom: Sub-atomic particles: Electrons, protons and neutrons, Models of atom; Valency, Atomic Number and Mass Number, Isotopes and Isobars.
Theme: The World of the Living
Unit II: Organization in the Living World
Cell - Basic Unit of life: Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes - basic structure, number.
Tissues, Organs, Organ System, Organism:
Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).
The following topics are included in the syllabus but will be assessed only formatively to reinforce understanding without adding to summative assessments. The reduces academic stress while ensuring meaningful learning. Schools can integrate these with existing chapters as they align well. Relevant NCERT textual material is enclosed for reference.
Health and Diseases: Health and its failure. Infectious and Non-infectious diseases, their causes and manifestation. Diseases caused by microbes (Virus, Bacteria and Protozoans) and their prevention; Principles of treatment and prevention. Pulse Polio programmes.
Unit III: Motion, Force and Work
Theme: Moving Things, People and Ideas
Motion: Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, elementary idea of uniform circular motion.
Force and Newton’s laws: Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration.
The following topic is included in the syllabus but will be assessed only formatively to reinforce understanding without adding to summative assessments. The reduces academic stress while ensuring meaningful learning. Schools can integrate this with existing chapters as they align well. Relevant NCERT textual material is enclosed for reference.
Elementary idea of conservation of Momentum
Gravitation: Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall.
Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy.
Work, Energy and Power: Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy (excluding commercial unit of Energy).
Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.
Theme: Food
Unit IV: Food Production
Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.
Note for Teachers: The NCERT text books present information in boxes across the book. These help students to get conceptual clarity. However, the information in these boxes would not be assessed in the year-end examination.
PRACTICALS
Practicals should be conducted alongside the concepts taught in theory classes. (LIST OF EXPERIMENTS)
1. Preparation of: Unit-I
(a) a true solution of common salt, sugar and alum
(b) a suspension of soil, chalk powder and fine sand in water
(c) a colloidal solution of starch in water and egg albumin/milk in water and distinguish between these on the basis of
• transparency
• filtration criterion
• stability
2. Preparation of
(a) A mixture
(b) A compound
using iron filings and sulphur powder and distinguishing between these on the basis of:
• appearance, i.e., homogeneity and heterogeneity
• behaviour towards a magnet
• behaviour towards carbon disulphide as a solvent
• effect of heat
3. Perform the following reactions and classify them as physical or chemical changes:
(a) Iron with copper sulphate solution in water
(b) Burning of magnesium ribbon in air
(c) Zinc with dilute sulphuric acid
(d) Heating of copper sulphate crystals
(e) Sodium sulphate with barium chloride in the form of their solutions in water
4. Preparation of stained temporary mounts of (a) onion peel, (b) human cheek cells & to record observations and draw their labeled diagrams
5. Identification of Parenchyma, Collenchyma and Sclerenchyma tissues in plants, striped, smooth and cardiac muscle fibers and nerve cells in animals, from prepared slides. Draw their labeled diagrams. Unit-II
6. Determination of the melting point of ice and the boiling point of water.
7. Verification of the laws of reflection of sound.
8. Determination of the density of solid (denser than water) by using a spring balance and a measuring cylinder.
9. Establishing the relation between the loss in weight of a solid when fully immersed in
(a) Tap water
(b) Strongly salty water with the weight of water displaced by it by taking at least two different solids.
10. Determination of the speed of a pulse propagated through a stretched string/ slinky (helical spring). Unit-III
11. Verification of the law of conservation of mass in a chemical reaction. Unit-III
PRESCRIBED BOOKS:
• Science-Textbook for class IX-NCERT Publication
• Assessment of Practical Skills in Science-Class IX – CBSE Publication
QR Code: Provides access to the digital content of each chapter.
Chapter Number and Chapter Name
Concept Map: A visual tool outlining the concepts covered in a chapter.
Chapter at a Glance: A short and crisp summary of a chapter with all the important ideas/concepts/relations.
NCERT Zone: Detailed model answers to NCERT questions (Intext and Exercises) given in simple, easy to understand language.
HOTS: “High Order Thinking Skills” Questions.
Competency Based Questions: A set of questions (Multiple choice, Assertion-Reason and Case-based questions), that are based on competencies defined by CBSE.
Practice Questions: A set of 20 unsolved question, covering the entire chapter for practice.
Brain Charge: A unique set of questions that are designed to be fun and thought-provoking. Try these to supercharge your brain.
Challenge Yourself: A unique set of higher order questions that are designed to challenge you and your understanding of the chapter. Additional Questions can be found on scanning the QR code.
Answers to Unsolved Problems: Solution to objective/numerical questions from the Practice Questions, Brain Charge and Challenge Yourself sections. Complete solutions can be accessed on scanning the QR code.
Complete Exemplar Solutions: Scan the QR code at end of each chapter for complete NCERT Exemplar solutions.
Self-Assessment: Assessment at the end of the each chapter to help learners prepare for tests and exams. Complete Solutions will be available on QR code.
1 Matter in Our Surroundings
The chapter ʻMatter in Our Surroundingsʼ introduces the basic concepts of matter, explaining how everything in our environment, from air to water, comprises matter that occupies space and has mass. It explores various states of matter—solids, liquids, and gases and explains how these different forms are influenced by particle movement and arrangement. The chapter further delves into the nature of these particles, kinetic energy, diffusion, and factors such as temperature and pressure that influence changes in state. Additionally, it covers key processes like evaporation, melting, and sublimation, enhancing our understanding of the material world through relatable examples and experiments.
Matter
Anything that possesses mass, occupies space and of which all physical substances are made of
Particles
• Particles are very small in size
• Particles have mass
Characteristics
• Particles have space between them
• Particles are continuously moving
• Particles attract each other
Solids
• Have definite shape, distinct boundaries and fixed volume
• Maintain their shapes
• They are rigid and have negligible compressibility
Liquids
• Have no fixed shape but have a fixed volume
• Take up the shape of the container in which they are kept
• They are not rigid and have slight compressibility.
Gases
• Have neither definite shape nor definite volume.
• Take up the shape of the container in which they are kept.
• Highly compressible, diffusible and have low density.
States of matter
Arise due to variation in characteristics of particles
Change of States
All the three states are interconvertible
Solid into Vapour/Sublimation
Conversion of solid state into gaseous state without passing through liquid state is sublimation
Vapour into Solid/Deposition
The direct change of gas to solid without changing into liquid is deposition
Solid into Liquid/Melting
• Conversion of solid state of matter into liquid state by application of heat
• Increase in temperature increases the kinetic energy of particles, they start moving freely and at one point change to liquid state
• Temperature at which a solid melts to become liquid at atmospheric pressure is its melting point
Liquid into Vapour/Evaporation
• Conversion of liquid into vapour without reaching the boiling point
• Supply of heat energy results in faster movement of particles and at one point they break free from forces of attraction of each other and change to vapour
Liquid into Solid/Freezing
• Conversion of liquid state into solid state at 0°C
• Lowering the temperature decreases the kinetic energy of particles, they start moving slowly and at one point change into solid state
• Temperature at which a liquid changes to a solid at atmospheric pressure is its freezing point
Vapour into Liquid/Condensation
• Conversion of gaseous state into liquid state
• Occurs when pressure of vapour becomes equal to the maximum pressure of the liquid at that temperature
Chapter at a Glance
• Matter is everything around us that occupies space and has mass, including air, water, and physical objects.
• Matter is found in three states: solids, liquids, and gases, each demonstrating unique particle arrangements.
• In solids, particles are tightly packed, causing objects to maintain a fixed shape and definite volume.
• Liquids can flow and take the form of their container yet have a fixed volume due to flexible particle positioning.
• Gases have no fixed shape or volume; particles move freely filling up all available space, due to weak forces.
• The kinetic energy of particles in matter rises with temperature, causing changes in state and increasing movement.
• Diffusion occurs when particles move from high to low concentration areas, with speed increasing as temperature rises.
• Forces of attraction are: strong in solids, moderate in liquids, and weak in gases.
• Evaporation is a surface phenomenon which causes liquid to gas transformation at temperatures below boiling, leading to cooling effects.
• The rate of evaporation increases with increase in surface area, temperature, humidity and wind speed.
• Sublimation transforms a solid directly into gas, as seen in dry ice; deposition is the reverse, from gas to solid.
• Boiling is a bulk process where particles from the entire volume of a liquid gain enough energy to convert into the vapour state at a specific temperature.
• Melting is a process in which a solid absorbs heat to transform into liquid.
• Freezing is a process in which a liquid releases heat to transform into solid.
• During vaporisation, a liquid converts into gas by absorbing heat.
• During Condensation, a gas converts into liquid by releasing heat.
• The latent heat is absorbed or released during state transitions like melting and vaporization, without any temperature change.
• Latent heat of vaporisation is the heat energy required to change 1 kg of a liquid to gas at atmospheric pressure at its boiling point.
• Latent heat of fusion is the amount of heat energy required to change 1 kg of solid into liquid at its melting point.
• Compressibility is the ability of matter to be compressed. Order of compressibility: Gases > liquids > solids.
• The density of matter is generally highest in solids state and lowest in gasesous state.
• Kinetic energy is the energy possessed by particles due to their motion; it increases with temperature, causing particles to move faster and potentially change their state.
• The balance of kinetic energy and attraction between particles defines the state of matter and influences the transformation of one state into another.
• Key measurable quantities and their corresponding units:
Temperature kelvin K
Length metre m
Mass kilogram kg
Weight newton N
Volume cubic metre m3
Density kilogram per cubic metre kg m-3
Pressure pascal Pa
Formulae
• Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273.15
• Temperature in Celsius (°C) = Temperature in Kelvin (K) - 273.15
Ans. The items that are matter include Chair, air, almonds, and lemon water because they occupy space and have mass. Items like love, smell, hate, thought, and cold are not considered as matter because they do not have mass or volume.
2. Give reasons for the following observation: The smell of hot sizzling food reaches you several metres away, but to get the smell from cold food you have to go close.
Ans. The smell of hot sizzling food reaches you from several metres away because higher temperature increases the kinetic energy of aroma particles, causing them to move faster and diffuse more quickly through the air. In contrast, cold food releases particles at a slower rate, requiring closer proximity to detect its smell due to the decreased kinetic energy and slower diffusion.
3. A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?
Ans. This observation shows that particles of matter have spaces between them. In liquids like water, the particles are loosely packed, allowing a diver to cut through it.
4. What are the characteristics of the particles of matter?
Ans. The characteristics of the particles of matter are:
• Particles of matter are very small in size.
• Particles of matter have spaces between them, which allows substances to mix or dissolve in each other.
• Particles of matter are constantly moving, and their motion increases with a rise in temperature.
• Particles of matter attract each other, and the strength of this attraction differs in solids, liquids, and gases.
5. The mass per unit volume of a substance is called density (density = mass/volume). Arrange the following in order of increasing density – air, exhaust from chimneys, honey, water, chalk, cotton and iron.
Ans. The order of increasing density is:
Air < Exhaust from chimneys < Cotton < Water < Honey < Chalk < Iron
This arrangement is based on the concept of density, where gases like air have the lowest density and solids like iron have the highest.
6. (a) Tabulate the differences in the characteristics of states of matter.
(b) Comment upon the following: rigidity, compressibility, fluidity, filling a gas container, shape, kinetic energy and density.
Ans. (a) Characteristic
Solids
Liquids Gases
Shape Solids have a definite shape. Liquids take the shape of their container. Gases do not have a definite shape.
Volume Solids have a fixed volume. Liquids have a fixed volume. Gases have neither fixed shape nor volume.
Particle Motion Particles vibrate in fixed positions. Particles move freely but are closely packed. Particles move rapidly in all directions.
Compressibility Solids are not compressible. Liquids are slightly compressible. Gases are highly compressible.
Density Solids have high density. Liquids have moderate density. Gases have low density.
Diffusion Diffusion occurs very slowly in solids. Liquids diffuse at a moderate rate. Gases diffuse quickly and easily.
(b) Rigidity: Solids are rigid, meaning they retain their shape and do not flow. This is due to strong intermolecular forces and fixed particle positions.
Compressibility: Gases are highly compressible because of large intermolecular spaces. Liquids are slightly compressible, while solids are nearly incompressible.
Fluidity: Liquids and gases exhibit fluidity—they can flow and change shape. Solids lack this property.
Filling a Gas Container: Particles of a gas possess minimal intermolecular forces and move freely in all the directions occupying the complete space available to them.
Shape: The geometry of an object is called its shape. Solids have a definite shape, whereas liquids and gases take the shape of the container they occupy.
Kinetic Energy: Kinetic energy is the energy possessed by a body due to its motion. It depends on the mass of the object and the square of its velocity. It increases from solids to gases.
Density: It is defined as mass per unit volume of a substance. Generally, solids have the highest density, followed by liquids. Gases have the lowest density because of widely spaced particles. The S.I. unit of density is kg m–3.
Density = Mass Volume
7. (a) A gas fills completely the vessel in which it is kept.
(b) A gas exerts pressure on the walls of the container.
(c) A wooden table should be called a solid.
(d) We can easily move our hand in air but to do the same through a solid block of wood we need a karate expert.
Ans. (a) Particles of gas have least forces of attraction between them therefore, they move freely in all directions and occupy all the space available to them. Hence, a gas fills the vessel completely in which it is kept.
(b) Gas particles possess high kinetic energy and move randomly at high speeds. As they move, they frequently collide with each other and with the walls of the container, exerting force per unit area, which is experienced as the pressure.
(c) A wooden table is called a solid because it has a definite shape, fixed boundaries, and a fixed volume. It maintains its shape unless a significant force is applied and cannot be compressed due to the closely packed particles, and hence it possess all the characteristics of a solid.
(d) We can easily move our hand through air because the particles in gases are loosely arranged with large spaces between them, allowing free movement. In contrast, a solid block of wood has tightly packed particles with strong forces of attraction, making it rigid and difficult to pass through without applying significant force, such as with karate.
8. Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why.
Ans. Though, ice is a solid, but it has a cage-like structure in which some spaces are present between the particles of water (these spaces are left when water solidifies). These spaces trap air particles. Infact these spaces are larger as compared to the spaces present between the particles of liquid water. Thus, the volume of ice is greater than that of water. Hence, the density of ice is less than that of water. A substance with lower density than water can floats on water. Thus, ice floats on water.
9. Convert the following temperature to Celsius scale:
(a) 300 K
(b) 573 K
Ans. To convert Kelvin to Celsius, the formula is (K–273)°C
(a) 300 – 273 = 27°C
10. What is the physical state of water at:
(a) 250°C
(b) 573 – 273 = 300°C
(b) 100°C
Ans. (a) At 250°C, water is in a gaseous state (steam) since it is above its boiling point.
(b) At 100°C, water is at its boiling point and can exist as both gas and a liquid. If heat continues to be added all the liquid will change into gas, otherwise the water will exist as a liquid at 100°C.
11. For any substance, why does the temperature remain constant during the change of state?
Ans. During the change of state, temperature remains constant because the absorbed or released heat energy is used in breaking or forming intermolecular forces, this energy is known as latent heat. This energy is not utilized for increasing kinetic energy, hence the temperature does not change.
12. Suggest a method to liquefy atmospheric gases.
Ans. Atmospheric gases can be liquefied by increasing pressure and decreasing temperature. This process reduces the distance between molecules and allows gases to transition into a liquid state.
13. Why does a desert cooler cool better on a hot dry day?
Ans. A desert cooler cools better on a hot dry day because the low humidity and high temperature increase the rate of evaporation. As more water evaporates, it absorbs more heat from the surroundings, resulting in a greater cooling effect.
14. How does the water kept in an earthen pot (matka) become cool during summer?
Ans. Water in an earthen pot becomes cool during summer because the pot’s porous walls allow some water to seep out and evaporate. During evaporation, this water absorbs heat from the remaining water inside the pot, reducing its temperature. This process helps naturally cool the water in hot weather.
15. Why does our palm feel cold when we put some acetone or petrol or perfume on it?
Ans. Our palm feels cold when acetone, petrol, or perfume is applied because they evaporate quickly, absorbing heat energy from the skin. The evaporation process decreases the skinʼs temperature, creating a cooling sensation.
16. Why are we able to sip hot tea or milk faster from a saucer rather than a cup?
Ans. We are able to sip hot tea or milk faster from a saucer because it spreads the liquid over a larger surface area, increasing the rate of evaporation. This process cools the tea or milk more quickly than in a cup, which has a smaller surface area and slower evaporation.
17. What type of clothes should we wear in summer?
Ans. In summer, we should wear cotton clothes because they absorb sweat effectively and allow it to evaporate quickly. This evaporation removes heat from the body, producing a cooling effect, making it ideal for hot, humid conditions.
NCERT Exercises
1. Convert the following temperatures to the celsius scale.
(a) 293 K
(b) 470 K
Ans. To convert Kelvin to Celsius, the formula is: K – 273 = °C
(a) 293 – 273 = 20°C
(b) 470 – 273 = 197°C
2. Convert the following temperatures to the kelvin scale.
(a) 25°C
(b) 373°C
Ans. To convert Celsius to Kelvin, the formula is: °C + 273 = K
(a) 25 + 273 = 298 K
3. Give reason for the following observations.
(b) 373 + 273 = 646 K
(a) Naphthalene balls disappear with time without leaving any solid.
(b) We can get the smell of perfume sitting several metres away.
Ans. (a) Naphthalene balls disappear over time due to sublimation, a process in which a solid directly changes into gas at room temperature without passing through the liquid state.
(b) The smell of perfume can be detected from several metres away due to diffusion, where perfume particles mix with air particles and spread through the air, allowing the fragrance to reach distant places.
4. Arrange the following substances in increasing order of forces of attraction between the particles— water, sugar, oxygen.
Ans. oxygen < water < sugar.
5. What is the physical state of water at— (a) 25°C (b) 0°C (c) 100°C?
Ans. (a) At 25°C, water is in the liquid state.
(b) At 0°C, water exists as both solid and liquid, since it is at its melting/freezing point.
(c) At 100°C, water is in the gaseous state (steam) at atmospheric pressure, as it reaches its boiling point.
6. Give two reasons to justify—
(a) water at room temperature is a liquid.
(b) an iron almirah is a solid at room temperature.
Ans. (a) Water at room temperature (27°C) is a liquid because it has a definite volume but no fixed shape, and it flows to take the shape of its container.
(b) An iron almirah is a solid at room temperature because it has a definite shape and volume, and its particles are tightly packed in a rigid structure.
7. Why is ice at 273 K more effective in cooling than water at the same temperature?
Ans. Ice at 273 K is more effective in cooling than water at the same temperature because it absorbs extra heat from the surroundings to transition from solid to liquid. This additional heat absorption results in greater cooling compared to water, which doesnʼt require additional energy to maintain its state.
8. What produces more severe burns, boiling water or steam?
Ans. Steam produces more severe burns compared to boiling water because it contains additional energy in the form of latent heat of vaporization. When steam condenses on the skin, it releases this extra heat, causing deeper and more severe burns than boiling water.
9. Name A, B, C, D, E and F in the following diagram showing change in its state.
Increase heat and decrease pressure
Decrease heat and increase pressure
Ans. A = Melting or fusion, here the solid changes into liquid.
B = Evaporation or vaporisation, here the liquid changes into gas.
C = Condensation or liquefaction, here the gas changes into liquid.
D = Freezing or solidification, here the liquid changes into solid.
E = Sublimation, here solid directly changes into gas without coming in liquid state.
F = Deposition, here gas changes into solid without coming in liquid state.
Multiple Choice Questions
1. Which of the following statements is not correct?
(a) Matter is continuous in nature.
(b) Interparticle spaces are maximum in the gaseous state of a substance.
(c) Solid state is the least compact state of a substance.
(d) Particles which constitute the matter follow a zig-zag path.
2. Which of the following is the correct order of interparticle spaces present in the matter?
(a) Solid > Liquid > Gas
(c) Solid < Liquid < Gas
(b) Solid > Liquid < Gas
(d) Solid < Liquid > Gas
3. The property to flow is unique to fluids. Which one of the following statements is correct?
(a) Only gases behave like fluids.
(NCERT Exemplar)
(b) Gases and solids behave like fluids.
(c) Gases and liquids behave like fluids. (d) Only liquids behave like fluids.
4. Which of the following is TRUE for solids?
(a) They can be compressed easily. (b) They have a definite shape and volume.
(c) They do not have a fixed volume.
(d) The particles are very far apart.
5. Which of the following is not a characteristic of gas?
(a) High rigidity
(c) Low density
6. Forces of attraction between particles is
(a) maximum in solids and minimum in gases.
(b) maximum in gases and minimum in solids.
(c) maximum in liquids and minimum in gases.
(d) maximum in gases and minimum in liquids.
7. Gases can be converted to liquid by
(b) High compressibility
(d) High rate of diffusion
(a) decreasing temperature and decreasing pressure
(b) decreasing temperature and increasing pressure
(c) increasing temperature and decreasing pressure
(d) increasing temperature and increasing pressure
8. When a gas is compressed under constant temperature, the volume of the gas:
(a) Increases (b) Decreases (c) Remains the same (d) Changes into liquid
9. In the following flow chart showing the change in state, the processes A and B are Solid Liquid Gas A B
(a) A = Boiling, B = Melting
(c) A = Evaporation, B = Condensation
(b) A = Melting, B = Boiling
(d) A = Condensation, B = Evaporation
10. Aquatic animals use dissolved oxygen present in water for respiration. This uptake of oxygen occurs due to which phenomenon?
11. When an incense stick (agarbatti) is lit, its fragrance spreads across the room. This observation best demonstrates which characteristic of the particles of matter?
(a) Particles of matter have spaces between them.
(b) Particles of matter are attracted to each other.
(c) Particles of matter are continuously moving.
(d) Particles of matter have a fixed shape.
12. During summer, water kept in an earthen pot becomes cool because of the phenomenon of (NCERT Exemplar) (a) diffusion. (b) transpiration. (c) osmosis. (d) evaporation.
13. Which condition out of the following will increase the evaporation of water? (NCERT Exemplar)
(a) Increase in temperature of water (b) Decrease in temperature of water
(c) Less exposed surface area of water (d) Adding common salt to water
14. During the rainy season, the rate of evaporation may decrease due to which of the following factors?
(a) Increase in surface area
(b) Increase in temperature (c) Increase in humidity (d) Decrease in humidity
15. Which phenomenon causes naphthalene balls to disappear over time without leaving any residue?
1. (a) A substance has no mass. Can it be considered matter?
(b) A given substance X has a definite volume but no definite shape, and it can diffuse easily. What is the physical state of substance X?
Ans. (a) No, it cannot be considered matter because matter must have both mass and volume.
(b) The physical state of substance X is liquid, as liquids have a definite volume, no fixed shape, and they can diffuse.
2. (a) Rubber band changes its shape on stretching. Is it solid?
(b) Why are wet clothes spread out while drying?
Ans. (a) Rubber band changes its shape under force and regains its shape after removing the force, however during the process, its mass remains constant. Thus, it is a solid.
(b) Spreading wet clothes increases the surface area exposed to air, which speeds up evaporation, allowing the clothes to dry faster.
3. Define evaporation and list the factors that affect its rate.
Ans. The process of conversion of liquid into its vapour state at any temperature below its boiling point is called evaporation. The rate of evaporation depends on surface area, temperature, humidity, and wind speed.
4. What characteristic of matter do the following statements demonstrate?
(a) When 50 g of sugar is dissolved in 100 mL of water, there is no increase in volume.
(b) A drop of blue ink slowly spreads throughout a glass of water without stirring.
Ans. (a) This demonstrates that particles of matter have spaces between them, allowing the sugar particles to fit into these spaces without causing an increase in volume.
(b) This shows that particles of matter are constantly moving. The spreading of ink in water is due to diffusion, where particles move and mix on their own.
5. Give reasons.
(a) Cotton is a solid, yet it floats on water.
(b) Gases exert more pressure on the walls of a container than solids.
Ans. (a) Cotton is made up of tiny pores that trap air. This trapped air reduces the overall density of cotton, making it less dense than water, which is why it floats on water.
(b) Gas particles move randomly at high speeds and collide frequently with each other and the walls of the container. These continuous collisions exert a greater pressure compared to solids, where particles are tightly packed and vibrate in fixed positions.
6. Define density and give its SI unit.
Ans. Density is the physical property of matter that refers to the amount of mass present in a unit volume of a substance. It indicates how tightly the particles are packed within the material.
It is given by the formula: Density = Mass Volume
The SI unit of density is kilogram per cubic metre (kg/m³).
7. (a) Why do gases apply more pressure on the walls of a container than solids do?
(b) What property of gases allows oxygen to be stored in cylinders for hospital use?
Ans. (a) Gas particles move quickly and randomly in all directions. As they move, they collide with one another and with the walls of the container. These constant and forceful collisions cause gases to exert more pressure on the container walls than solids, whose particles are tightly packed and hardly move.
(b) Gases can be easily compressed and turned into liquids under pressure. This ability to compress gases into smaller volumes makes it possible to store oxygen in cylinders, which is useful for supplying hospitals.
8. (a) Define Diffusion.
(b) How do aquatic animals breathe underwater?
Ans. (a) Diffusion is the process of intermixing of particles of two or more substances on their own, from a region of higher concentration to a region of lower concentration, due to the random motion of particles. It occurs more rapidly in gases, slower in liquids, and slowest in solids.
(b) Aquatic animals breathe underwater by using the dissolved oxygen present in water, which they absorb through specialized organs like gills.
9. Explain the following.
(a) Why do we see water droplets on the outer surface of a glass containing ice-cold water?
(b) Why do we sweat more on a humid day?
Ans. (a) The water vapour present in the air comes into contact with the cold surface of the glass, loses heat, and condenses to form tiny droplets of water on the outer surface.
(b) On a humid day, the air already contains a high amount of water vapour. This reduces the rate of evaporation of sweat from our skin, as the air cannot absorb more moisture. As a result, sweat remains on the body, making us feel sweatier.
10. (a) What is sublimation?
(b) Why does the temperature remain constant during sublimation?
Ans. (a) Sublimation is the process in which a solid directly changes into a gas without passing through the liquid state.
(b) During sublimation, the heat supplied is used to overcome the forces holding the solid particles together. This energy goes into the change of state, not into raising the temperature, so the temperature remains constant.
11. When water is cooled to a temperature T, it gets converted into ice by a process called P.
(a) What is the value of temperature T in Kelvin?
(b) What is the process P known as?
Ans. (a) Water changes into ice at 0°C. To convert this temperature into the Kelvin scale, the formula is: K = °C + 273
So, T = 0 + 273 = 273 K
Therefore, the temperature at which water freezes is 273 K.
(b) The process of conversion of a liquid into solid is known as freezing.
12. (a) Conversion of solid to vapour is called sublimation. Name the term used to denote the conversion of vapour to solid.
(b) Conversion of solid state to liquid state is called fusion; what is meant by latent heat of fusion?
(NCERT Exemplar)
Ans. (a) The conversion of vapour directly to solid is called deposition. It is the reverse process of sublimation.
(b) Latent heat of fusion is the amount of heat energy required to convert one kilogram of a solid into liquid at its melting point, without any change in temperature. This energy is used to overcome the forces of attraction between the particles.
13. A piece of chalk can be broken down into small pieces when hammered, but it is not possible to do so with an iron bar. Why?
Ans. The particles in the iron bar are closely packed together, making it stronger and more rigid. In contrast, the particles in chalk are more loosely arranged, allowing them to easily separate when force is applied. This is why chalk breaks easily, while iron does not.
14. Alka was making tea in a kettle. Suddenly she felt intense heat from the puff of steam gushing out of the spout of the kettle. She wondered whether the temperature of the steam was higher than that of the water boiling in the kettle. Comment. (NCERT Exemplar)
Ans. The temperature of both boiling water and steam is 100°C, but steam has more energy than boiling water because it contains latent heat of vaporisation. This extra energy is released when steam condenses, which is why the puff of steam feels much hotter and can cause more severe burns than boiling water.
15. What would be the effect of the following on the rate of diffusion of liquids?
(a) temperature
(b) density
Ans. (a) On increasing temperature, the rate of diffusion of liquids increases because particles move faster with higher kinetic energy.
(b) The rate of diffusion is higher for a liquid with lower density, as its particles are more loosely packed and can move more freely.
Short Answer Questions (50-80 words)
1. (a) Explain the interconversion of the three states of matter in terms of force of attraction and kinetic energy of the particles.
(b) Arrange the three states of matter in the increasing order of:
(i) Rate of diffusion
(ii) Particle motion
Ans. (a) The change in states of matter depends on the particlesʼ kinetic energy and the force of attraction between them. On heating, particles gain energy and move faster, weakening attractive forces and changing solids to liquids and then to gases. On cooling, particles lose energy, move slower, and attractive forces pull them closer, turning gases into liquids and then into solids. Thus, heating and cooling control state changes through energy and attraction balance.
(b) (i) Rate of diffusion: Solid < Liquid < Gas
(ii) Particle motion: Solid < Liquid < Gas
2. (a) Dry ice is compressed under high pressure. What happens to it when the pressure is released?
(b) State one difference between gas and vapour.
(c) Why do we feel comfortable under a fan when we are sweating on a hot summer day?
Ans. (a) On releasing the pressure, dry ice sublimes—it directly changes from solid to vapour without converting into the liquid state.
(b) A gas exists in its natural gaseous state at room temperature, while a vapour is formed when a liquid or solid is heated and can easily condense back to its original state upon cooling.
(c) A fan makes us feel comfortable on a hot summer day because it increases the rate of evaporation of sweat. The moving air helps sweat evaporate faster, and since evaporation causes cooling, our body feels cooler.
3. Look at the following figures and suggest in which of the vessels A, B, C, or D the rate of evaporation will be the highest? Explain. (NCERT Exemplar)
Moving Fan
A B C D
Ans. The rate of evaporation will be highest in vessel C. This is because vessel C has a large surface area and is placed closer to the fan, which increases the wind speed, further enhancing evaporation. Although vessels A and D are of similar size, A is farther from the fan and D is covered with a lid, which prevents evaporation. Vessel B has a smaller surface area, so its rate of evaporation is also lower than vessel C.
4. Write the three differences between evaporation and boiling. Ans.
Evaporation
Evaporation is a natural process where a liquid changes into vapour at any temperature, usually below its boiling point.
Boiling
Boiling is a process where a liquid changes into vapour at a specific temperature (boiling point) with continuous heating.
It occurs only on the surface of the liquid. It occurs throughout the entire liquid mass.
Evaporation is a slow and gradual process. Boiling is a fast and vigorous process.
5. Comment on the following statements:
(a) Evaporation produces cooling.
(b) Rate of evaporation of an aqueous solution decreases with increase in humidity.
(c) Sponge, though compressible, is a solid. (NCERT Exemplar)
Ans. (a) Evaporation produces cooling because the liquid absorbs heat energy from its surroundings to convert into vapour. This loss of heat from the surroundings results in a cooling effect.
(b) The rate of evaporation decreases with increase in humidity because when the air already contains a large amount of water vapour, it cannot accommodate more vapour easily. Hence, evaporation slows down in humid conditions.
(c) A sponge is a solid even though it can be compressed. This is because it has a definite shape and volume. The presence of tiny pores filled with air makes it compressible, but it still retains the fundamental properties of a solid.
6. In the given graph, the physical state of water is shown with respect to temperature and time.
(a) Which region contains only solid?
(b) Which region contains the liquid form of water?
(c) Which region shows latent heat of vaporisation?
Ans. (a) Since temperature in region AB is less than 0°C, the water will be in its solid state(ice).
(b) Since temperature in region CD is between 0–100°C, the water will be in its liquid state.
(c) Region DE shows the latent heat of vaporisation. During this phase, temperature remains constant at 100°C while water changes from liquid to vapour.
7. With the help of an example, explain how diffusion of gases in water is important.
Ans. Gases from the atmosphere, such as oxygen and carbon dioxide, diffuse and dissolve in water, making them available to aquatic life. For example, fish absorb dissolved oxygen through their gills to survive, and aquatic plants use dissolved carbon dioxide for photosynthesis. Without this diffusion, aquatic organisms would not get the gases they need to live and grow.
8. Substance ʻAʼ has high compressibility and can be easily liquefied. It can take up the shape of any container. Predict the nature of the substance. Enlist four properties of this state of matter.
Ans. Substance ʻAʼ is a gas.
Properties of gases:
(i) Gases have a high tendency to diffuse rapidly into the available space.
(ii) They are highly compressible, meaning their volume can be reduced under pressure.
(iii) Gas particles move randomly at high speeds in all directions.
(iv) Gases have minimum density compared to solids and liquids.
9. Give reasons for the following:
(a) Butter is generally wrapped in a wet cloth during summer if no refrigeration is available.
(b) Ghee freezes at room temperature while mustard oil does not during winter.
(c) When sugar crystals dissolve in water, the water level does not rise appreciably.
Ans. (a) In summer, wrapping butter in a wet cloth helps keep it cool through evaporation. As water from the cloth evaporates, it absorbs heat from the butter, causing a cooling effect and preventing the butter from melting quickly.
(b) Ghee has a higher melting point than mustard oil. So, at lower temperatures (like during winter), ghee solidifies, while mustard oil, which has a lower melting point, remains in the liquid state.
(c) This is because the particles of sugar occupy the spaces between water molecules. Water has intermolecular spaces, and sugar particles settle into these spaces without significantly increasing the volume.
10. (a) It is a hot summer day. Priyanshi and Ali are wearing cotton and nylon clothes respectively. Who would be more comfortable and why?
(b) You want to wear your favourite shirt to a party, but it is still wet after a wash. What steps would you take to dry it faster? (NCERT Exemplar)
Ans. (a) Priyanshi would be more comfortable because cotton absorbs sweat and allows it to evaporate quickly, producing a cooling effect. Nylon, on the other hand, does not absorb sweat well and can stick to the skin, making Ali feel hotter and uncomfortable.
(b) To dry the shirt faster, one would:
(i) Spread it out to increase the surface area exposed to air, as evaporation is faster over a larger surface.
(ii) If possible, place it in sunlight, since higher temperature increases the rate of evaporation.
(iii) Choose a spot with good air circulation or cross-ventilation, because evaporation occurs faster in moving air.
By using these methods together—larger surface area, higher temperature, and better airflow—the shirt will dry in the shortest possible time.
11. (a) How do you differentiate between, liquids and gases on the basis of their melting and boiling points?
(b) Kinetic energy of particles of water in three vessels A, B and C are EA, EB and EC respectively and EA > EB > EC. Arrange the temperature TA, TB and TC of water in the three vessels in increasing order.
Ans. (a) Liquids have a melting point below room temperature and a boiling point above room temperature. Gases have both melting and boiling points below room temperature, which is why they exist in gaseous form under normal conditions.
(b) Since the kinetic energy of particles increases with temperature, the temperature order is: Thus, the order is: TA > TB > TC.
12. (a) What is matter? Write two properties of solids and liquids each.
(b) Why does ice at 0°C appear colder in the mouth than water at 0°C?
(c) A patient suffering from fever is advised to put the strips of wet cloth on the forehead.
Ans. (a) Matter is a substance which has mass and occupies space.
(i) In solids, force of attraction between particles is strongest and space between its particles is very less.
(ii) In liquids, force of attraction is relatively weaker and space between the particles is more in comparison to solids.
(b) Ice at 0°C feels colder than water at the same temperature because it absorbs additional latent heat of fusion from the mouth to melt, resulting in a greater cooling effect.
(c) During fever, the body temperature is high. When wet cloth strips are placed on the forehead, the water evaporates, taking away heat from the body. This lowers the body temperature, providing relief.
Long Answer Questions (80–120 words)
1. (a) Draw the arrangement of particles in solids, liquids and gases.
(b) Differentiate between solids, liquids, and gases on the basis of their general properties.
Ans. (a)
Shape
Volume
Compressibility
Solids have a definite shape. Liquids do not have a definite shape and take the shape of the container they are in. Gases do not have a definite shape and fill the entire container they occupy.
Solids have a definite volume. Liquids have a definite volume. Gases do not have a definite volume and can expand or compress easily.
Solids are almost incompressible due to tightly packed particles. Liquids are slightly compressible. Gases are highly compressible as there is a lot of space between particles.
Interparticle Space The particles in solids are very closely packed with very small spaces.
Force of Attraction
The force of attraction between particles in solids is very strong.
Liquids have moderately spaced particles. Gases have large spaces between particles.
Liquids have moderate force of attraction between their particles. Gases have very weak force of attraction between particles.
2. (a) What determines the state of a substance?
(b) Convert 40°C into Kelvin.
(c) Water droplets are observed on the outer surface of a glass tumbler containing ice-cold water. Give reason.
(d) What is the physical state of water at:
(i) 250°C?
(ii) 90°C?
Ans. (a) The state of a substance is determined by its temperature and pressure.
(b) To convert Celsius to Kelvin:
K = °C + 273
So, (273 + 40)°C = 313 K
(c) Water droplets are observed on the outer surface of a glass tumbler containing ice-cold water because the air around the glass contains water vapour. When this moist air comes in contact with the cold surface of the glass, the temperature of the air drops, and the water vapour loses energy and condenses into liquid water. This process is called condensation, and it forms droplets on the outer surface of the tumbler.
(d) (i) At 250°C, water is in the gaseous state as it is well above its boiling point (100°C).
(ii) At 90°C, water is in the liquid state, as it is still below the boiling point.
3. The graph in Fig. 1.1 shows the variation in the temperature of a substance(wax) with time. Based on the graph, answer the following:
(a) What is the physical state of the substance at points A, B, C, and D?
(b) What is the melting point of wax?
(c) What is the boiling point of wax?
(d) Which portions of the graph indicate that change of state is taking place?
(e) Name the terms used for the heat absorbed during the change of states involved in the above processes.
Ans. (a) • At point A: Wax is in the solid state.
• At point B: Wax is in solid + liquid state (melting is taking place).
• At point C: Wax is in the liquid state.
• At point D: Wax is in liquid + vapour state (boiling is taking place).
(b) Melting point of wax = 15°C
(c) Boiling point of wax = 110°C
Fig. 1.1
(d) The horizontal portions of the graph parallel to the time axis indicate change of state:
• A₁ to B₁: Solid to liquid (melting)
• D₁ to D₂: Liquid to vapour (boiling)
(e) Latent heat of fusion is the amount of heat required to convert a unit mass of solid into liquid at its melting point. Latent heat of vaporisation is the amount of heat required to convert a unit mass liquid into gas at constant temperature.
4. (a) A balloon bursts when kept in the sun. Why?
(b) Why do people sweat a lot on a hot, humid day?
(c) Between 1 kg cotton and 1 kg sand, which is heavier?
(d) A small volume of water in a kettle can fill a kitchen with steam. Explain why.
Ans.
(a) When a balloon is kept in the sun, the air inside it gets heated. As the temperature rises, the air molecules move faster and exert more pressure on the inner walls of the balloon. This causes the balloon to expand. If the pressure becomes too much for the balloon to handle, it bursts.
(b) On a hot day, our body sweats to cool down through evaporation. However, on a humid day, the air is already full of moisture and cannot absorb more. As a result, sweat doesnʼt evaporate easily, and we feel sweatier and more uncomfortable.
(c) Both have the same weight—1 kg.
However, sand is denser than cotton, so it takes up less space. Cotton is bulkier because it has more air spaces and needs more volume to make up the same weight.
(d) When water is heated in a kettle, it changes from liquid to steam (gaseous state). In the gaseous state, water particles have high kinetic energy and move freely in all directions. Unlike liquids, gases do not have a fixed volume—they expand to fill any available space. Since gas particles are far apart and have negligible intermolecular forces, even a small amount of water can produce a large volume of steam, which spreads throughout the kitchen.
Competency Based Questions
Multiple Choice Questions (Choose the most appropriate option/options)
1. A diver is able to cut through water in a swimming pool. The property shown by the matter is
(a) the particles are of very small size.
(b) the matter has space between its particles.
(c) the particles are in solid state.
(d) the particles are running here and there, have no space between them.
2. In which of the following conditions, the distance between the molecules of hydrogen gas would increase?
(NCERT Exemplar)
(i) Increasing pressure on hydrogen contained in a closed container.
(ii) Some hydrogen gas leaking out of the container.
(iii) Increasing the volume of the container of hydrogen gas.
(iv) Adding more hydrogen gas to the container without increasing the volume of the container.
(a) (i) and (iii) (b) (i) and (iv) (c) (ii) and (iii) (d) (ii) and (iv)
3. Seema visited a Natural Gas Compressing Unit and found that the gas can be liquefied under specific conditions of temperature and pressure. While sharing her experience with friends she got confused. Help her to identify the correct set of conditions. (NCERT Exemplar)
(a) Low temperature, low pressure
(c) Low temperature, high pressure
(b) High temperature, low pressure
(d) High temperature, high pressure
4. The melting point of three solids X, Y and Z are 280 K, 1670 K and 380 K respectively. The forces of attraction are in order of
(a) X > Y > Z
(c) Y > Z > X
(b) X > Z > Y
(d) Z > Y > X
5. Which one of the following sets of phenomena would increase on raising the temperature? (NCERT Exemplar)
(a) Evaporation, diffusion, compression of gases
(b) Evaporation, compression of gases, solubility
(c) Evaporation, diffusion, expansion of gases
(d) Evaporation, solubility, diffusion, compression of gases.
6. The boiling points of diethyl ether, acetone and n-butyl alcohol are 35°C, 56°C and 118°C, respectively. Which one of the following correctly represents their boiling points in Kelvin scale? (NCERT Exemplar)
(a) 306 K, 329 K and 391 K
(c) 308 K, 329 K and 391 K
(b) 308 K, 329 K and 392 K
(d) 329 K, 392 K and 308 K
7. Particles in steam at 373 K have more energy than water at the same temperature. This is because the particles in steam have
(a) absorbed extra energy in the form of latent heat of vaporisation.
(b) released extra energy in the form of latent energy.
(c) absorbed extra energy in the form of latent heat of fusion.
(d) absorbed extra energy in the form of kinetic energy.
8. When ice is given heat, it reaches a temperature of 0°C and starts melting. The more heat is gained, the more it changes to water. However, the temperature remains constant at 0°C until all ice changes to water. Why does the temperature remain constant? (CBSE QB)
(a) Because of the latent heat of fusion.
(b) Because of the kinetic energy of ice particles.
(c) Because of the latent heat of vaporisation.
(d) Because of the kinetic energy of water particles.
9. Which of the following has highest kinetic energy?
(a) Particles of ice at 0°C
(c) Particles of water at 100°C
(b) Particles of water at 0°C
(d) Particles of steam at 100°C
10. Kabir came back home after playing football. He felt hot and sweaty. He stood under the fan. He noticed he started feeling cool and dry. What explains his observation? (CBSE QB)
(a) When sweat evaporates, warm air moves away from the body.
(b) When sweat evaporates, it gains heat energy from the surroundings.
(c) When sweat evaporates, it absorbs energy from the body making it cooler.
(d) When sweat evaporates, cooler air from the surroundings reaches the body.
11. When a teaspoon of solid sugar is dissolved in a glass of liquid water, what phase or phases are present after mixing?
(a) Still solid and liquid only
(c) Liquid only
(b) Solid only
(d) None of these
12. Which of the following changes represents the process of deposition?
(a) Solid Liquid
(c) Gas Liquid
(b) Liquid Gas
(d) Gas Solid
13. In which of the following conditions, the distance between the molecules of hydrogen gas would increase?
(i) Increasing pressure on hydrogen contained in a closed container
(ii) Some hydrogen gas leaking out of the container
(iii) Increasing the volume of the container of hydrogen gas
(iv) Adding more hydrogen gas to the container without increasing the volume of the container
(a) (i) and (iii) (b) (i) and (iv) (c) (ii) and (iii) (d) (ii) and (iv)
14. A solid substance possess
(a) rigidity, fluidity and weak force of attraction between the particles.
(b) rigidity, fluidity and fixed volume.
(c) rigidity, fixed volume and high attractive forces of attraction.
(d) rigidity, fixed shape and large intermolecular space between particles.
15. Which of the following does not undergo sublimation?
(a) Ammonium chloride
(b) Sodium chloride
(c) Solid carbon dioxide (d) Iodine
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): On applying pressure, liquids convert into gases.
Reason (R): On applying pressure, intermolecular space decreases.
2. Assertion (A): The rate of diffusion of solids is higher than that of liquids.
Reason (R): In liquid state, particles move freely and have greater space between each other as compared to solid state.
3. Assertion (A): Naphthalene does not leave a residue when kept in open for some time.
Reason (R): The conversion of a gas directly into solid is called condensation.
4. Assertion (A): Solids may break by applying force but it is difficult to change their shape.
Reason (R): Solids are rigid due to tightly packed particles.
5. Assertion (A): Ice cubes float on water.
Reason (R): Ice cubes are water in solid form. So, ice and water have same density.
6. Assertion (A): There is no change in the temperature of a substance when it undergoes a change in its physical state.
Reason (R): Heat supplied is absorbed as latent heat.
Reason (R): In the molten state, the ions are free to move.
8. Assertion (A): Smell of burning incense sticks spreads all around due to the diffusion of its fumes into air.
Reason (R): Increased temperature results in increased kinetic energy of the molecules.
1. A student performed an experiment in the chemistry lab. He placed ice cubes in a conical flask, suspended a thermometer in it, and placed the flask on a burner. He recorded the temperature at regular intervals and wrote the observations in his notebook:
Based on the above observations, answer the following:
(a) State the change(s) observed between 2–4 minutes.
(b) Name the process involved during the reading 2–4 minutes.
(c) Define the heat involved during 30–35 minutes and state its SI unit.
(d) At what time interval does water start to heat up after all ice has melted? How do you know?
2. The phenomenon of evaporation refers to the change of a liquid into vapour at any temperature below its boiling point. The rate of evaporation depends on various factors—it increases with surface area, temperature, and wind speed, and decreases with humidity. Evaporation causes cooling, as the particles with higher kinetic energy escape first, lowering the average energy of the remaining liquid. Also, liquids with lower boiling points evaporate faster because their molecules require less energy to escape into the vapour phase.
(a) Why are synthetic clothes uncomfortable to wear in summers?
(b) Why does water spilled on the floor evaporate faster than water in a glass?
(c) How does wind speed affect the rate of evaporation?
(d) Two liquids are placed in separate open containers. One liquid boils at 60°C, while the other boils at 100°C. Both containers are set beneath a fan. Which liquid will evaporate more quickly, and why?
3. Matter is anything that occupies space and has mass. It exists mainly in three physical states—solid, liquid, and gas. The properties of matter such as shape, volume, and compressibility differ in each state due to differences in particle arrangement and movement.
In solids, particles are tightly packed in a fixed arrangement. They only vibrate in place, which gives solids a definite shape and volume. In liquids, particles are close but can slide past one another, allowing liquids to flow and take the shape of the container while retaining a fixed volume. In gases, particles are far apart and move freely in all directions. They have neither a fixed shape nor a fixed volume and are highly compressible.
Changes in temperature or pressure can cause matter to change its state (e.g., melting, boiling, condensation). These changes happen because particles gain or lose energy, altering their movement and arrangement.
(a) How are particles arranged and how do they behave in a solid?
(b) What happens to the movement and energy of particles when ice is heated and begins to melt? Relate it to the change in state.
(c) Solids have a fixed shape, but liquids and gases do not. Explain why.
(d) A sealed syringe is filled with air and another identical syringe is filled with water. When equal pressure is applied to both plungers, one moves more easily than the other. Which syringe shows more movement and why? Explain using the particle nature of matter.
4. The common states of matter are solid, liquid, and gas. Solids form when the attraction between individual particles (atoms or molecules) is stronger than the energy (mainly kinetic energy or heat) that causes them to move apart. In solids, the particles are locked in positions near each other, giving solids a definite shape and volume. Although the particles still move, they only vibrate in place.
Liquids are formed when the particle energy increases and the rigid structure of solids breaks down. Liquid particles can slide past each other and collide, but they remain close together. As a result, liquids can flow and take the shape of a container but cannot be compressed easily. Therefore, they have a fixed volume but no fixed shape.
Gases are formed when particle energy exceeds the intermolecular attraction. Gas particles move freely and rapidly in all directions, spreading out to fill the entire container. They are highly compressible and have neither fixed shape nor fixed volume.
(Source: ʻStates of Matter (Part-1)ʼ by David Tin Win, Faculty of Science and Technology, Assumption University, Bangkok, Thailand)
(a) A student builds a model of a substance showing its particles fixed in position and closely packed. What can you conclude about the state of matter being represented? Give a reason for your answer.
(b) Why do liquids retain a fixed volume but not a fixed shape? Use particle arrangement and movement to explain.
(c) Gases can be compressed easily and can fill both small and large containers. Explain how the particle arrangement in gases makes this possible.
(d) A drop of food colouring spreads slowly throughout a glass of water without stirring. What property of water particles causes this to happen?
1. (a) Between 2 and 4 minutes, the ice starts melting and changes into water at 0°C. This is a change of state from solid to liquid.
(b) The process is called fusion or melting.
(c) The heat involved is called latent heat of vaporisation. It is the amount of heat required to convert 1 kg of liquid into vapour at its boiling point without a rise in temperature. The SI unit is Joules per kilogram (J/kg).
(d) Water starts to heat up after 4 minutes. This is evident from the rise in temperature at 6 minutes (12°C), which indicates that the ice has fully melted and heat is now being used to raise the temperature of the water.
2. (a) Synthetic clothes are uncomfortable in summers because they are not porous and do not absorb sweat. As sweat doesnʼt evaporate easily, the body cannot cool down properly, making us feel hot and sticky.
(b) Water spilled on the floor has a larger surface area exposed to air compared to water in a glass. A larger surface area allows more water molecules to evaporate at once, so the evaporation is faster.
(c) Increased wind speed removes the water vapour from the surface more quickly. This allows more liquid molecules to escape into the air, thereby increasing the rate of evaporation.
(d) The liquid with a boiling point of 60°C will evaporate faster. A lower boiling point means weaker intermolecular forces, so its molecules require less energy to evaporate — especially when assisted by airflow from a fan.
3. (a) In solids, particles are tightly packed in a fixed, orderly arrangement. They cannot move freely but only vibrate in their fixed positions, which gives solids a definite shape and volume.
(b) When ice is heated, the particles absorb heat energy and begin to vibrate more vigorously. As the temperature reaches the melting point, the intermolecular forces weaken, and the particles gain enough energy to move past each other, changing the state from solid to liquid.
(c) Solids have a fixed shape because their particles are held together by strong intermolecular forces in fixed positions. Liquids and gases have weaker forces of attraction, allowing particles to move more freely. Liquids can flow and take the shape of a container, and gases can expand to fill the entire space available.
(d) The syringe filled with air (gas) shows more movement. This is because gas particles are far apart and easily compressible, so the plunger moves down easily. Water, being a liquid, has particles that are more closely packed and not compressible, so the plunger resists movement. This demonstrates the compressibility difference between gases and liquids based on their particle arrangement.
4. (a) The model represents a solid. In solids, particles are closely packed and fixed in place, which explains their definite shape and volume. They can only vibrate and do not move freely.
(b) Liquids have loosely packed particles that can slide past each other, allowing the liquid to flow and take the shape of its container. However, since the particles remain close together, liquids retain a fixed volume.
(c) Gas particles are widely spaced and move freely, which makes them easily compressible when pressure is applied. Their free movement also enables them to expand and fill any container, regardless of size.
(d) The particles of water are in constant, random motion, which allows the food colouring to spread throughout the water. This phenomenon is known as diffusion, and it occurs due to the motion of particles in liquids.
Practice Questions
Multiple Choice Questions
1. If during the determination of the melting point of ice, the ice is contaminated with a non-volatile impurity such as common salt, what will happen to its freezing point?
(a) It will increase
(c) It will remain unchanged
(b) It will decrease
(d) It may increase or decrease
2. When heat is continuously supplied to boiling water using a gas burner, what happens to the temperature of the water during vaporisation?
(a) It rises slowly
(c) It rises until steam is produced
(b) It does not rise at all
(d) It rises and then becomes constant
3. The melting points of four solids A, B, C, and D are 388 K, 54 K, 290 K, and 600 K, respectively. Based on this data, the interparticle forces of attraction in these solids follow which of the following orders?
(a) A < B < C < D (b) B < C < A < D (c) C < B < A < D (d) B < D < C < A
4. Mohit has a bowl filled with a substance P. He observes the following:
(I) A pencil can run through the substance.
(II) It takes the shape of the bowl.
(III) Its volume remains 100 mL when placed in different measuring cylinders.
(IV) Its mass remains 100 g in different containers.
Based on these observations, how can substance P be classified? (CBSE QB)
(a) A gas because it has a volume
(c) A liquid because its shape changes
(b) A solid because it has a fixed mass
(d) A gas because objects can pass through it
5. A student performed an activity where he dropped a small amount of blue ink into water. He observed that the ink spread throughout the water without stirring. Why does diffusion happen in liquids? (CBSE QB)
(a) Because liquids have a fixed volume
(b) Because liquids do not have a fixed shape
(c) Because the particles of liquids can move around
(d) Because the particles of liquids are closely spaced
Assertion-Reason Based Questions
In the following two questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
6. Assertion (A): Gases completely fill the container they are placed in.
Reason (R): The particles of gases are closely packed and vibrate in place.
7. Assertion (A): The temperature of a substance remains constant during its change of state.
Reason (R): During a change of state, all the heat supplied is used to increase the kinetic energy of particles.
Very Short Answer Questions (30-50 words)
8. When a drop of blue ink is put in water, the blue colour spreads and the whole solution becomes blue.
(a) Explain the phenomenon in brief due to which this happens.
(b) Give one everyday example where this phenomenon is used or observed.
9. (a) How is the high compressibility property of gases useful to us?
(b) The melting point of ice is 273.16 K. What does this mean?
10. A certain solid has a melting point of 140°C and a boiling point of 250°C.
(a) In which state will the substance be at 240°C?
(b) If the substance is gradually cooled from 240°C, at what temperature will it solidify?
11. Give reasons for the following:
(a) After a hot sunny day, people sprinkle water on the roof or open ground.
(b) A desert cooler works better on a dry summer day.
12. Convert the following temperatures:
(a) 300 K into Celsius
(b) 25°C into Kelvin
13. Neena took some ammonium chloride in a china dish and covered it with an inverted funnel, plugging the stem with cotton. She then heated the dish slowly.
Answer the following questions:
(a) What would she observe during the experiment?
(b) Name and define the phenomenon that takes place.
(c) Name any two other substances with which she can make a similar observation.
14. You are given the following substances along with their melting and boiling points:
Identify the physical states of substances X, Y, and Z at room temperature (30°C). Explain your answer.
15. The following setup is kept in a room on a hot summer day to observe the melting of ice:
Rahul observes the changes in temperature at different intervals as follows:
(i) For the first time interval t₁ to t₂ seconds, the temperature gradually drops from 30°C to 0°C.
(ii) For the next time interval t₂ to t₃ seconds, the temperature remains constant until all the ice melts.
(iii) In the next time interval from t₃ onwards, the temperature increases gradually from 0°C to 100°C.
(a) Draw the graph to show the above-mentioned changes.
(b) Name the property by virtue of which the temperature remained constant in observation (ii).
(c) What should be subsequent reading of temperature on further heating after t4 seconds of Rahul's observation?
16. (a) How do pressure and temperature determine the state of a substance?
(b) What is the chemical formula of dry ice?
(c) Dry ice is compressed under high pressure. What happens to it when the pressure is released?
17. Write three characteristics of particles of matter. Give one example of each.
Long Answer Questions (80-120 words)
18. (a) What is evaporation?
(b) Discuss the various factors which affect the rate of evaporation.
(c) Latent heat of evaporation of two liquids A and B is 100 J/kg and 150 J/kg respectively. Which one can produce more cooling effect and why?
19. (a) Describe an activity to demonstrate that the rate of evaporation increases with surface area.
(b) Explain why clothes dry faster when spread out than when folded.
(c) Why do puddles of water dry up faster on a sunny, windy day compared to a humid, still day?
20. Define latent heat. Explain the differences between latent heat of fusion and latent heat of vaporization giving one example each.
Brain Charge
1. Crossword Puzzle
Across
1. The physical state of matter with the least intermolecular force.
4. The SI unit of temperature.
6. The amount of space occupied by matter.
7. The process of changing from gas to liquid.
8. A change in which the identity of the substance remains the same.
Down
2. The process of changing directly from solid to gas.
3. The temperature at which a liquid changes to gas at atmospheric pressure.
5. The process by which a liquid changes into gas on heating.
9. The state of matter that has a definite volume but no definite shape.
2. Word Puzzle
Use the first alphabet of each answer to the puzzle and combine them to spell a word.
1. The process by which a substance changes from solid to liquid.
2. A gaseous substance present all around us that supports life.
3. It is a factor that affects the rate of evaporation.
4. The process in which a solid directly changes into gas without becoming liquid.
5. The process by which a liquid changes into vapour at any temperature below its boiling point.
6. The characteristic of solids due to which they retain their shape and do not flow.
Challenge Yourself
1. If a gas is compressed in a cylinder, what changes occur in the arrangement and behavior of its particles? What happens if we keep compressing this gas indefinitely?
2. (a) Why do liquids flow but solids do not, even though both have definite volumes? Relate your answer to particle arrangement and movement.
(b) If the same amount of heat is supplied to equal masses of ice and water, which will show a rise in temperature first and why?
3. (a) In a pressure cooker, food cooks faster. How does increasing pressure inside the cooker affect the boiling point of water and what role does this play in cooking?
(b) Dry ice is used to keep food cold during transport. Why does it not leave any liquid behind as it changes state, and what property of dry ice is responsible for this?
4. (a) The temperature of a substance is recorded as 25°C. Convert this temperature into the Kelvin scale. If the substance is heated further by 50°C, what will be the final temperature in Kelvin?
(b) The zero on the Celsius scale corresponds to the freezing point of water. Find out what the zero on the Kelvin scale corresponds to. Answers
Practice Questions
Substance X is a gas. Substance Y is a solid. Substance Z is a liquid.
Brain Charge
1. GAS
Challenge Yourself
(a)
SELF-ASSESSMENT
Time: 1.5 Hour
Multiple Choice Questions
Scan me for Solutions
Max. Marks: 50
(10 × 1 = 10 Marks)
1. Which of the following has the highest rate of diffusion? (a) Solid (b) Liquid (c) Gas (d) Plasma
2. The boiling point of water on the Kelvin scale is: (a) 0 K (b) 100 K (c) 273 K (d) 373 K
3. Which of these processes involves change of state from gas to liquid? (a) Evaporation (b) Condensation (c) Sublimation (d) Melting
4. Which of the following factors affect the rate of evaporation?
(a) Temperature (b) Surface area (c) Humidity (d) All of these
5. When a solid melts into a liquid, the temperature: (a) Increases (b) Decreases (c) Remains constant (d) Becomes zero
6. What is the SI unit of pressure? (a) Newton (b) Pascal (c) Joule (d) Watt
7. What happens to the kinetic energy of water particles at 100°C during boiling? (a) It decreases. (b) It becomes zero. (c) It increases. (d) It remains the same.
8. If the volume of a gas is halved at constant temperature, what happens to its density? (a) It becomes zero (b) It doubles (c) It remains the same (d) It becomes half
9. Which one is a sublimating substance? (a) Ice (b) Salt (c) Naphthalene (d) Sugar
10. Which process allows the smell of perfume to spread throughout a room even without direct contact? (a) Sublimation (b) Evaporation (c) Diffusion (d) Condensation
Assertion–Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
11. Assertion (A): A gas can be converted to a liquid by increasing pressure and decreasing temperature. Reason (R): Increasing pressure and lowering temperature bring particles closer, allowing intermolecular forces to take effect.
12. Assertion (A): Latent heat increases the temperature of a substance during a phase change.
Reason (R): Latent heat adds energy to particles to overcome inter-particle forces.
13. Assertion (A): Solids cannot be compressed.
Reason (R): Solids have negligible space between particles.
14. Assertion (A): Diffusion is faster in gases than in liquids.
Reason (R): Gas particles have greater kinetic energy and more space between them.
15. Riya conducted an experiment to study the rate of evaporation. She poured equal amounts of water into two shallow dishes. She placed one dish near a sunny window and the other in a shaded, humid corner of the room. After a few hours, she noticed that the dish near the window had much less water remaining than the one in the corner. She also observed that when she increased the surface area by spreading the water, it evaporated even faster.
(a) Which factor caused the faster evaporation in the dish near the window?
(i) Smaller surface area
(iii) Increased humidity
(ii) Higher temperature due to sunlight
(iv) Lower wind speed
(b) Why did the water in the humid corner evaporate slowly?
(i) The water was frozen.
(ii) The dish was smaller.
(iii) The air was already saturated with water vapour.
(iv) There was too much sunlight.
(c) How does increasing the surface area affect evaporation?
(i) Slows it down (ii) Speeds it up (iii) Stops it completely (iv) Has no effect
(d) Which of the following would NOT increase the rate of evaporation?
(i) Increasing temperature
(iii) Decreasing surface area
(ii) Decreasing humidity
(iv) Increasing wind speed
16. Matter is anything that occupies space and has mass. Solid, liquid, and gas are the three states of matter. Water can be a solid (ice), a liquid (water), and a gas (water vapour or steam). Its state can be changed by heating or cooling.
Observation Summary:
(I) Ice slowly changes to water when placed in a warm area. This process occurs at 0°C.
(II) Water heats up, and as its temperature rises, it changes to water vapour through evaporation. (III) Once water reaches 100°C, bubbles form, and it boils to become steam.
Also, when steam is cooled, it condenses to form water, and if cooled further, it freezes into ice as shown below.
Condenses to from water Cool below 100°C
(a) The image shows two changes. Identify the process X and Y.
0°C
below 0°C
Freezens or solidifies to form ice
(b) The diagram below shows the change in state of water. What is the process Y called? Define it.
(c) In the following diagram, identify the states X, Y and Z.
Very Short Answer Questions (30-50 words)
17. (a) Define Kinetic energy?
(3 × 2 = 6 Marks)
(b) How do the differences in kinetic energy and inter-particle force affect the states of matter?
18. Why does increasing surface area speed up evaporation? Explain with an example.
19. Why do mountaineers use pressure cookers while cooking at high altitudes?
Short Answer Questions (50-80 words)
20. (a) Define sublimation and give two examples of substances that undergo this process.
(4 × 3 = 12 Marks)
(b) Why is sublimation considered a physical change?
21. (a) What is latent heat of fusion?
(b) Why does the temperature remain constant when ice melts at 0°C despite continuous heating?
22. (a) Describe the effect of pressure on the boiling point of a liquid.
(b) How is this principle used in pressure cookers?
23. (a) Why do naphthalene balls become smaller with time even at room temperature?
(b) Which property of matter does this illustrate?
Long Answer Questions (80-120 words)
24. (a) Explain the concept of latent heat.
(b) Differentiate between latent heat of fusion and latent heat of vaporisation.
(c) Why does temperature remain constant during these phase changes?
25. (a) Why can a piece of chalk be broken easily but not a piece of iron?
(b) Why do liquids take the shape of the container but solids do not?
(c) Which state of matter has the least intermolecular space and why?
(d) Describe an activity to prove that matter occupies space.
(2 × 5 = 10 Marks)
2 Is Matter Around Us Pure?
This chapter delves into the fundamental nature of matter, focusing on the distinction between pure substances and mixtures. It explores the classification of mixtures, including homogeneous and heterogeneous types, while providing insights into solutions, suspensions, and colloidal systems. Through practical examples, such as water, sugar, and everyday consumables, students will learn about these basic concepts. The chapter encourages readers to observe and understand the substances around them, using clear definitions and relevant examples to explain the nature of matter.
Matter
Anything that has mass and occupies space
Pure substances
Matter with a fixed composition that cannot be separated by physical means.
Elements
Consist of one type of atom and cannot be broken down further
Metals
Shiny, malleable, ductile, and good conductors (e.g., iron, copper).
Matter that are that are formed when two or more elements chemically combine.
Organics
Contain carbon, often found in living organisms (e.g., glucose).
Inorganic
Non-carbon based (e.g., water, NaCl).
Mixtures
A combination of two or more substances that retain their individual properties
Homogeneous Mixtures
Mixtures with a uniform composition throughout and can not be seperated easily.
Solutions
Uniform composition, solute dissolved in solvent (e.g., air, alloys).
Hetrogeneous Mixtures
Mixtures which do not have a uniform composition and can be separated easily.
Colloids
Mixture in which particles don't settle (e.g., milk).
Suspension
Mixture with large particles settle on standing (e.g., muddy water).
Show intermediate properties (e.g., silicon, boron).
Chapter at a Glance
• Matter is anything that has mass and occupies space.
• Matter is classified into two pure substances and mixtures.
• Pure Substance is a matter with a fixed composition that cannot be separated by physical means.
• Pure substances can be elements or compounds, exhibiting consistent properties throughout.
• Mixtures is a combination of two or more substances that retain their individual properties.
• Mixtures can be classified as homogeneous or heterogeneous based on their composition and appearance.
• Elements are the substances that consist of one type of atom and cannot be broken down further.
• Elements are pure substances that make up all matter. They are represented in the periodic table.
• Compounds are substances that are formed when two or more elements chemically combine.
• Compounds can be broken down into their components through chemical reactions.
• Homogeneous mixtures are the mixtures with a uniform composition throughout.
• Heterogeneous mixtures are the mixtures which do not have a uniform composition and can be separated easily.
• Solutions are homogeneous mixtures where one substance is dissolved in another, primarily consisting of solvent and solute.
• The concentration is the amount of solute (mass or volume) in a specified amount of solution.
• A solution is a homogeneous mixture with particles smaller than 1 nm not visible to the naked eye.
• Solutions can have varied concentrations, such as mass by mass, mass by volume, and volume by volume percentages.
• Mass by Mass Percentage is expressed as (Mass of solute / Mass of solution) × 100.
• Mass by Volume Percentage is expressed as (Mass of solute / Volume of solution) × 100.
• Volume by Volume Percentage is expressed as (Volume of solute / Volume of solution) × 100.
• Suspensions are heterogeneous mixtures containing solid particles that are sufficiently large to settle out over time.
• Colloids are heterogeneous mixtures where the particle size is too small to settle out or be seen with the naked eye.
• Tyndall Effect is the scattering of light by colloidal particles, making the path of light visible in certain conditions.
Formulae
• Mass by Mass Percentage = Mass of solute Mass of solution × 100
• Mass by Volume Percentage = Mass of solute Volume of solution × 100
• Volume by Volume Percentage = Volume of solute Volume of solution × 100
NCERT Zone
Intext Questions
1. What is meant by a substance?
Ans. A substance is a form of matter that has a uniform and definite composition. It consists of one type of particle and exhibits the same chemical properties throughout. Elements and compounds are both considered substances. For instance, sugar and sodium chloride are substances that consist of only one kind of pure matter, with uniform composition.
2. List the points of differences between homogeneous and heterogeneous mixtures.
Ans.
Homogeneous Mixture
Heterogeneous Mixture
1. It has a uniform composition throughout 1. It has a non-uniform composition
2. Contains a single phase (only one visible layer)
3. Its components cannot be easily separated by physical means
4. Examples: Salt water, air, vinegar
2. Contains two or more phases (distinct layers or components)
3. Its components can often be separated by physical means
4. Examples: Salad, sand and iron filings, oil and water
3. Differentiate between homogenous and heterogeneous mixtures with examples.
Ans. Homogeneous mixtures such as air or saltwater have uniform composition and properties throughout. They appear as a single phase and are also called solutions. Heterogeneous mixtures like soil or a fruit salad contain distinct, physically separate components, and their composition varies from one part to another, showing multiple phases.
4. How are sol, solution and suspension different from each other?
Ans.
Feature
Definition
Particle Size
Uniformity
Solution
A homogeneous mixture of two or more substances where the solute is completely dissolved in the solvent.
Very small particles (less than 1 nanometer)
Completely uniform throughout; appears as a single phase
Sol (a type of Colloid) Suspension
A colloidal system in which solid particles are dispersed in a liquid medium. A heterogeneous mixture in which solid particles are suspended in a liquid or gas.
Intermediate-sized particles (1–100 nanometers)
Appears uniform to the naked eye but particles are not truly dissolved
Large particles (greater than 100 nanometers)
Not uniform; different components are visible
Appearance Transparent and clear Cloudy or translucent Opaque and cloudy
Stability Stable; solute does not settle at the bottom
Tyndall Effect Does not scatter light; no Tyndall effect
Separation Method Cannot be separated by filtration
Stable; particles do not settle easily Unstable; particles settle over time if left undisturbed
Scatters light; shows Tyndall effect
Cannot be separated by ordinary filtration
Examples Salt in water, sugar in water, air Paint, blood, gelatin, milk of magnesia
May scatter light weakly depending on particle size
Can be separated by filtration or decantation
Sand in water, muddy water, flour in water
5. To make a saturated solution, 36 g of sodium chloride is dissolved in 100 g of water at 293 K. Find its concentration at this temperature.
Ans. Given,
Mass of solute (sodium chloride) = 36 g
Mass of solvent (water) = 100 g
Mass of solution = Mass of solute + Mass of solvent = 36 g + 100 g = 136 g
Concentration = Mass of solute Mass of solution × 100 = 36 g 136 g × 100 = 26.47%.
6. Classify the following as chemical or physical changes: cutting of trees, melting of butter in a pan, rusting of almirah, boiling of water to form steam, passing of electric current, through water and the water breaking down into hydrogen and oxygen gases, dissolving common salt in water, making a fruit salad with raw fruits, and burning of paper and wood.
Ans. • Cutting of trees: Physical change
• Melting of butter: Physical change
• Rusting of almirah: Chemical change
• Boiling of water: Physical change
• Breaking water into hydrogen and oxygen: Chemical change
• Dissolving salt in water: Physical change
• Making fruit salad: Physical change
• Burning paper and wood: Chemical change.
NCERT Exercises
1. Which separation techniques will you apply for the separation of the following?
(a) Sodium chloride from its solution in water
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride
(c) Small pieces of metal in the engine oil of a car
(d) Different pigments from an extract of flower petals
(e) Butter from curd
(f) Oil from water
(g) Tea leaves from tea
(h) Iron pins from sand
(i) Wheat grains from husk
(j) Fine mud particles suspended in water
Ans. (a) Sodium chloride from its solution in water: Evaporation
(b) Ammonium chloride from a mixture containing sodium chloride and ammonium chloride: Sublimation
(c) Small pieces of metal in the engine oil of a car: Filtration
(d) Different pigments from an extract of flower petals: Chromatography
(e) Butter from curd: Centrifugation
(f) Oil from water: Separating funnel
(g) Tea leaves from tea: Filtration
(h) Iron pins from sand: Magnetic separation
(i) Wheat grains from husk: Winnowing
(j) Fine mud particles suspended in water: Sedimentation and decantation.
2. Write the steps you would use for making tea. Use the words solution, solvent, solute, dissolve, soluble, insoluble, filtrate and residue.
Ans. The following steps are used for making tea.
• Boil water, which acts as the solvent in the tea-making process.
• Add tea leaves to the boiling water. The soluble parts of the tea leaves (like flavor and color) dissolve in the hot water, making them the solute.
• This creates a solution of tea.
• Add sugar, which is also a soluble substance. It dissolves completely in the hot solution.
• Add milk to the tea, which mixes uniformly with the solution.
• Strain the tea using a sieve or strainer. The liquid that passes through is called the filtrate—this is the final tea you drink.
• The insoluble part of the tea leaves that does not dissolve is left behind in the strainer as the residue
3. Pragya tested the solubility of three different substances at different temperatures and collected the data as given below (results are given in the following table, as grams of substance dissolved in 100 grams of water to form a saturated solution).
(a) What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
(b) Pragya makes a saturated solution of potassium chloride in water at 353 K and leaves the solution to cool at room temperature. What would she observe as the solution cools? Explain.
(c) Find the solubility of each salt at 293 K. Which salt has the highest solubility at this temperature?
(d) What is the effect of change of temperature on the solubility of a salt? What mass of potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 grams of water at 313 K?
Ans. (a) At 313 K, the solubility of potassium nitrate is 62 g per 100 g of water.
For 50 g of water:
Let x be the mass of potassium nitrate needed.
Then, x 50 g = 62 g 100 g , so x = 62 g × 50 g 100 g = 31 g.
Therefore, 31 g of potassium nitrate is needed for a saturated solution.
(b) As Pragya’s saturated solution of potassium chloride cools, she would observe the formation of potassium chloride crystals. This happens because the solubility of potassium chloride decreases with decreasing temperature, causing the excess solute to come out of the solution as solid crystals.
(c) At 293 K, the solubility of salts are:
Potassium nitrate – 32 g per 100 g of water
Sodium chloride – 36 g per 100 g of water
Potassium chloride – 35 g per 100 g of water
Ammonium chloride – 37 g per 100 g of water.
Ammonium chloride has the highest solubility at this temperature.
(d) The solubility of a salt typically increases with temperature. It is clear from the given data that as temperature increases, more solute can dissolve in the solvent, forming a saturated solution. However, this behavior can vary depending on the specific nature of the salt and the solvent.
4. Explain the following giving examples.
(a) Saturated solution (b) Pure substance (c) Colloid (d) Suspension
Ans. (a) A saturated solution is a solution in which maximum amount of solute has been dissolved at a particular temperature. For e.g a saturated solution of potassium nitrate has 32 grams of potassium nitrate per 100 grams of water at 290 K.
(b) A pure substance is made of only one kind of particle and has uniform properties, e.g., distilled water, copper, iron, zinc etc.
(c) A colloid is a heterogeneous mixture that has dispersed particles of size (1nm–100nm) which is larger than a solution(< 1 nm) but smaller than in a suspension (100 nm). These particles are not visible to eyes but are able to scatter light. Examples include milk
(d) A suspension is a heterogeneous mixture with visible solute particles. Here the size of the solute particles are more than 100 nm and are visible to naked eyes. e.g., sand in water.
5. Classify each of the following as a homogeneous or heterogeneous mixture. soda water, wood, air, soil, vinegar, filtered tea.
Ans. Homogeneous Mixture Heterogeneous Mixture
6. How would you confirm that a colourless liquid given to you is pure water?
Ans. To confirm if a colorless liquid is pure water, check if it boils at 100°C under standard atmospheric pressure and freezes at 0°C. Additionally, its density should be approximately 1 g/cm³. Conducting electrical conductivity tests should show no conductivity, indicating absence of dissolved ions, confirming purity.
7. Which of the following materials fall in the category of a �pure substance�?
(a) Ice (b) Milk (c) Iron (d) Hydrochloric acid (e) Calcium oxide (f) Mercury (g) Brick (h) Wood (i) Air
Ans. (a) Ice, (c) Iron, (d) Hydrochloric acid, (e) Calcium oxide, and (f) Mercury are pure substances.
8. Identify the solutions among the following mixtures.
(a) Soil (b) Sea water (c) Air (d) Coal (e) Soda water
Ans. (b) Sea water, (c) Air, and (e) Soda water are solutions.
9. Which of the following will show �Tyndall effect�?
(a) Growth of a plant (b) Rusting of iron (c) Mixing of iron filings and sand (d) Cooking of food (e) Digestion of food (f) Freezing of water (g) Burning of a candle
Ans. Chemical changes: (a) Growth of a plant, (b) Rusting of iron, (d) Cooking of food, (e) Digestion of food, (g) Burning of a candle.
Multiple Choice Questions
1. Which of the following is not a compound? (a) Water (b) Carbon dioxide (c) Iron (d) Sodium chloride
2. Which of the following is a solution? (a) Milk (b) Muddy water (c) Salt water (d) Smoke
3. Which of the following is an example of a pure substance? (a) Brass (b) Bronze (c) Oxygen gas (d) Sea water
4. Two substances, A and B were made to react to form a third substance, A2B according to the following reaction:
2A + B → A2B
Which of the following statements concerning this reaction are incorrect?
(i) The product AB shows the properties of substances A and B
(ii) The product will always have a fixed composition
(iii) The product so formed cannot be classified as a compound
(iv) The product so formed is an element
(a) (i), (ii) and (iii) (b) (ii), (iii) and (iv) (c) (i), (iii) and (iv) (d) (i), (ii) and (iv)
5. What is the main difference between a suspension and a colloid?
(a) Particle size
(c) Type of solute
6. Which of the following are chemical changes?
(i) Decaying of wood
(iii) Sawing of wood
(a) (i) and (ii)
(c) (iii) and (iv)
7. What kind of mixture is blood?
(a) Homogeneous mixture
(c) Compound
(b) Type of solvent
(d) Method of separation
(ii) Burning of wood
(NCERT Exemplar)
(iv) Hammering of a nail into a piece of wood
(b) (ii) and (iii)
(d) (i) and (iv)
(b) Heterogeneous mixture
(d) Element
8. Which of the following is a characteristic of mixtures?
(a) Fixed melting and boiling points
(c) Always homogeneous
(b) Components retain their individual properties
(d) Chemically combined
9. Which of the following is a mixture of elements only?
(a) air
(c) chalk
(b) brass
(d) water
10. Which of the following statements is true about a homogeneous mixture?
(a) The components are easily distinguishable.
(b) The components are uniformly distributed.
(c) The components can be separated by filtration.
(d) It always contains water.
11. Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions? (NCERT Exemplar)
(a) 1.00 g of NaCl + 100 g of water
(b) 0.11 g of NaCl + 100 g of water
(c) 0.01 g of NaCl + 99.99 g of water
(d) 0.10 g of NaCl + 99.90 g of water
12. Which of the following statements are true for pure substances? (NCERT Exemplar)
(i) Pure substances contain only one kind of particles
(ii) Pure substances may be compounds or mixtures
(iii) Pure substances have the same composition throughout
(iv) Pure substances can be exemplified by all elements other than nickel
(a) (i) and (ii) (b) (i) and (iii)
13. The particle size in a colloid is:
(a) Larger than 100 nm
(c) Between 1 nm and 100 nm
(c) (iii) and (iv) (d) (ii) and (iii)
(b) Less than 1 nm
(d) Does not affect the state of the colloid
14. Which one of the following is not a property of a compound?
(a) It has a definite composition.
(b) It can be separated into its components by physical methods.
(c) It has properties different from its constituent elements.
(d) It is a pure substance.
15. A mixture of sulphur and carbon disulphide is (NCERT Exemplar)
(a) heterogeneous and shows Tyndall effect
(b) homogeneous and shows Tyndall effect
(c) heterogeneous and does not show Tyndall effect
(d) homogeneous and does not show Tyndall effect
16. Tincture of iodine has antiseptic properties. This solution is made by dissolving (NCERT Exemplar) (a) iodine in potassium iodide
(b) iodine in vaseline
(c) iodine in water (d) iodine in alcohol
17. Which of the following applications does not depend on water as the solvent?
(a) Making alcoholic drinks
(c) Making detergent
18. Which of the following are physical changes?
(i) Melting of iron metal
(ii) Rusting of iron
(iii) Bending of an iron rod
(iv) Drawing a wire of iron metal
(a) (i), (ii) and (iii)
(b) Making cooking oil
(d) Making shampoo
(b) (i), (ii) and (iv) (c) (i), (iii) and (iv)
(d) (ii), (iii) and (iv)
(NCERT Exemplar)
19. Which of the following is an impure substance?
(a) Mercury (b) Lemonade (c) Sugar (d) Silicon
20. Rusting of an article made up of iron is called (NCERT Exemplar) (a) corrosion and it is a physical as well as chemical change (b) dissolution and it is a physical change
(c) corrosion and it is a chemical change (d) dissolution and it is a chemical change
Answers
Constructed Response Questions
Very Short Answer Questions (30-50 words)
1. The ‘sea-water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment. Ans. Sea-water is a homogeneous mixture when considering dissolved salts and minerals, giving a uniform appearance. However, it is heterogeneous due to suspended materials like sand and organic debris, making the overall nature of sea-water depend on the presence or absence of such suspended particles.
2. All mixtures are homogeneous. Comment upon this statement.
Ans. The statement is incorrect because mixtures can be either homogeneous or heterogeneous. A homogeneous mixture has a uniform composition throughout—for example, a solution of sugar or salt in water. Alloys, such as brass (which contains 70% copper and 30% zinc), are also homogeneous mixtures. In contrast, heterogeneous mixtures have visibly different components or phases, like a salad or a mixture of sand and iron filings.
3. Paints are often stirred before being put to use. Why?
Ans. Paint is a colloidal solution, where fine particles of pigment are dispersed in a liquid medium (like water or oil). Over time, the dispersed particles may lose charge and begin to settle at the bottom—this process is called sedimentation. Stirring the paint helps to redistribute these particles evenly throughout the liquid medium, restoring the uniformity of the colloid. This ensures that the paint has the right color, consistency, and texture when applied.
4. Can physical and chemical change occur together? Illustrate your answer.
Ans. Yes, physical and chemical changes can occur together in the same process. This means that while the substance is undergoing a chemical change, it may also go through a physical change at the same time. Example: Burning of a Candle When a candle burns, both physical and chemical changes happen:
• The wax melts from solid to liquid due to heat. This is a change in state, and it can be reversed by cooling, so it�s a physical change.
• The wax near the wick burns, reacting with oxygen to form carbon dioxide, water vapor, heat, and light. This is a chemical change which results into formation of new substance.
5. Why do we apply alum (fitkari) on a cut? What is this effect called?
Ans. Alum (fitkari) is applied on a cut because it helps in stopping bleeding and preventing infection. When dissolved in water, alum forms a true solution, and the positive charge of Al+3 ions in this solution help in coagulating blood, by neutralising negative charge of blood. This effect is called coagulation or precipitation of colloids.
6. Is milk a pure substance? Comment.
Ans. Main compounds of milk are lactose and casein. And it is also called a colloidal mixture (i.e. in which one substance of dispersed insoluble or soluble particles is suspended throughout another substance). Therefore milk is regarded as a mixture not as a pure substance.
7. Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do. (NCERT Exemplar)
Ans. Colloidal particles are small and light enough to remain suspended due to Brownian motion, which keeps them distributed throughout the medium. Suspensions, however, contain larger particles that can overcome Brownian motion and settle due to gravity when undisturbed.
8. Classify the following as physical or chemical properties.
(NCERT Exemplar)
(a) The composition of a sample of steel is: 98% iron, 1.5% carbon and 0.5% other elements.
(b) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
(c) Metallic sodium is soft enough to be cut with a knife.
(d) Most metal oxides form alkalies on interacting with water.
Ans. (a) Physical property – it describes the composition.
(b) Chemical property – dissolving zinc and evolving gas shows reactivity.
(c) Physical property – describes a physical characteristic (softness).
(d) Chemical property – alkali formation is a chemical change.
9. The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50% (mass by volume) solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 mL of water, ‘B’ dissolved 50 g of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 mL of solution. Which one of them has made the desired solution and why?
(NCERT Exemplar)
Ans. Student �C� has made it correctly because 50% (mass by volume) means 50 g of solute for every 100 mL of solution and not in 100 mL of solvent.
Mass by volume percentage = mass of solute (in g) volume of solution (in mL) × 100 = 50 100 × 100 = 50%
Student �A� dissolved 50 g of NaOH in 100 mL of water (solvent) which is incorrect.
Student �B� dissolved 50 g of NaOH in 100 g of water (solvent), which is incorrect.
10. Smoke and fog both are aerosols. In what way are they different?
(NCERT Exemplar)
Ans. Smoke is an aerosol consisting of solid particles dispersed in a gas (air), often from combustion processes. Fog, on the other hand, consists of tiny liquid droplets (water) suspended in air, formed typically by condensation of water vapor under cooler conditions than smoke.
Short Answer Questions (50-80 words)
1. Which of the following are not compounds?
(a) Chlorine gas
(NCERT Exemplar)
(b) Potassium chloride
(c) Iron (d) Iron sulphide
(e) Aluminium (f) Iodine
(g) Carbon (h) Carbon monoxide (i) Sulphur powder
Ans. The correct answers are (a) Chlorine gas, (c) Iron, (e) Aluminium, (f) Iodine, (g) Carbon, and (i) Sulphur powder. These are elements, not compounds. Compounds like potassium chloride and iron sulphide consist of two or more elements chemically combined.
2. Classify the following into elements, compounds and mixtures.
(i) Pure sand (ii) Air
(iii) Ammonia gas (iv) Ice
(v) Glass (vi) CaO
Ans. Elements – Nil
Compounds – Pure sand, Ice, CaO, Ammonia gas
Mixture – Air, Glass.
3. On heating, calcium carbonate gets converted into calcium oxide and carbon dioxide.
(a) Is this a physical or a chemical change?
(b) Can you prepare one acidic and one basic solution by using the products formed in the above process? If so, write the chemical equation involved.
Ans. Calcium Carbonate Calcium oxide + Carbon dioxide heat
(a) It is a chemical change in which new substances are formed.
(b) Calcium oxide when dissolved in water, forms a basic solution. CaO + H2O Ca(OH)2
Carbon dioxide when dissolved in water, forms an acidic solution CO2 + H2O H2CO3
4. Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, nonsonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
(b) Name a non-metal which exists as a liquid at room temperature.
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
(d) Name a non-metal which is known to form the largest number of compounds.
(e) Name a non-metal other than carbon which shows allotropy.
(f) Name a non-metal which is required for combustion.
Ans. (a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Sulphur, phosphorus
(f) Oxygen
5. Classify the substances given in figure into elements and compounds.
Ans. Elements – Cu, Zn, F2, O2, Diamond, Hg Compounds – CaCO3, H2O, NaCl, Wood
6. A scientist tests three unknown liquids. Only one shows the Tyndall effect, the second is transparent but has particles under a microscope, and the third scatters no light. Classify each.
Ans. The first liquid shows the Tyndall effect, indicating the presence of dispersed particles that scatter light—this is a characteristic of a colloid.
The second liquid appears transparent but has particles visible under a microscope, suggesting relatively large, undissolved particles that may settle over time, typical of a suspension.
The third liquid shows no light scattering, meaning the particles are completely dissolved and uniformly distributed, which are properties of a true solution.
Thus, based on observations, the first liquid is a colloid, the second is a suspension, and the third is a true solution.
7. A student describes air as a pure substance. Do you agree? Justify your answer with scientific reasoning.
Ans. No, air is not a pure substance. In scientific terms, a pure substance contains only one type of particle.
• Air is a mixture of gases like nitrogen, oxygen, carbon dioxide, etc., with no fixed composition.
• These gases retain their individual properties and can be separated physically. Hence, air is a homogeneous mixture, not a pure substance.
8. An unknown liquid is tested. It is colorless, does not show the Tyndall effect, boils at a constant temperature, and cannot be separated by physical means. Is it a mixture or a compound? Justify using three scientific points.
Ans. The unknown liquid is a compound based on the following scientific observations.
• First, it does not show the Tyndall effect, which means it is not a colloid or suspension.
• Second, it boils at a constant temperature, a property of pure substances like compounds, unlike mixtures which boil over a range.
• Third, it cannot be separated by physical means, indicating that the components are chemically combined.
Therefore, the liquid is a compound, likely a pure substance such as distilled water.
9. Elucidate the different types of solutions by discussing dilute, concentrated, and saturated solutions with appropriate examples.
Ans. Solutions are classified based on the amount of solute dissolved in a solvent.
• A dilute solution contains a small amount of solute compared to the solvent; for example, lightly mixed lemon juice.
• A concentrated solution has a large amount of solute, like thick sugar syrup.
• A saturated solution holds the maximum amount of solute that can dissolve at a given temperature; for example, salt water in which no more salt dissolves.
10. Classify each of the following, as a physical or a chemical change. Give reasons.
(a) Drying of a shirt in the sun.
(b) Rising of hot air over a radiator.
(c) Burning of kerosene in a lantern.
(d) Change in the colour of black tea on adding lemon juice to it.
(e) Churning of milk cream to get butter.
Ans. (a) Physical change – drying involves water evaporating without changing the fabric�s chemical identity.
(b) Physical change – rise of hot air is due to temperature change without altering air�s chemical makeup.
(c) Chemical change – burning kerosene forms new compounds (CO₂, H₂O) from a reaction.
(d) Chemical change – adding lemon alters tea chemically, often due to acid-base reactions.
(e) Physical change – churning separates cream by mechanical means, not altering milk chemically.
11. Differentiate between compound and element.
Ans.
Element
Compound
1. A pure substance made of only one type of atom. 1. A substance made of two or more different atoms chemically combined.
2. Cannot be broken down into simpler substances by chemical means
3. Represented by a single symbol (e.g., H, O, Fe).
4. Properties are uniform throughout the element.
5. Hydrogen (H), Oxygen (O), Gold (Au), Iron (Fe).
2. Can be broken down into simpler substances (elements) by chemical means.
3. Represented by a chemical formula (e.g., H2O, CO2).
4. Properties are different from the individual elements that make them up.
5. Water (H2O), Carbon dioxide (CO2), Sodium chloride (NaCl).
12. Find out the mass by volume percentage of 15% solution of sulphuric acid (density = 1.02 g mL-¹)
Ans. 15% solution of H2SO4 means 15 g of H2SO4 is present in 100 g of solution.
The density of the solution is 1.02 g/mL.
Volume of solution = mass of solution density = 100 g 1.02 g/mL = 98.04 mL
To find the mass by volume percentage of a solution, we use the formula:
Mass by volume percentage = mass of solute volume of solution × 100
Mass by volume percentage = 15 g 98.04 g/mL ×100
Mass by volume percentage = 15.30%
13. Give two reasons to justify that water is a compound and not a mixture.
Ans. (a) Water (H₂O) is made up of hydrogen and oxygen in a fixed ratio of 2:1 by atoms (or 1:8 by mass). This definite proportion never changes, which is a characteristic of compounds. In mixtures, components can be present in any ratio.
(b) Water is formed by a chemical reaction between hydrogen and oxygen gases. In this reaction, the original properties of hydrogen and oxygen are lost, and a new substance with different properties (water) is formed. In mixtures, no chemical reaction occurs and the substances retain their original properties.
14. Explain why alloys are considered mixtures and not compounds, even though they appear to be homogeneous.
Ans. Alloys like brass (copper + zinc) or steel (iron + carbon) are considered mixtures because:
• Their components are not chemically combined.
• The proportion of metals can vary.
• The individual properties of metals are often retained (e.g., conductivity).
• They can sometimes be separated by physical means (like melting point differences).
15. Differentiate between mixture and compound.
Ans. Mixture
1. A combination of two or more substances physically mixed together.
2. Components can be mixed in any proportion.
3. Components can be separated by physical means (e.g., filtration, evaporation).
4. Each substance retains its own properties.
5. Usually no energy change when components are mixed.
6. Examples: Air, salad, saltwater
Compound
1. A substance formed when two or more elements chemically bond.
2. Elements are combined in a definite ratio.
3. Components can only be separated by chemical methods.
4. The compound has different properties from its elements.
5. Energy is usually absorbed or released during formation.
6. Examples: Water (H₂O), carbon dioxide (CO₂), sodium chloride (NaCl)
Long Answer Questions (80–120 words)
1. Iron filings and sulphur were mixed together and divided into two parts, �A� and �B�. Part �A� was heated strongly while Part �B� was not heated. Dilute hydrochloric acid was added to both the Parts and evolution of gas was seen in both the cases. How will you identify the gases evolved? (NCERT Exemplar)
Ans. In Part ‘A’, when iron and sulphur are heated, they react to form iron sulphide (FeS). Upon adding dilute hydrochloric acid (HCl), hydrogen sulphide (H₂S) gas is evolved, which has a rotten egg smell. In Part ‘B’, since neither reacted chemically, adding dilute HCl primarily reacts with iron to produce hydrogen gas (H₂), which is colorless, odorless, and flammable; this can be identified by the �pop� sound when a burning splint is brought near. Thus, gas evolution in Part ‘A’ suggests hydrogen sulphide from the chemical compound, while Part ‘B’ suggests hydrogen due to unreacted iron.
2. A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/tumbler in the box as shown in the figure. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it? (NCERT Exemplar)
Shoe box
Glass tumbler containing sample Eye
(a) Explain why the milk sample was illuminated. Name the phenomenon involved.
(b) Same results were not observed with a salt solution. Explain.
(c) Can you suggest two more solutions which would show the same effect as shown by the milk solution?
Ans. (a) The milk sample was illuminated due to the Tyndall effect, where colloidal particles scatter the beam of light and make its path visible. Milk, being a colloid, demonstrates this phenomenon.
(b) The salt solution did not scatter light because it is a true solution with particles that are too small to cause scattering, hence the light passed through without illumination.
(c) Two more solutions that would show the Tyndall effect are starch solution and diluted soap solution. Both are colloids and will scatter light under the right conditions.
3. During an experiment the students were asked to prepare a 10% (Mass/ Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.
(a) Are the two solutions of the same concentration
(b) Compare the mass % of the two solutions.
Ans. (a) No, the two solutions do not have the same concentration. Ramesh and Sarika used different methods to prepare the solutions, leading to different total masses and thus different mass percentages.
(b) Ramesh’s Solution:
Sugar = 10 g, Water = 100 g
Total mass of solution = 10 g + 100 g = 110 g
mass % of sugar = 10 110 � 100% = 9.09%
Sarika’s Solution:
Sugar = 10 g, Total mass of solution = 100 g
Water = 100 g − 10 g = 90 g
mass % of sugar = 10 100 � 100% = 10%
Conclusion:
Sarika’s solution has exactly 10% mass/mass concentration, while Ramesh’s solution has a lower concentration of 9.09%.
Therefore, Sarika followed the instruction correctly.
4. (a) How can you determine whether a given solution is saturated or unsaturated?
(b) What changes would occur if a saturated solution of sodium chloride is (i) heated or (ii) cooled?
Ans. (a) To test whether a solution is saturated or unsaturated, add a small amount of solute to it and stir well. If the solute dissolves completely, the solution is unsaturated, meaning it can still
accommodate more solute. If the added solute does not dissolve and settles at the bottom, it indicates the solution is saturated, as it has reached its maximum solubility at that temperature.
(b) If a saturated solution of sodium chloride is heated, it becomes capable of dissolving more salt because solubility increases with temperature. However, if the same solution is cooled, some of the dissolved salt may crystallize out, as the solubility decreases.
5. How do metals differ from non-metals in terms of their physical and chemical properties?
Ans. Metals
1. Metals are lustrous and have a shiny surface.
2. Metals can be hammered into sheets and drawn into wires.
3. Metals generally have high density and high melting and boiling points.
4. Most metals are solids at room temperature, except mercury and gallium.
5. Metals are sonorous—they produce a ringing sound when struck.
6. Metals are good conductors of heat and electricity.
7. Metals are generally hard (exceptions: sodium, potassium).
8. Metals have high tensile strength.
Competency Based Questions
Multiple
Non-metals
1. Non-metals are dull and lack lustre.
2. Non-metals are neither malleable nor ductile.
3. Non-metals usually have low density and low melting and boiling points.
4. Non-metals can exist as solids, liquids, or gases at room temperature.
5. Non-metals are not sonorous.
6. Non-metals are poor conductors of heat and electricity.
7. Metals are generally hard (exceptions: sodium, potassium).
8. Non-metals have low tensile strength.
Choice Questions (Choose the most appropriate option/options)
1. Which property distinguishes a colloidal solution from a true solution?
(a) Uniform appearance throughout
(b) Complete dissolution of solute
(c) Scattering of light (Tyndall effect)
(d) Constant boiling point
2. Which statement best describes a compound?
(a) An element with variable composition.
(b) A pure form of matter with uniform properties but no chemical bonding.
(c) A substance formed by chemical combination of elements in a fixed ratio.
(d) A mixture of different substances.
3. Suspensions are best characterized as:
(a) Homogeneous mixtures with dissolved particles.
(b) Heterogeneous mixtures with larger particles that settle over time.
(c) Solutions with uniformly distributed solute.
(d) Colloidal mixtures with stable particles.
4. Which example best illustrates a suspension?
(a) Sugar solution (b) Air (c) Milk (d) Muddy water
5. Which of the following differentiates a compound from a mixture?
(a) Fixed composition and chemical bonding
(b) Variable composition and physical blending
(c) Uniform particle size and non-separation by filtration
(d) Presence of solute and solvent
6. Which set correctly classifies elements into metals, non-metals, and metalloids?
(a) Iron (Metal), Oxygen (Non-metal), Silicon (Metalloid)
(c) Non-uniform composition with visible distinct parts
(d) Stable homogeneous structure
9. Which of the following is an example of a heterogeneous mixture?
(a) Sand and water.
(c) Sugar solution.
(b) Alloy of copper and zinc
(d) Salt dissolved in water
10. If 40 g of salt is dissolved in 320 g of water forming a saturated solution, what is the concentration (mass percentage)?
(a) 12.5%. (b) 15%. (c) 11.1%. (d) 10%
11. What is a common misconception about the term �pure� among consumers?
(a) That pure substances are mixtures
(b) That pure substances can be easily separated
(c) That pure means free from adulteration
(d) That purity refers to varied chemical composition
12. In a solution, the solvent is the component that:
(a) remains undissolved in the mixture.
(b) is always a liquid.
(c) dissolves the solute and is usually present in greater amount.
(d) can be chemically changed to solute.
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): A solution of salt in water is a homogeneous mixture.
Reason (R): The components of a homogeneous mixture are uniformly distributed throughout the mixture.
2. Assertion (A): Milk is a colloid.
Reason (R): Colloids are heterogeneous mixtures in which the particle size is intermediate between those in true solutions and suspensions.
3. Assertion (A): When a beam of light is passed through a colloidal solution, its path becomes visible.
Reason (R): Light do not gets scattered by colloidal particles.
4. Assertion (A): Milk is an aerosol.
Reason (R): Milk contains liquid as dispersed phase in liquid as dispersion medium.
5. Assertion (A): Rusting of iron is a chemical change.
Reason (R): During rusting, iron is converted into hydrated ferric oxide which cannot be converted back to iron.
6. Assertion (A): Water and methane are known as compounds.
Reason (R): Water and methane are heterogenous in nature.
7. Assertion (A): Every pure substance is either an element or a compound.
Reason (R): All mixtures contain multiple types of pure substances.
8. Assertion (A): Chemical changes result in the formation of new substances with distinct properties.
Reason (R): All chemical changes are accompanied by a change in physical state.
Case-Based/Source-Based/Passage-Based Questions
1. A chemical factory specializes in manufacturing synthetic dyes used in textiles. In the dyeing process, dye powders are first dissolved in water to prepare solutions before being applied to fabrics. The company noticed that some dyes dissolved instantly, while others left undissolved residue, leading to patchy colouring. To solve this issue, a quality control team was assigned to study the properties of different types of solutions. They observed the following:
• When Dye A was mixed with water, it dissolved completely and gave a clear, coloured solution.
• Dye B, when mixed, remained suspended for a while and settled over time.
• A laser beam passed through Dye A’s solution did not scatter, but with Dye B, a visible path was seen.
• The team also tried filtering both solutions. Dye A passed through, but Dye B left particles on the filter paper.
(a) What does the inability of Dye A to scatter light indicate?
(i) Large particle size
(iii) Particle size is less than 1 nm
(ii) It’s a colloidal solution
(iv) It’s an opaque solution
(b) The presence of particles on filter paper after filtering Dye B’s solution proves that:
(i) Dye B is a soluble substance
(iii) Dye B is forming a suspension
(ii) Dye B is forming a colloid
(iv) Dye B is forming a true solution
(c) Which of the following properties is not applicable to Dye A’s solution?
(i) Homogeneity
(iii) Particle size > 100 nm
(ii) Transparency
(iv) Stability
(d) If a solution remains stable for hours and cannot be filtered but scatters light faintly, it is likely:
(i) A suspension
(iii) A true solution
(ii) A colloid
(iv) A pure substance
2. The Modern Engineering Corporation (MEC) specializes in building heavy machinery. For years, the company used pure iron for making machine parts. However, they noticed that over time, the iron components rusted easily and were not strong enough to handle heavy loads. So it was suggested that alloys often have better properties than the pure metals they’re made from. After switching to alloybased parts, the machinery’s life improved drastically, rusting was minimized, and performance increased.
(a) Which of the following correctly defines an alloy?
(i) A pure metal
(ii) A compound of metals
(iii) A heterogeneous mixture
(iv) A homogeneous mixture of metals or metals with non-metals
(b) Bronze, used in making medals, is an alloy of:
(i) Copper and zinc
(iii) Copper and iron
(ii) Copper and tin
(iv) Copper and lead
(c) Why are alloys considered mixtures and not compounds?
(i) They are formed by chemical reactions
(ii) The components can be separated chemically
(iii) They retain the properties of their individual elements
(iv) They have a fixed ratio of components
(d) Which of the following properties makes alloys useful in industries?
(i) They are easily broken
(ii) They conduct electricity poorly
(iii) They are stronger, more resistant to rust, and last longer
(iv) They are chemically unstable
3. While helping in the kitchen, someone noticed salt being added to water while boiling rice and wondered if it formed a new substance. It was explained that salt water is not a new substance, as salt can be recovered by evaporation. To explore further, salt was dissolved in water and left in the sun—salt crystals reappeared after evaporation. Later, it was found that sodium (a reactive metal) and chlorine (a poisonous gas) chemically combine to form sodium chloride (NaCl), the safe, edible table salt. This demonstrated that salt water is a mixture, while table salt is a compound with entirely new properties.
(a) Why can salt be recovered from salt water by evaporation?
(i) Because it is a compound
(ii) Because it has a low melting point
(iii) Because salt water is a mixture and salt retains its identity
(iv) Because water and salt react chemically
(b) Which of the following is true about compounds?
(i) They can be separated by physical means
(ii) They have variable composition
(iii) They retain the properties of elements
(iv) They have a fixed composition and new properties
(c) Which of these statements is correct about mixtures?
(i) They are always solid
(ii) Their components lose their properties
(iii) Their components are chemically combined
(iv) Their components retain their individual properties
(d) Which pair correctly shows a mixture and a compound?
(i) Iron & sulphur (compound), air (mixture)
(ii) Salt water (mixture), NaCl (compound)
(iii) Water (compound), milk (compound)
(iv) Soil (compound), carbon dioxide (mixture)
4. While adding milk to tea, a student noticed the mixture appeared cloudy but uniform. Curious about this, they shone a torch through the tea and observed the light beam becoming visible. Intrigued by the unusual effect, the student decided to explore further and discovered that such mixtures could behave in interesting ways. This observation sparked a deeper interest in understanding how certain mixtures form and how they behave in different situations.
(a) What type of mixture is formed when milk is added to tea?
(i) Homogeneous
(iii) Colloidal
(ii) Heterogeneous
(iv) Supercritical
(b) Why does the tea appear cloudy yet uniform after adding milk?
(i) It’s a chemical reaction
(ii) Milk disperses as large suspended particles
(iii) Colloidal particles scatter light uniformly
(iv) A supersaturated solution forms
(c) Which of the following is not a characteristic of colloids?
(i) Exhibit Tyndall effect
(ii) Settle on standing
(iii) Appears homogeneous to naked eye
(iv) Have dispersed phase and dispersion medium
(d) Which component acts as the dispersed phase in the milk-tea colloid?
(i) Tea (ii) Water
(iii) Milk fat/proteins
Answers
Multiple Choice Questions
Assertion-Reason Based Questions
1.
(iv) Sugar
Case-Based/Source-Based/Passage-Based Questions
1. (a) (iii) (b) (iii) (c) (iii) (d) (ii)
2. (a) (iv) (b) (ii) (c) (iii) (d) (iii)
3. (a) (iii) (b) (iv) (c) (iv) (d) (ii)
4. (a) (iii) (b) (iii) (c) (ii) (d) (iii)
Numerical Questions
1. If 15 g of a solute is dissolved to make 85 g of solution, what is the mass percentage of the solute?
Ans. Given,
Mass of solute = 15 g
Mass of solution = 85 g
Mass % = Mass of solute
Mass of solution � 100 = 15 85 × 100 17.65%.
2. A solution is made by dissolving 25 g of sugar in 100 g of water. Calculate the mass percentage of sugar in the resulting solution.
Ans. Given,
Mass of solute (sugar) = 25 g
Mass of solvent (water) = 100 g
Mass of solution = 25 + 100 = 125 g
Mass % = Mass of solute
Mass of solution � 100 = 25 125 × 100 = 20%.
3. Using the volume by volume (V/V) percentage method, if 30 mL of a solute is mixed to make 300 mL of solution, what is the V/V percentage of the solute?
Ans. Given,
Volume of solute = 30 mL
Volume of solution = 300 mL
Volume % = Volume of solute
Volume of solution � 100 = 30 300 × 100 = 10%.
4. In a saturated solution, 36 g of NaCl is dissolved in 100 g of water. Calculate the mass percentage of the NaCl in the saturated solution.
Ans. Given,
Mass of NaCl = 36 g
Mass of water = 100 g
Total mass of solution = 136 g
Mass % = Mass of solute
Mass of solution � 100 = 36 136 × 100 = 26.47%
5. For a solution of iodine in alcohol, 5 g of iodine is dissolved in 45 mL of alcohol. Given that the density of the resulting solution is 0.95 g/mL, calculate the mass by volume percentage of iodine (assume the final volume remains approximately 45 mL).
Ans. Given,
Mass of solute = 5 g
Volume of solution = 45 mL
Mass/Vol % = Mass of solute
Volume of solution � 100 = 5 45 × 100 = 11.11%.
6. A solution has a mass percentage of 20% and a total mass of 500 g. Determine the mass of the solute and the solvent.
Ans. Given,
Mass % = 20%
Total mass = 500 g
Mass of solute = Mass % 100 � 100 = 20 100 × 500 = 100 g
Mass of solvent = Total mass – Solute = 500 – 100 = 400 g
7. A solution is prepared by dissolving 75 g of solute in 225 g of water. If 30 g of water evaporates, calculate the new mass percentage of the solute.
Ans. Given,
Initial solute = 75 g
Initial water = 225 g
Water evaporated = 30 g
New total mass = 75 + 195 = 270 g
Mass % = Mass of solute
Mass of solution � 100 = 75 270 × 100 = 27.78%
8. An unknown solute is dissolved in 150 g of water to yield a saturated solution with an 18% mass percentage. Calculate the mass of the solute used.
Ans. Given,
Water = 150 g
Mass % = 18%
Let mass of solute be x g
Total mass of solution = (x + 150) g
Mass % = Mass of solute
Mass of solution � 100 = x x + 150 � 100 = 18.
100x = 18x + 2700 x = 2700 88 g = 32.93 g
9. Two solutions are mixed: one of 200 g at 15% solute and another of 300 g at 30% solute. Calculate the overall mass percentage of the solute in the mixture.
Ans. Given,
First solution: 200 g, mass % = 15%
Second solution: 300 g, mass % = 30%
Total mass = 500 g
Total solute = (0.15 × 200) + (0.30 × 300) = 30 + 90 = 120 g
Mass % = Total solute
Total mass � 100 = 120 500 × 100 = 24.0%
10. Calculate the mass of sodium sulphate required to prepare a 20% (w/w) solution in 100 g of water. Ans. Given,
Water = 100 g
Let mass of sodium sulphate be x g,
Total mass = x + 100
Mass % = Total solute
Total mass � 100 = 20 = x x + 100 � 100 = 20
100x = 20x + 2000 x = 2000 80 = 25 g
Practice Questions
Multiple Choice Questions
1. Choose the correct option.
Type of Colloid
A. Sol (solid in liquid)
B. Emulsion (liquid in liquid)
C. Foam (gas in liquid)
D. Aerosol (solid in gas)
E. Gel (liquid in solid)
(a) A-l, B-ll, C-lV, D-lll, E-V
(c) A-lV, B-ll, C-lll, D-l, E-V
Example
l. Smoke
ll. Milk
lll. Shaving cream
lV. Paint
V. Jelly
(b) A-ll, B-V, C-l, D-lll, E-lV
(d) A-lll, B-V, C-l, D-ll, E-lV
2. A solution that cannot dissolve any more solute at a given temperature is said to be?
(a) Unsaturated (b) Saturated
(c) Supersaturated
(d) Dilute
3. Which of the following statements about solutions is true?
(a) Solutions are always homogeneous mixtures.
(b) Solutions can be separated into their components by filtration.
(c) The solute is present in larger quantity than the solvent in a solution.
(d) Solutions cannot be formed with gases.
4. Which of the following is a chemical change?
(a) Freezing water to form ice
(c) Digesting food in the stomach
(b) Cutting a piece of wood into smaller pieces
(d) Melting chocolate to form a liquid
5. A solution contains 40 g of common salt in 320 g of water. The mass by mass percentage of the solution is
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
6. Assertion (A): In a solution, the solute is present in a larger amount.
Reason (R): The solvent dissolves the solute.
7. Assertion (A): Mixtures have fixed melting and boiling points.
Reason (R): A mixture contains two or more substances in any proportion.
Very Short Answer Questions (30-50 words)
8. What is the difference between a solid solution and a solid sol? Explain with an example.
9. Define an alloy. What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments?
10. How can a saturated solution be made unsaturated?
11. Give an example of:
(a) Solid- solid homogeneous mixture
(b) Solid- liquid homogenous mixture
(c) Liquid- liquid heterogeneous mixture
(d) Solid- solid heterogeneous mixture
12. Classify the following as physical or chemical change:
(a) Sublimation of ammonium chloride
(b) Freezing of mercury
(c) Glowing of a bulb
(d) Magnetizing a piece of iron
Short Answer Questions (50-80 words)
13. (a) How can a saturated solution be converted into super saturated solution?
(b) A saturated solution of potassium nitrate at 313 K contains 62 g of solute per 100 g of water. How many grams of potassium nitrate are required to prepare a saturated solution in 50 g of water?
14. Differentiate between physical and chemical change.
15. A teacher shows you two clear liquids in separate beakers. One is a solution of salt in water, and the other is a colloid of milk. Without tasting or smelling, how can you identify which beaker contains the solution and which contains the colloid? Provide a detailed explanation.
16. A saturated solution of a salt is prepared at a temperature where its solubility is 36 g per 100 g of water. If 250 g of water is used and upon cooling 15 g of salt crystallizes out, calculate the mass percentage of salt remaining in the solution.
17. Raman put iron filings and sulphur powder in a china dish, then he heated it. Explain how can he determine the difference between the two. Also give two points of difference between the substances formed.
Long Answer Questions (80-120 words)
18. (a) Name the solute and solvent in the given solutions:
(i) Aerated drinks (ii) Tincture of iodine
(b) What is the practical application of tincture of iodine?
(c) Name the type of colloid to which face cream belongs? Also mention dispersing phase and dispersing medium present in it.
(d) What is the size of colloidal particles?
19. (a) Give one example of each.
(i) emulsion (ii) gel (iii) sol (iv) Solid sol
(b) Give one real life situation where you observe Tyndall effect.
(c) How are the colloidal particles able to scatter light?
20. (a) A chemist prepares a saturated solution with a mass concentration of 35%. If 120 g of water is used, calculate the mass of solute required.
(b) What do you mean by concentrated and dilute solution?
(c) What do you mean by solubility of a solution?
Brain Charge
1. Unscramble the following words related to pure substances and mixtures:
(a) EUMXITR _______________
(c) GUEHNOMEONO _______________
(e) ROTEHENOGEENU ______________
2. Circle the odd one out and explain why.
(b) NOOMCEUPD _______________
(d) LEMENTE _______________
(a) Water (H₂O), Air, Salt (NaCl), Carbon dioxide (CO₂)
(b) Iron, Copper, Sugar solution, Gold
(c) Lemonade, Blood, Milk, Water
3. Find who am I
(a) I have two or more substances, not chemically joined, My composition may vary, and my nature is kind. You can find me in your kitchen, on your plate, What am I? Don’t hesitate!
(b) Hydrogen and oxygen, when they unite, They form something that’s colorless and right. Essential for life, from sky to shore, Guess who I am—you’ve used me before!
(c) Though I’m a mix, I look the same, Milk and air can share my name.
I hide my parts so neat and tight, You’ll need no stir, I’m just right.
(d) I’m cloudy and messy, not very shy, You can see my bits with the naked eye. Sand in water or salad on your plate, Guess who I am before it’s too late!
Challenge Yourself
1. You are given a sample of water from a river and a sample of distilled water. How would you determine which one is the pure substance and which one is a mixture? Explain the reasoning behind your method.
2. Discuss how is the smoke precipitated?
3. Why do some food products, like salad dressings, require emulsifiers to remain stable, while others, like milk, remain stable without any added stabilizers?
4. In the preparation of certain types of paints, pigments are often dispersed in liquids to form a colloidal system. Why is it crucial to maintain the colloidal nature of paint for its application, and what happens if the colloid becomes unstable?
5. How does a delta form?
6. How is artificial rain induce?
Answers
Practice Questions
1.
Brain Charge
1. (a) MIXTURE
(b) COMPOUND
(c) HOMOGENEOUS
(d) ELEMENT
(e) HETEROGENEOUS
2. (a) Air – it�s a mixture; others are compounds
(b) Sugar solution – it�s a mixture; others are elements
(c) Water – it�s a pure compound; others are mixtures
3. (a) Mixture (b) Water (H₂O)
(c) Homogeneous mixture (d) Heterogeneous mixture
Challenge Yourself
1. Visual Observation:
• Distilled Water: Clear and colorless
• River Water: May look cloudy or have particles
Boiling Point Test:
• Distilled Water: Boils at 100°C (fixed point)
• River Water: Boiling point may vary due to impurities
• River Water: Conducts electricity (due to dissolved salts and minerals)
Evaporation Test:
• Distilled Water: Leaves no residue
• River Water: Leaves solid residues (impurities)
2. Smoke is a colloid in which solid particles are dispersed in a gas (usually air). These particles often carry similar electrical charges, which keep them suspended by repelling each other.
To precipitate smoke, this charge stability must be disrupted. In devices like electrostatic precipitators, oppositely charged plates are used. When smoke passes through, the charged particles are attracted to the oppositely charged plates, neutralizing their charges. This causes the particles to lose repulsion, aggregate, and settle out of the air, effectively removing the smoke from the gas.
3. Emulsifiers are substances that help mix liquids that normally don�t blend together, like oil and water. In things like salad dressings, oil and vinegar (or water) don’t naturally mix. Emulsifiers like mustard or lecithin work by reducing the surface tension between the oil and water, helping them form a stable mixture called a colloid.
Milk, on the other hand, is already a natural mixture where tiny fat particles are spread out in water. Milk doesn�t need extra emulsifiers because it has natural proteins called casein. These proteins act like an invisible coat around the fat particles, stopping them from clumping together. This is why milk stays stable without needing added emulsifiers, unlike some other food products.
4. In paints, pigments are dispersed as colloidal particles in a liquid medium. Maintaining this colloidal nature is crucial because the small, evenly distributed particles ensure a smooth, uniform finish when applied. If the colloid becomes unstable, the particles may aggregate or settle due to loss of repulsive forces (like charge stabilization). This leads to clumping, uneven color, and poor coverage. Stability in the colloid prevents sedimentation and keeps the paint ready for use without constant stirring, ensuring consistent quality and application.
5. A delta forms due to colloidal particles (like clay and silt) carried by river water. These particles often carry similar charges, which keep them suspended. When river water meets the still water of a sea or lake, the speed slows, and dissolved salts neutralize the charges on the particles, causing them to aggregate and settle. This process, known as coagulation, leads to the deposition of sediments, gradually forming a delta.
6. Artificial rain, or cloud seeding, involves introducing substances like silver iodide into moist clouds. These substances have charged particles that act as nuclei around which water droplets form. The charged particles attract water molecules, causing them to cluster around them. As the droplets grow in size, they become heavy enough to fall as rain. This method is used in areas experiencing drought to increase rainfall, but its success depends on the charge interactions and the moisture content of the clouds, as well as weather conditions.
SELF-ASSESSMENT
Time: 1.5 Hour
Scan me for Solutions
Max. Marks: 50
Multiple Choice Questions (10 × 1 = 10 Marks)
1. Which of the following is not a physical change?
(a) Glowing of a flouroscent tube.
(b) Burning of a magnesium wire
(c) Adding water to anhydrous copper sulphate
(d) Dissolving sulphur in carbon disulphide
2. Blood can be best coagulated by:
(a) barium chloride
(b) alcohol
(c) sodium chloride (d) alum
3. Which statement correctly differentiates between types of mixtures?
(a) Homogeneous mixtures have uniform composition while heterogeneous mixtures do not
(b) Mixtures only consist of compounds
(c) Mixtures have fixed composition
(d) All mixtures are non-separable by physical methods
4. Brass is made by mixing copper and zinc. What type of mixture is brass?
(a) Heterogeneous solid mixture
(c) Heterogeneous compound
(b) Homogeneous compound
(d) Homogeneous solid solution
5. Which of the following alloys is used in making aircraft bodies due to its high strength and low density?
(a) Brass
(b) Bronze
(c) Solder (d) Duralumin
6. What property distinguishes mixtures from pure substances?
(a) Mixtures can vary in composition
(b) Pure substances are physical blends
(c) Mixtures are always chemically bonded (d) Pure substances have variable properties
7. Which of the following best differentiates a physical change from a chemical change?
(a) Physical change alters form but not composition, chemical change alters composition
(b) Physical change always involves a colour change
(c) There is no energy change in physical change
(d) Chemical change is reversible, physical change is irreversible
8. Blood can be best coagulated by:
(a) Barium chloride
(b) Alcohol
(c) Sodium chloride (d) Alum
9. Which of the following is not a suspension?
(a) Paints
(b) Bleaching Powder
(c) Lime Water (d) Potassium Permanganate
10. Which of the following solutions show Tyndall effect?
(a) Sulphur dissolved in carbon disulphide (b) Sugar dissolved in water
(c) Iodine dissolved in alcohol
Assertion–Reason Based Questions
(d) Milk
(4 × 1 = 4 Marks)
In each of the following questions, a statement of assertion is given by the corresponding statement of reason. Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): In a solution, the solvent dissolves the solute resulting in a uniform mixture throughout. Reason (R): The extremely small size of solute particles prevents them from settling or being separated by filtration.
12. Assertion (A): Brass is homogeneous mixture of Cu and Zn even though its composition can be variable. Reason (R): Cu and Zn cannot be separated by any physical method.
13. Assertion (A): A saturated solution cannot dissolve any additional solute at a given temperature.
Reason (R): At the saturation point, equilibrium is established between the dissolving solute and precipitating solute.
14. Assertion (A): Suspensions are heterogeneous mixtures in which the dispersed particles eventually settle over time.
Reason (R): The particles in a suspension are small enough to remain permanently suspended in the medium.
15. In a laboratory experiment, two samples of a substance are heated. One sample shows a change in state and color, while the other only changes state without altering color.
(a) What does the change in color indicate in one of the samples?
(b) How can you differentiate between a physical change and a chemical change?
(c) Which of the following is a sign of a chemical change?
(i) Melting
(iii) Dissolving
(ii) Color change
(iv) Evaporation
(d) Why are simultaneous physical and chemical changes significant in everyday phenomena?
16. In a village in Rajasthan, well water used for drinking was found to be cloudy and salty. A group of school students, during a science project, tested the water and discovered it was a mixture of salts, sand, and impurities. They explained that mixtures, unlike compounds such as NaCl and H₂O, can be separated using physical methods. The project helped villagers understand the difference between mixtures and compounds, showing how simple science can solve real-life problems.
(a) Which of the following can be separated by physical methods?
(i) Water (H₂O)
(iii) Sand and salt mixture
(ii) Carbon dioxide (CO₂)
(iv) Glucose
(b) Match the following:
Example
Type
A. Air 1. Compound
B. Water 2. Mixture
C. Carbon dioxide 3. Mixture
D. Milkshake 4. Compound
(i) A-4, B-2, C-3, D-1
(iii) A-2, B-3, C-4, D-1
(ii) A-3, B-1, C-4, D-2
(iv) A-2, B-1, C-3, D-4
(c) Why can�t a compound be separated by physical methods?
(d) Which method would best separate sand from water?
(i) Distillation
(iii) Filtration
(ii) Evaporation
(iv) Crystallization
Very Short Answer Questions (30-50 words) (3 × 2 = 6 Marks)
17. What is tincture of iodine? Why is it important?
18. You are given two samples of water labelled as ‘A’ and ‘B’. Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment.
19. If a saturated solution of sodium chloride is prepared by dissolving 36 g of salt in 100 g of water at 293 K, calculate its mass by mass percentage concentration.
20. Is air a mixture or compound? Give three reasons.
21. What gas is produced when a mixture of 7 g iron filings and 4 g of sulphur powder is treated with dilute sulphuric acid at room temperature? what gas would be produced, if the same mixture is first heated strongly, cooled and then treated with dilute sulphuric acid? what is the cause of this difference in behaviour?
22. How will you distinguish the following:
(a) sea water from pure water
(b) metals from non-metals
(c) two different compounds having same boiling point.
23. Milk in the market is sold in different varieties like full fat, 2% fat and 1% fat. How are these varieties of milk prepared?
Long Answer Questions (80-120 words)
24. (a) Under which category of mixtures will you classify alloys and why?
(b) A solution is always a liquid. Comment.
(c) Can a solution be heterogeneous?
(2 × 5 = 10 Marks)
25. (a) A non-metal combines with oxygen to form two gases �A� and �B� . Gas �A� is poisonous but extinguishing fire. The non-metal also exist in two allotropic �C� and �D� where �C� is the good conductor of electricity and �D� is the bad conductor of electricity. Name the non-metal and identify A, B, C, D.
(b) Classify the following as pure substances or mixtures. Separate the pure substances into elements, compounds and divide the mixtures into homogenous and heterogenous:
(i) Air (ii) Milk (iii) Graphite (iv) Gasoline (v) Diamond (vi) Tap water (vii) Distilled water (viii) Oxygen (ix) Brass (x) 22 Carat gold (xi) Steel (xii) Iron (xiii) Sodium chloride (xiv) Iodised table salt.
3 Atoms and Molecules
This chapter explores the foundational concepts of atoms and molecules, delving into their historical and scientific development. It highlights the laws of chemical combination and provides insights into the structure and composition of matter, emphasizing the significance of symbols and atomic masses in representing elements. The contributions of notable figures such as Lavoisier and Dalton are discussed, alongside Indian philosophers like Maharishi Kanad, who introduced early notions of atomic concepts. The chapter aims to build a foundation for understanding chemical relationships, properties, and formulae.
Laws of chemical combination
Law of conservation of mass
• Mass can neither be created nor be destroyed in a chemical reaction.
Law of constant proportions
• A compound consists of same elements combined in a fixed ratio by mass.
Atom
• Smallest and basic unit of matter.
• Consists of electrons, protons & neutrons.
• Incapable of independent existence except noble gas atom.
Atomic Mass
• Mass of one atom of an element.
• Measured in atomic mass unit (u).
• One atomic mass unit (u) = 1/12th the mass of one atom of carbon-12.
Molecule
• Consists of two or more atoms.
• Capable of independent existence
Ions
• Charged species
• In ions, number of protons ≠ number of electrons
Molecule of element
• Consists of two or more atoms of same kind
Molecule of Compound
• Consists of two or more atoms of different kind.
Cation
• Positively charged ion
• Na+, Ca2+ etc.
Anion
• Negatively charged ion
• Cl–, O2– etc.
Atomicity
• Number of atoms constituting a molecule
Molecular Mass
• Sum of the atomic masses of all atoms in a molecule of the substance.
• Expressed in atomic mass unit (u).
Chapter at a Glance
• Atoms and Molecules are the fundamental units of matter that make up all substances; atoms combine to form molecules.
• Maharishi Kanad, Democritus and Leucippus postulated the existence of smallest unit of matter. Democritus called these indivisible particles atoms (meaning indivisible).
• An atom is the smallest particle of the element that cannot usually exist independently and retain all its chemical properties.
• A molecule is the smallest particle of an element or a compound capable of independent existence under ordinary conditions. It shows all the properties of the substance.
• Maharishi Kanad introduced the concept of Paramanu as the smallest units of matter in Indian philosophy.
• The law of conservation of mass states that mass is conserved in chemical reactions, as established by Antoine Lavoisier.
• The law of constant proportions states that in a pure chemical elements are always present in a fixed ratio by mass, as stated by Joseph Proust.
• John Dalton further developed atomic theory, explaining how matter is composed of small particles (atoms) that partake in chemical reactions.
• Symbols for elements derived from names, history, and language reflect specific characteristics. Dalton’s theory introduced systematic naming of elements.
• The International Union of Pure and Applied Chemistry (IUPAC) plays a crucial role in standardizing chemical symbols and nomenclature worldwide.
• The atomic mass unit (u) is defined with reference to the carbon-12 isotope, standardizing measurements.
• Chemical Formula is a representation of a compound using symbols for its constituent elements and numerical subscripts.
• Chemical formula expresses the proportions of elements in compounds, crucial for understanding their composition.
• Atomic radii are measured in nanometres where 1 nm equals 10–9 m.
• Polyatomic Ions are the ions consisting of two or more bonded atoms that function as a single charged entity.
• Common polyatomic Ions include sulphate (SO₄²⁻), nitrate (NO₃⁻), phosphate (PO₄³⁻), and carbonate (CO₃²⁻).
• Valency is the combining capacity of an element, crucial for writing chemical formulas accurately.
• Molecular mass is the sum of atomic masses of all atoms in a molecule, expressed in atomic mass units (u).
• To calculate molecular mass, add atomic masses of each atom in the molecule, e.g., water (H2O) is 18u (2 hydrogen + 1 oxygen). Atomic Masses of Few Elements
Symbols for Some Elements
Element
Aluminium Al Copper Cu Nitrogen N
Argon Ar Fluorine F Oxygen O
Barium Ba Gold Au Potassium K
Boron B Hydrogen H Silicon Si
Bromine Br Iodine I Silver Ag
Calcium Ca Iron Fe Sodium Na
Carbon C Lead Pb Sulphur S
Chlorine Cl Magnesium Mg Uranium U
Cobalt Co Neon Ne Zinc Zn
NCERT Zone
Intext Questions
1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.
Total mass of reactants = Mass of sodium carbonate + Mass of acetic acid = 5.3 g + 6 g = 11.3 g
Total mass of products = Mass of sodium acetate + Mass of carbon dioxide + Mass of water. = 8.2 g + 2.2 g + 0.9 g = 11.3 g
Since mass of reactants is equal to mass of products, it is in agreement with the law of conservation of mass.
2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
Ans. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass, this means that each gram of hydrogen will require 8 grams of oxygen to form water.
Therefore, 3 g of hydrogen gas will react completely with 3 × 8 = 24 g of oxygen gas.
3. Which postulate of Dalton’s atomic theory is a result of the law of conservation of mass?
Ans. Dalton’s postulate that “Atoms can neither be created nor destroyed,” is a result of law of conservation of mass.
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Ans. “Atoms combine in fixed ratio to form compounds” can explain the law of definite proportions.
5. Define the atomic mass unit.
Ans. Atomic mass unit is a mass unit equal to exactly 1 12 th of mass of 1 atom of carbon–12.
6. Why is it not possible to see an atom with naked eyes?
Ans. Atoms cannot be seen with naked eyes because they are very small in size (radius ~10–9 m).
7. Write down the formulae of
(a) Sodium oxide
(b) Aluminium chloride
(c) Sodium sulphide (d) Magnesium hydroxide
Ans. (a) Sodium oxide: Na2O
(c) Sodium sulphide: Na2S
(b) Aluminium chloride: AlCl3
(d) Magnesium hydroxide: Mg(OH)2
8. Write down the names of compounds represented by following formulae.
(a) Al2(SO4)3
(c) K2SO4
(e) CaCO3
Ans. (a) Al2(SO4)3 is aluminium sulphate
(b) CaCl2
(d) KNO3
(b) CaCl2 is calcium chloride (c) K2SO4 is potassium sulphate (d) KNO3 is potassium nitrate
(e) CaCO3 is calcium carbonate
9. What is meant by the term chemical formula?
Ans. Chemical formula is the formula of a compound which is represented with the help of symbol of elements. It tells us about the number and type of each element present in the compound.
10. How many atoms are present in a (a) H2S molecule and (b) PO3–4 ion?
Ans. (a) In H2S molecule, 2 atoms of hydrogen and 1 atom of sulphur is present. Therefore, in total, there are three atoms.
(b) In PO43– ion, there is one atom of phosphorus and four atoms of oxygen. Hence, total atoms are 5.
11. Calculate the molecular masses of
(a) H2 (b) O2
(d) CO2 (e) CH4
(c) Cl2
(f) C2H6
(g) NH3 (h) C2H4 (i) CH3OH
Ans. (a) Molecular mass of H2 = 2 × 1 = 2u
(b) Molecular mass of O2 = 2 × 16 = 32u
(c) Molecular mass of Cl2 = 2 × 35.5 = 71u
(d) Molecular mass of CO2 = 1 × 12 + 2 × 16 = 12 + 32 = 44u
12. Calculate the formula unit masses of ZnO, Na2O and K2CO3. [Atomic mass of Zn = 65u, Na = 23u, K = 39u, C = 12u, O = 16u]
Ans. (a) Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65 + 16 = 81u
(b) Formula unit mass of Na2O = 2Na + 1O = 2 × 23 + 16 = 46 + 16 = 62u
(c) Formula unit mass of K2CO3 = 2K + 1C + 3O = 2 × 39 + 1 × 12 + 3 × 16 = 78 + 12 + 48 = 138u
NCERT Exercises
1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Ans. Percentage of boron Mass of boron in compound 100 Mass of compound
0.096 g 10040% 0.24 g
Percentage of oxygen Mass of oxygen in compound 100 Mass of compound 0.144 g 10060% 0.24 g
2. When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
Ans. C + O2 → CO2
If 3 g of carbon are burned in 50 g of oxygen, only 8 g of oxygen will combine with the 3 g of carbon. The remaining 42 g of oxygen will remain unreacted. As a result, only 11 g of carbon dioxide will be produced, since 3 g of carbon react with 8 g of oxygen to form it.
The above answer is governed by the law of constant proportions.
3. What are polyatomic ions? Give examples.
Ans. Polyatomic ions are ions that consist of two or more atoms chemically bonded together, carrying an overall positive or negative charge. These atoms act as a single unit and do not break apart during chemical reactions. 22 CO(carbonate344 ion),SO(sulphate ion), and NH(ammonium ion)
are examples of polyatomic ions.
4. Write the chemical formulae of the following:
(a) Magnesium chloride
(c) Copper nitrate
(e) Calcium carbonate
Ans. (a) MgCl2
(c) Cu(NO3)2
(e) CaCO3
(b) Calcium oxide
(d) Aluminium chloride
(b) CaO
(d) AlCl3
5. Give the names of elements present in the following compounds.
(a) Quick lime
(c) Baking powder
Ans. (a) Quick lime (CaO) contains calcium and oxygen
(b) Hydrogen bromide
(d) Potassium sulphate
(b) Hydrogen bromide (HBr) contains hydrogen and bromine.
(c) Baking powder (NaHCO3) contains sodium, hydrogen, carbon and oxygen.
(d) Potassium sulphate (K2SO4) contains potassium, sulphur and oxygen.
6. Calculate the molar mass of the following substances:
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
Ans. (a) Ethyne, C2H2 = 12 × 2 + 1 × 2 = 26 g mol–1
2. A student wants to make a homogeneous mixture of salt, sugar, and water that weighs 300 g. The student has 50 g sugar and 70 g salt. How much water should he add to the mixture? (CBSE QB)
(a) 70 g (b) 120 g (c) 150 g (d) 180 g
3. One atomic mass unit is defined as (a) 1 12 of the mass of one carbon-12 atom (b) mass of one hydrogen atom (c) 12 g (d) 1 g
4. Who proposed the method on which modern atomic symbols are based upon?
(a) Dalton (b) Berzilius (c) Bohr (d) Mendeleev
5. While performing experiment to verify the law of conservation of mass in a chemical reaction, we generally use 5% solutions of two salts. 5% salt solution means 5 g salt is added to (a) 100 g of water. (b) 95 g of water. (c) 105 g of water. (d) 100 mL of water.
6. The same proportion of carbon and oxygen in CO2 obtained from different sources proves the law of (a) reciprocal proportion. (b) multiple proportion. (c) constant proportion. (d) conservation of mass.
7. Who proposed the law of constant proportion?
(a) Dalton (b) Berzilius (c) Joseph Proust (d) A. Lavoisier
8. A student makes two solutions using components as listed in the table.
The student notices that sand settles at the bottom of the beaker in Solution 1. Which solution would be heavier?
(a) Solution 1, because sand is heavy.
(b) Solution 2, because it has more water.
(c) Solution 2, because the more salt dissolves in water.
(d) Both solutions are equally heavy.
9. Which of the following is correct symbol of chloride ion?
(a) Cl (b) Cl+ (c) Cl- (d) Cl2+
10. A student learns that aluminium forms compound with chlorine and oxygen. She records the valencies of the three elements.
Element Valency
Which option gives the correct formula of aluminium oxide and aluminium chloride?
16. Four students A, B, C and D verified the law of conservation of mass in a chemical reaction of barium chloride and sodium sulphate. All of them took 107.2 g barium chloride solution and 116.1 g of sodium sulphate solution and mixed them in a beaker of mass 150 g. They reported their results as follows:
Student Colour of Reaction Mixture after Mixing
A White precipitate
B Brown precipitate
C White precipitate
D Brown precipitate
Mass of Reaction Mixture in the Beaker (including mass of beaker)
g
g
g
g
The correct observation is that of student (a) A (b) B (c) C (d) D
17. A student has a sample of 200 g of table sugar (sucrose). For 200 g of sucrose, there are 84 g of carbon. Based on Dalton’s atomic theory, how much carbon will be there in 300 g of sucrose?
(Sucrose = C12H22O11) (CBSE QB) (a) 42 g (b) 23.8 g (c) 56 g (d) 126 g
18. The table lists some compounds and their mass ratio.
Which compound does NOT support the law of constant proportions? (CBSE QB) (a) H2O (b) CO2 (c) NO2 (d) MgS
19. Which of the following is/are diatomic molecules? (i) Oxygen (ii) Hydrogen (iii) Nitrogen (iv) Chlorine (a) (ii) and (iii) (b) (iii) and (iv) (c) (ii), (iii) and (iv) (d) (i), (ii), (iii) and (iv)
20. Which of the following symbols is incorrect? (a) Helium (He) (b) Aluminium (Al) (c) Carbon (C) (d) Cobalt (CO)
Answers
Constructed Response Questions
Very Short Answer Questions (30–50 words)
1. Write down the names of compounds represented by the following formulae: (a) K2SO4 (b) MgCl2 (c) Na2SO4 (d) Pb(NO3)2
(c) Na2SO4: Sodium sulphate (d) Pb(NO3)2: Lead nitrate
2. Determine the formula unit mass of ammonium nitrate (NH₄)NO₃.
(Given, atomic masses: N = 14u, H = 1u, and O = 16u).
Ans. Mass of NH₄ = 1 N + 4 H = 14 + (4 × 1) = 18u.
Mass of NO₃ = 1 N + 3 O = 14 + (3 × 16) = 14 + 48 = 62u.
Therefore, total mass of (NH₄)NO3 = 18 + 62 = 80u.
3. Dalton’s atomic theory is contradicted by the formula of Sucrose (C12H22O11). Justify the statement.
Ans. Dalton’s atomic theory states that atoms of different elements combine together in simple whole number ratio. In the formula of sucrose, C12H22O11, the carbon, hydrogen and oxygen combine in whole number ratio but the ratio is not simple.
4. State two examples in each case and write their chemical formulae:
(a) Molecules containing only one type of atom
(b) Molecules containing two different types of atoms
Ans. (a) Oxygen (O₂), Chlorine (Cl₂)
(b) Carbon dioxide (CO₂), Water (H₂O)
5. Calculate the molecular mass of methanol (CH₃OH), given the atomic masses: C = 12u, H = 1u, and O = 16u.
Ans. One molecule of CH3OH contains 1 carbon atom, 1 oxygen atom and 4 hydrogen atoms.
Mass of Carbon (C): 1 atom × 12u = 12u
Mass of Hydrogen (H): 4 atoms × 1u = 4u
Mass of Oxygen (O): 1 atom × 16u = 16u
Therefore, molecular mass of methanol = 12u + 4u + 16u = 32u
6. Write down the formulae of
(a) sodium hydroxide
(c) magnesium sulphide
(b) aluminium oxide
(d) potassium hydroxide
Ans. The chemical formulae for the given compounds are:
(a) Sodium hydroxide: NaOH
(c) Magnesium sulphide: MgS
(b) Aluminium oxide: Al2O3
(d) Potassium hydroxide: KOH
7. Write the cations and anions present (if any) in the following compounds (NCERT Exemplar)
(a) CH3COONa (b) NaCl (c) H2 (d) NH4NO3
Ans. (a) CH3COONa: Cation = Na+, Anion = CH3COO–
(c) H2: No ions
(b) NaCl: Cation = Na+, Anion = Cl–
(d) NH4NO3: Cation = NH4+, Anion = NO3-
8. Calcium Carbonate (CaCO3) contains 40% calcium, 12% carbon and 48% oxygen by mass. Knowing that the law of constant proportions holds good, calculate the mass of the constituent elements in 2 g of calcium carbonate.
Ans. Calcium Carbonate (CaCO₃) contains 40% calcium (Ca), 12% carbon (C) and 48% oxygen (O)
We need to find the mass of each element in 2 g of CaCO₃, using the law of constant proportions.
Mass of calcium = 40% of 2 g
= 40 100 × 2 = 0.8 g
Mass of carbon = 12% of 2 g
= 12 100 × 2 = 0.24 g
Mass of oxygen = 48% of 2 g
= 48 100 × 2 = 0.96 g
9. Explain why a chlorine atom is electrically neutral while a chloride ion carries a charge.
Ans. A chlorine atom is neutral as it has 17 protons and 17 electrons, balancing positive and negative charges. When it gains one electron to complete its octet, it becomes a chloride ion (Cl⁻). Now with 17 protons and 18 electrons, the extra electron gives it a –1 charge, making the chloride ion negatively charged.
10. What is the percentage of sulphur in sulphuric acid?
[At. mass: H = 1u, S = 32u, O = 16u]
Ans. Molecular mass of H2SO4
= (2 × 1) + (1 × 32) + (4 × 16)
= 2 + 32 + 64 = 98u
Percentage of S in H2SO4 = 32 98 × 100 = 32.65%
Short Answer Questions (50–80 words)
1. Write the molecular formulae for the following compounds (NCERT Exemplar)
(a) Copper (II) bromide
(c) Calcium (II) phosphate
(e) Mercury (II) chloride
Ans. (a) Copper (II) bromide: CuBr2
(c) Calcium (II) phosphate: Ca3(PO4)2
(e) Mercury (II) chloride: HgCl2
(b) Aluminium (III) nitrate
(d) Iron (III) sulphide
(f) Magnesium (II) acetate
(b) Aluminium (III) nitrate: Al(NO3)3
(d) Iron (III) sulphide: Fe2S3
(f) Magnesium (II) acetate: Mg(CH3COO)2
2. Differentiate between cations and anions, providing suitable examples for each.
Ans. Ions are charged particles that form when atoms lose or gain electrons. Cations are positively charged species, formed when electrons are lost; for example, the sodium ion (Na+). Anions, on the other hand, are negatively charged due to an electron gain, such as the chloride ion (Cl-). These ions behave differently in chemical reactions, with cations attracting anions to form ionic compounds, which are crucial for a wide range of chemical processes and structures.
3. State the number of atoms present in each of the following chemical species (NCERT Exemplar)
(a) CO23– (b) PO34– (c) P2O5 (d) CO
Ans. (a) CO23–: 1 Carbon and 3 Oxygens = 4 atoms
(b) PO34–: 1 Phosphorus and 4 Oxygens = 5 atoms
(c) P2O5: 2 Phosphorus and 5 Oxygens = 7 atoms
(d) CO: 1 Carbon and 1 Oxygen = 2 atoms
4. Write the molecular formulae of all the compounds that can be formed by the combination of following ions Cu2+, Na+, Fe3+, Cl–, SO42–, PO34–
Ans. We have three cations and three anions, with these we can form 9 possible compounds.
For copper (Cu): CuCl2, CuSO4, Cu3(PO4)2
For sodium (Na): NaCl, Na2SO4, Na3PO4
For iron (Fe): FeCl3, Fe2(SO4)3, FePO4
5. (a) Define polyatomic ion.
(b) Write the name of the compound (NH4)2SO4 and mention the ions present in it.
Ans. (a) A polyatomic ion is a group of two or more atoms that are bonded together and function as a single charged particle (either positive or negative) in a chemical reaction.
(b) Chemical name of (NH4)2SO4 is ammonium sulphate and the ions present in ammonium sulphate are ammonium ion (NH₄⁺) and sulphate ion (SO₄²⁻).
6. If 100 g of calcium carbonate on heating produces 44 g of carbon dioxide, how much quicklime will be formed? which law is followed for solving this problem?
Ans. Given:
Mass of calcium carbonate (CaCO₃) = 100 g
Mass of carbon dioxide (CO₂) produced = 44 g
When calcium carbonate is heated, it decomposes as:
So, Mass of CaO = Mass of CaCO3 − Mass of CO2 = 100 g − 44 g = 56 g
Therefore, 56 g of quicklime will be formed on heating 100 g of calcium carbonate.
Law of Conservation of Mass is followed here. It states that mass can neither be created nor destroyed in a chemical reaction.
7. Define ions and valency. Explain with examples.
Ans. Ions are atoms or groups of atoms that carry an electric charge because they have either lost or gained electrons.
If an atom loses electrons, it becomes a positively charged ion and is called a cation.
If an atom gains electrons, it becomes a negatively charged ion and is called an anion.
Examples:
Sodium cation (Na⁺) is formed when a sodium atom loses one electron.
Chloride anion (Cl⁻) is formed when a chlorine atom gains one electron.
Valency is the combining capacity of an element, or the number of electrons an atom can lose, gain, or share to become stable. It tells us how an atom will bond with other atoms.
Examples:
Sodium (Na) has a valency of 1 (it loses 1 electron).
Oxygen (O) has a valency of 2 (it gains 2 electrons).
8. Calculate the molecular mass of potassium ferrocyanide K4[Fe(CN)6].
Given, atomic masses: K = 39.0u, Fe = 56.0u, C = 12.0u, N = 14.0u.
Ans. potassium ferrocyanide K4[Fe(CN)6] conatins 4 potassium ion (k+), 1 iron ion (Fe2+) and 6 cyanide ions (CN–). each cyanide ion contains 1 carbon atom and one nitrogen atom. Mass of 4 K atoms = 4 × 39.0 = 156.0u
Mass of 1 Fe atom = 1 × 56.0 = 56.0u
Mass of 6 C atoms = 6 × 12.0 = 72.0u
Mass of 6 N atoms = 6 × 14.0 = 84.0u
Total molecular mass = 156.0 + 56.0 + 72.0 + 84.0 = 368.0u
The molecular mass of potassium ferrocyanide is 368u.
9. Classify each of the following on the basis of their atomicity. (NCERT Exemplar) (a) F2 (b) NO2 (c) N2O (d) C2H6 (e) P4 (f) H2O2 (g) P4O10 (h) O3 (i) HCl (j) CH4 (k) He (l) Ag
Ans. (a) 2 (b) 3 (c) 3 (d) 8
(e) 4 (f) 4 (g) 14 (h) 3
(i) 2 (j) 5
(k) 1 (Noble gases do not combine and exist as monoatomic gases)
(l) Polyatomic: It is difficult to talk about the atomicity of metals as any measurable quantity will contain millions of atoms bound by metallic bond.
10. Explain the difference between molecular mass and formula unit mass, providing examples to illustrate your explanation.
Ans. Molecular mass is the total of all the atomic masses in one molecule. It is used for compounds made of molecules, like water (H₂O), where atoms are joined by covalent bonds.
Formula unit mass is the total of atomic masses in the simplest ratio of ions in an ionic compound, like salt (NaCl). Ionic compounds don’t have separate molecules; they have a big network of ions.
So, the main difference is: molecular mass is for molecules, and formula unit mass is for ionic compounds. Both are found by adding up atomic masses of constituent atoms.
11. An element X shows a variable valency of 3 and 5. What are the formulae of the oxides formed by it.
Ans. Formula of oxide when X has valency 3
Symbol
Valency
Formula = X2O3
Formula of oxide when X has valency 5
Symbol
Valency
Formula = X2O5
12. What mass of silver nitrate will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrate if the law of conservation of mass is true?
Ans. Given:
Mass of sodium chloride (NaCl) = 5.85 g
Mass of silver chloride (AgCl) produced = 14.35 g
Mass of sodium nitrate (NaNO₃) produced = 8.5 g
Reaction: NaCl + AgNO3 → AgCl + NaNO3
By the law of conservation of mass, the total mass of reactants = total mass of products.
So, Mass of NaCl + Mass of AgNO3 = Mass of AgCl + Mass of NaNO3
Substitute the known values:
5.85 g + Mass of AgNO3 = 14.35 g + 8.5 g
Mass of AgNO3 = 22.85 g − 5.85 g
Mass of AgNO3 = 22.85 g − 5.85 g
Mass of AgNO3 =17.0 g
The mass of silver nitrate (AgNO₃) required is 17.0 g.
13. Give the formulae of the compounds formed from the following sets of elements: (NCERT Exemplar)
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen (d) Carbon and chlorine
(e) Sodium and oxygen (f) Carbon and oxygen
Ans. (a) Calcium and fluorine: CaF2
(c) Nitrogen and hydrogen: NH3
(b) Hydrogen and sulphur: H2S
(d) Carbon and chlorine: CCl4
(e) Sodium and oxygen: Na2O (f) Carbon and oxygen: CO2
Long Answer Questions (80–120 words)
1. Write down the postulates of Dalton’s atomic theory.
Ans. The postulates of this Dalton’s atomic theory are as follows:
(a) All matter is made of very tiny particles called atoms, which participate in chemical reactions.
(b) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
(c) Atoms of a given element are identical in mass and chemical properties.
(d) Atoms of different elements have different masses and chemical properties.
(e) Atoms combine in the ratio of small whole numbers to form compounds.
(f) The relative number and kinds of atoms are constant in a given compound.
2. What are the limitations of Dalton’s Atomic theory?
Ans. (a) Subatomic particles: Dalton said atoms cannot be divided. But later, protons, electrons, and neutrons were discovered, showing atoms can be further divided into smaller particles.
(b) Isotopes: Dalton said all atoms of an element are the same. But isotopes of the same element have different masses (like hydrogen, deuterium, and tritium).
(c) Isobars: Dalton said atoms of different elements must have different masses. But some atoms of different elements have the same mass number (like ⁴⁰Ar and ⁴⁰Ca).
(d) Complex compounds: Dalton said elements combine in simple ratios. But some compounds, like sugar (C₁₁H₂₂O₁₁), have complex ratios.
(e) Allotropes: Dalton couldn’t explain why the same element can form substances with different properties, like diamond and graphite (both made of carbon).
3. Using the valency method, write down the formulae of the following compounds:
(a) Ferric oxide (b) Potassium hydroxide
(c) Hydrogen sulphide
(e) Mercury (II) chloride
Ans. (a) Formula of ferric oxide
Symbol
Valency
Formula = Fe2O3
(d) Magnesium (II) acetate
(b) Formula of Potassium hydroxide
Symbol
Valency
Formula = KOH Fe 3 O 2 K 1 OH 1
(c) Formula of Hydrogen sulphide
Symbol
Valency
Formula = H2S
(d) Formula of Magnesium II acetate
Symbol
Valency
Formula = Mg(CH3COO)2 H 1 S 2 Mg 2 CH3COO 1
(e) Formula of Mercury II chloride
Symbol
Valency
Formula = HgCl2
4. Calculate the mass percent of each element present in the molecule of calcium carbonate.
(a) Mass % of Ca in CaCO3 = 3 Mass of Calcium40 100 40% Mass of CaCO100
(b) Mass % of carbon in CaCO3 = 3 Mass of Carbon12 100 12% Mass of CaCO100
(c) Mass % of oxygen in CaCO3 = 3 Mass of Oxygen48 100 48% Mass of CaCO100
5. Calculate the percentage composition of all the elements present in ammonium sulphate, (NH4)2SO4 (where mass of N = 14u, H = 1u, S = 32u, O = 32u)
Ans. Mass of N = 2 × 14 = 28u
Mass of H = 8 × 1 = 8u
Mass of S = 1 × 32 = 32u
Mass of O = 4 × 16 = 64u
Total molecular mass = 28 + 8 + 32 + 64 = 132u
Mass % of N in (NH4)2SO4 = 424 Mass of N28 100 21.21% Mass of (NH)SO132
Mass % of H in (NH4)2SO4 = 424 Mass of H8 100 6.06% Mass of (NH)SO132
Mass % of S in (NH4)2SO4 = 424 Mass of S32 100 24.24% Mass of (NH)SO132
Mass % of O in (NH4)2SO4 = 424 Mass of O64 100 48.48% Mass of (NH)SO132
Competency Based Questions
Multiple Choice Questions (Choose the most appropriate option/options)
1. An ionic compound will be formed by the combination of one of the following pairs of elements. This pair of elements is:
(a) Barium and Oxygen (b) Sulphur and Carbon
(c) Hydrogen and hydrogen (d) Chlorine and chlorine
2. The correct molecular formula of silver iodide is:
(a) AgI2 (b) AgI (c) AgI3 (d) AgI6
3. Which of the following statements is not true about an atom? (NCERT Exemplar)
(a) Atoms are not able to exist independently
(b) Atoms are the basic units from which molecules and ions are formed
(c) Atoms are always neutral in nature
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch
4. What is the molecular mass of aluminium nitrate? (a) 214 u (b) 198 u (c) 213 u (d) 324 u
5. A student studies that magnesium (Mg) and iron (Fe) react with oxygen (O) to form MgO and FeO. Although both compounds contain two atoms each, their molecular masses are different. What is the reason for the difference in their molecular masses? (CBSE QB)
(a) Difference in the atomic mass of iron and magnesium
(b) Difference in the number of iron and magnesium atoms
(c) Different in the atomic number of iron and magnesium
(d) Difference in the number of O atoms in the two compounds
6. How would you correctly write the formula for calcium hydroxide?
(a) CaO₂H₂ (b) CaOH₂ (c) Ca2OH (d) Ca(OH)2
7. The correct chemical symbol for nitrogen gas is (NCERT Exemplar) (a) N (b) N2 (c) N+ (d) N–
8. If the molecular mass of HNO₃ is 63 u, what is the total mass contributed by oxygen in the molecule?
(a) 14 u (b) 63 u (c) 48 u (d) 16 u
9. Name the anion in K2SO4.
(a) sulphite ion (b) sulphide ion (c) sulphate ion (d) sulphur ion
10. What is the chemical name of AlPO4?
(a) Aluminium phosphor
(b) Aluminium phosphorus
(c) Aluminium phosphate (d) Aluminium phosphite
11. A change in the physical state can be brought about (NCERT Exemplar) (a) only when energy is given to the system (b) only when energy is taken out from the system (c) when energy is either given to, or taken out from the system (d) without any energy change
12. The percentage of hydrogen in water is (a) 1.11% (b) 11.11% (c) 8.89% (d) 88.9%
Assertion–Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
1. Assertion (A): The adoption of carbon-12 as the standard reference ensures precise determination of atomic masses.
Reason (R): Using carbon-12 minimizes discrepancies and yields near whole number atomic masses for most elements.
2. Assertion (A): Dalton’s postulate that atoms are indivisible is consistent with modern discoveries. Reason (R): Subatomic particles such as electrons, protons, and neutrons have been discovered within the atom.
3. Assertion (A): Ions are always positively charged. Reason (R): Ions are formed by losing or gaining of electrons.
4. Assertion (A): A molecule is the smallest particle of a compound that retains its distinct chemical properties. Reason (R): It is composed of atoms bonded together in any ratio.
5. Assertion (A): Polyatomic ions behave as a single unit in chemical reactions.
Reason (R): The atoms within a polyatomic ion maintain fixed ratios and do not dissociate under normal conditions.
6. Assertion (A): The law of constant proportions is universally applicable to pure compounds. Reason (R): Every sample of a pure compound exhibits the same fixed mass ratio of constituent elements.
7. Assertion (A): In water, the ratio of mass of hydrogen to the mass of oxygen is always 1:8 whatever is the source of water.
Reason (R): According to law of constant proportion, the elements are always present in definite proportion by mass in a chemical substance.
Case-Based/Source-Based/Passage-Based Questions
1. Read the given passage and answer the questions that follow:
Since ancient times, philosophers and scientists were curious about the smallest building blocks that make up everything around us. In the 19th century, experiments on chemical reactions showed that elements combined in specific and simple proportions, leading researchers to believe that there must be basic units behind these patterns. Inspired by these findings, some scientists suggested that matter is not continuous but made up of small, discrete particles. These ideas helped explain important laws like the Law of Conservation of Mass and the Law of Definite Proportions. One of the major figures of this period contributed a detailed explanation about the structure and behaviour of these tiny particles, creating a strong base for the development of modern chemistry. Although later studies discovered that these small particles were themselves made of even smaller parts, the early models played a critical role in shaping scientific thought.
(a) Dalton’s theory stated that atoms:
(i) Can be destroyed in chemical reactions (ii) Are indivisible and indestructible (iii) Are made up of protons and neutrons (iv) Continuously emit radiation
(b) According to Dalton’s Atomic Theory, atoms of the same element are:
(i) Different in mass (ii) Identical in mass and properties (iii) Different in properties (iv) Larger than molecules
(c) Compounds are formed by the combination of atoms of:
(i) Only one type. (ii) Different types in fixed ratios (iii) Random atoms (iv) Gases only
(d) Choose the correct feature of Dalton’s Atomic Theory:
(i) Atoms can split into electrons and neutrons naturally (ii) Atoms of the same element have the same chemical properties (iii) Atoms are visible under a normal microscope (iv) Atoms form liquids only
2. Read the given passage and answer the questions that follow:
Valency is the ability of an atom to combine with other atoms to form compounds. It shows how many bonds an atom can make during a chemical reaction. Valency is a basic property that helps determine how elements react with each other. Some elements have a fixed valency, while others can show more than one valency depending on the conditions. Valency plays a crucial role in writing correct chemical formulas and understanding the structure of molecules. It is one of the key concepts that connects atomic structure to chemical behaviour.
(a) The oxide of a metal ‘X’ has molecular formula X2O3. Find valency of X?
(i) 1 (ii) 3 (iii) 5 (iv) 2
(b) If an atom needs 1 electron to complete its octet, its valency is:
(i) 7 (ii) 1 (iii) 8 (iv) 0
(c) Match the following and choose the correct answer:
Column A Column B
1. Sodium (Na) A. Valency 4
2. Oxygen (O) B. Valency 2
3. Carbon (C) C. Valency 1
4. Nitrogen (N) D. Valency 3
5. Neon (Ne) E. Valency 0
(i) 1-A, 2-B, 3-C, 4-D, 5-E
(iii) 1-C, 2-E, 3-A, 4-D, 5-B
(ii) 1-C, 2-B, 3-A, 4-D, 5-E
(iv) 1-A, 2-B, 3-D, 4-C, 5-E
(d) Which of the following elements can show multiple valencies?
(i) Oxygen (ii) Hydrogen (iii) Sodium (iv) Phosphorous
3. Read the given passage and answer the questions that follow:
Dalton proposed that all substances—whether elements, compounds, or mixtures—are made up of tiny indivisible particles known as atoms. These atoms remain unchanged during chemical reactions; they can neither be created nor destroyed. Dalton’s ideas provided a straightforward explanation for the fundamental laws governing chemical combinations. His theory supported the Law of Conservation of Mass, the Law of Definite Proportions, and the Law of Multiple Proportions, based on different principles he outlined. Dalton was also the first to introduce the use of symbols for elements, assigning each symbol to represent a specific quantity—namely, a single atom of that element.
(a) Which principle of Dalton’s theory supports the Law of Conservation of Mass?
(i) Atoms cannot be created or destroyed.
(ii) Matter is made up of extremely tiny particles called atoms.
(iii) All atoms of a single element are identical in all respects.
(iv) Atoms of different elements combine in simple whole number ratios.
(b) Which postulate of Dalton’s theory justifies the Law of Definite Proportions?
(i) Atoms of elements stay unchanged during chemical reactions.
(ii) A compound is made of fixed numbers and types of atoms with constant masses.
(iii) Atoms of different elements are different from each other.
(iv) Atoms exist in a variety of types.
(c) Which of the following postulates of Dalton’s atomic theory proved wrong in respect of isotopes?
(i) Atoms cannot be created or destroyed.
(ii) All atoms of a single element are identical in all respects.
(iii) Matter is made up of extremely tiny particles called atoms.
(iv) Atoms of different elements combine in simple whole number ratios.
(d) When 5 g of calcium burns in 2 g of oxygen, 7 g of calcium oxide is formed. Even when 5 g of calcium is burnt in 20 g of oxygen, only 7 g of calcium oxide is produced. This situation demonstrates:
(i) Law of Conservation of Mass (ii) Law of Multiple Proportions
(iii) Law of Constant Proportions (iv) No law applies
4. Read the given passage and answer the questions that follow:
Atomic mass is the mass of a single atom, measured relative to the carbon-12 isotope, where one carbon-12 atom is defined to have exactly 12 atomic mass units (amu). Molecular mass, on the other hand, is the sum of the atomic masses of all atoms in a molecule. Both concepts are fundamental in chemistry for understanding the composition and behaviour of substances. The atomic mass of an element, usually given on the periodic table, reflects a weighted average based on the relative abundance of isotopes. Carbon-12 plays a special role as it serves as the standard for defining atomic mass units. Scientists use these measurements in stoichiometric calculations, helping predict the outcomes of chemical reactions.
(a) Which of the following molecules has the highest molecular mass?
(i) O₂ (ii) CO₂ (iii) H₂O (iv) CH₄
(b) Which of the following statements is true about isotopes?
(i) They have the same mass numbers but different atomic numbers. (ii) They have the same number of neutrons but different protons. (iii) They have the same atomic numbers but different mass numbers. (iv) They are different elements altogether.
(c) The molecular mass of ammonia (NH3) is approximately: (i) 12 amu (ii) 14 amu (iii) 17 amu (iv) 18 amu
(d) Match the Following:
Column A
A. Atomic mass unit
B. Oxygen molecule (O₂)
C. Molecular mass
D. Isotope
Column B
1. 32 amu
2. Based on carbon-12
3. Mass of a molecule
4. Same atomic number, different mass number (i) A-2, B-1, C-3, D-4 (ii) A-2, B-3, C-4, D-1 (iii) A-4, B-1, C-3, D-2 (iv) A-1, B-2, C-3, D-4
Answers
Multiple Choice Questions
Assertion-Reason Based Questions
1. (a) 2. (d) 3. (d)
Case-Based/Source-Based/Passage-Based Questions
1. (a) (ii) (b) (ii) (c) (ii) (d) (ii)
2. (a) (ii) (b) (i) (c) (ii) (d) (iv)
3. (a) (i) (b) (ii) (c) (ii) (d) (iii)
4. (a) (ii) (b) (iii) (c) (iii) (d) (i)
Practice Questions
Multiple Choice Questions
1. The ratio of H : O by mass in hydrogen peroxide H2O2 is (NCERT Exemplar) (a) 16 : 1 (b) 1 : 8 (c) 1 : 32 (d) 1 : 16
2. A student lists four compounds: PCl5, KBr, P4O10, and Na2CO3. How many types of atoms are present in these four compounds ? (a) 5 (b) 6 (c) 7 (d) 8
3. What is the formula unit mass of Na₂O, if the atomic masses of Na is 23u and O is 16u? (a) 62u (b) 76u (c) 69u (d) 46u
4. In a methanol molecule (CH3OH), what is the ratio by number of hydrogen to oxygen atoms?
(a) 3 : 1
(b) 1 : 3
(c) 2 : 1 (d) 4 : 1
5. Which limitation is associated with Dalton’s Atomic Theory?
(a) It proposes that atoms form mixtures randomly.
(b) It states that atoms are divisible.
(c) It cannot explain the existence of isotopes.
(d) It incorrectly describes molecule formation.
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
6. Assertion (A): For the formation of a molecule at least two atoms are needed.
Reason (R): Molecules are formed by chemical bonding between atoms of the same element or different elements without any exception.
7. Assertion (A): N2 and H2O are molecules.
Reason (R): A molecule can have only similar kinds of atoms.
Very Short Answer Questions (30-50 words)
8. Calculate the formula unit mass of CaCl2. (NCERT Exemplar)
9. Which of the following symbols of elements are incorrect? Give their correct symbols. (a) Cobalt (CO) (b) Carbon (C) (c) Aluminium (AL) (d) Sodium (So)
10. An element ‘X’ has molecular formula XCl2:
(a) Write formula of its oxide.
(b) Write the formula of its sulphate.
11. Fill in the blanks
(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called .
(b) A group of atoms carrying a fixed charge on them is called
(c) The formula unit mass of Ca3(PO4)2 is .
(d) Formula of sodium carbonate is and that of ammonium sulphate is .
12. Give the molecular formula of the oxides formed by elements ‘X’ and ‘Y’ having valencies as 1 and 3 respectively.
Short Answer Questions (50-80 words)
13. What is meant by the atomicity of elements? Provide examples of monoatomic, diatomic, and polyatomic molecules among elements.
14. A blue solid is thought to be a pure compound. Analysis of three samples of the material yield the following results.
Could the material be a pure compound? Justify.
15. A compound contains 40% boron and 60% oxygen by mass. In a 0.24 g sample, calculate the mass of boron and oxygen.
16. Calculate the molecular mass of glucose (C₆H₁₂O₆) and baking soda (NaHCO3). (Use atomic masses: C = 12u, Na = 23u, H = 1u, O = 16u)
17. Give the formulae of the compounds formed from the following sets of elements:
(a) Calcium and fluorine (b) Hydrogen and sulphur
(c) Nitrogen and hydrogen (d) Carbon and chlorine
(e) Sodium and oxygen (f) Carbon and oxygen
Long Answer Questions (80-120 words)
18. Find the ratio by mass of the combining elements in the following compounds.
(a) CaCO3
(b) MgCl2
(c) H2SO4
(d) C2H5OH
(e) Ca(OH)2
19. (a) Determine the molecular mass of ethanol (C₂H₅OH) and acetic acid (CH₃COOH). Use atomic masses: C = 12u, H = 1u, O = 16u.
(b) Write down the chemical formulae of the following compounds: (i) Magnesium Nitrate (ii) Sodium hydroxide (iii) Calcium sulphate
20. (a) You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting? (NCERT Exemplar)
(b) Calculate the percentage of oxygen in Al2(SO4)3.
(c) Name the elements present in chemical formula of egg shell?
Brain Charge
1. (a) In this crossword puzzle, names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle. (NCERT Exemplar)
1. Cl 2. H 3. Ar
(b) Identify the total number of inert gases, their names and symbols from this cross word puzzle.
2. Complete the following crossword puzzle by using the name of the chemical elements. Use the data given in table. (NCERT Exemplar)
Across
2. The element used by Rutherford during his α–scattering experiment
3. An element which forms rust on exposure to moist air
5. A very reactive non–metal stored under water
7. Zinc metal when treated with dilute hydrochloric acid produces a gas of this element which when tested with burning splinter produces a pop sound.
Down
1. A white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air
4. Both brass and bronze are alloys of the element
6. The metal which exists in the liquid state at room temperature
8. An element with symbol P
3. How can we find single atom of a substance? Explore and discuss way to capture an image of a single atom.
Challenge Yourself
1. If a molecule has a molecular mass of 44 amu, and it contains x carbon atoms and y oxygen atoms. Calculate the value of x + y and also name the molecule.
2. If 20 grams of hydrogen reacts with 160 grams of oxygen to form water, what will be the mass of hydrogen required to form 18 kg of water?
3. If a molecule has a molecular mass of 46 amu, and it contains x carbon atoms, y hydrogen atoms and z oxygen atoms. Calculate the value of y xz and also write the name and formula of the molecule.
4. Why was the postulate “Atoms are indivisible” of Dalton’s Atomic theory discarded with further discoveries?
Practice Questions
1. (d) 2. (c) 3. (a) 4. (d) 5. (c) 6. (b) 7. (c)
8. 111u
10. XO and XSO4
11. (a) Law of conservation of mass (b) Polyatomic ion (c) Calcium phosphate (d) Na2CO3, (NH4)2SO4
12. X2O, Y2O3
15. Mass of boron = 0.096 g, mass of oxygen = 0.144 g
1. (a) (i) Cl stands for Chlorine. (ii) H stands for Hydrogen.
(iii) Ar stands for Argon. (iv) O stands for Oxygen.
(v) Xe stands for Xenon. (vi) He stands for Helium.
(vii) F stands for Fluorine. (viii) Kr stands for Krypton.
(ix) Rn stands for Radon. (x) Ne stands for Neon. (b) Total number of inert gases in this crossword puzzle is 6. Their names and symbols are as follows.
Argon (Ar)
Helium (He)
Radon (Rn)
2. 2. Gold 1. Silver
3. Iron 4. Copper
5. Phosphorus 6. Mercury
7. Hydrogen 8. Lead
Xenon (Xe)
Krypton (Kr)
Neon (Ne)
3. Check out Avogadro’s number and Electron microscope.
Challenge Yourself
1. x + y = 3, carbon dioxide (CO₂).
2. 2 kg
3. 6 2 21 y xz , Ethyl alcohol (C2H6O)
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 1.5 Hour
Scan me for Solutions
Max. Marks: 50
Multiple Choice Questions (10 × 1 = 10 Marks)
1. Which unit is now accepted as the standard atomic mass unit according to IUPAC?
(a) The mass of a hydrogen atom.
(b) 1 16 th the mass of an oxygen atom.
(c) 1 12 th the mass of a carbon–12 atom.
(d) 1 gram per mole.
2. Which ancient Indian philosopher postulated the existence of indivisible particles known as ‘Parmanu’?
3. The chemical symbol for sodium is (NCERT Exemplar)
(a) So (b) Sd (c) NA (d) Na
4. Which of the following is correct statement ?
(a) Na2S is Sodium sulphide, Na2SO3 is Sodium sulphite and Na2SO4 is Sodium sulphate.
(b) Na2S is Sodium sulphite, Na2SO3 is Sodium sulphide and Na2SO4 is Sodium sulphate.
(c) Na2S is Sodium sulphite, Na2SO3 is Sodium sulphate and Na2SO4 is Sodium sulphide
(d) Na2S is Sodium sulphide, Na2SO3 is Sodium sulphate and Na2SO4 is Sodium thiosulphate.
5. Which of the following are incorrect for the mass of products in a chemical reaction?
(I) Mass of reactants is more than the mass of products in a chemical reaction.
(II) Mass of reactants or products can neither be created nor be destroyed.
(III) Mass of reactants before reaction is equal to the mass of products after reaction.
(IV) Mass of reactants decreases during reaction.
(a) (I) and (II) (b) (II) and (III) (c) (III) and (IV) (d) (I) and (IV)
6. From the following elements ozone, sulphur, argon and phosphorus, which has the highest and lowest atomicity?
(a) Ozone and Sulphur
(c) Sulphur and Argon
7. How many elements are there in C6H12O6?
(a) 1 (b) 2
8. Which of the following is a triatomic molecule?
(a) Oxygen (O₂)
(c) Nitrogen (N₂)
(b) Phosphorus and Argon
(d) Sulphur and Phosphorus
(c) 3 (d) 24
(b) Carbon dioxide (CO₂)
(d) Sodium chloride (NaCl)
9. What does the term ‘atomicity’ refer to in the context of molecules of elements?
(a) The energy level of electrons in an atom.
(b) The charge on an atom.
(c) The mass of an atom relative to carbon-12.
(d) The number of atoms that constitute a molecule of an element.
10. The molecular formula of zinc phosphate is
(a) Zn(PO4)3 (b) Zn2PO4
(c) Zn3(PO4)2
(d) Zn(PO4)
Assertion–Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): In ionic compounds, the formula unit mass is derived by summing the atomic masses of constituent ions without considering their charges.
Reason (R): Electrical charges do not contribute to the mass of the ions.
12. Assertion (A): When writing the chemical formula of an ionic compound, the total positive charge must balance the total negative charge.
Reason (R): Ionic compounds are neutral in nature because positive charge is balanced by negative charge.
13. Assertion (A): An atom is the smallest particle in an element that possesses the element’s properties. Reason (R): Molecules are made up of two or more atoms.
14. Assertion (A): Nitrogen has an atomic mass of 14.
Reason (R): Nitrogen atoms are 14 times heavier than carbon-12 atoms of the same mass.
15. Read the given passage and answer the questions that follow: In chemistry, charged particles play a crucial role in forming compounds and carrying out biological functions. These particles are found everywhere, from the air we breathe to the food we eat. They help with electrical conduction, chemical reactions, and processes like nerve transmission and muscle contractions. These particles can move freely in environments like water, allowing electricity to flow. Their interactions with other particles affect properties like melting and boiling points. They also form essential compounds such as salts, acids, and bases. The balance of these particles is vital for stability in chemical and biological systems.
When interacting with other substances, these particles can transform, leading to the formation of neutral or differently charged particles, influencing processes like crystal formation and pH regulation in living organisms.
(a) What happens when an atom loses an electron?
(i) It becomes an anion. (ii) It becomes a cation.
(iii) It becomes neutral. (iv) It forms a molecule.
(b) Ammonium ion is represented by
(c) Which of the following is an example of an anion?
(i) Na⁺ (ii) Cl⁻ (iii) H⁺ (iv) Mg²⁺
(d) Match the following ions with their corresponding charges and choose the correct option:
Ion Charge
1. Sodium (Na) A. –1
2. Chlorine (Cl) B. +1
3. Oxygen (O) C. +2
4. Magnesium (Mg) D. –2
(i) 1-A, 2-B, 3-D, 4-C
(iii) 1-D, 2-A, 3-B, 4-C
(ii) 1-B, 2-A, 3-C, 4-D
(iv) 1-B, 2-A, 3-D, 4-C
16. Read the given passage and answer the questions that follow:
In the study of substances, we often come across tiny groups of atoms that stay bonded together. Some of these groups are made of the same kind of atoms, while others are made of different kinds. They are the reason why substances have specific properties, like how they react, their color, and even how they taste or smell. In nature, many gases and solids exist because similar atoms cling tightly to one another in small groups. These tiny groups are stable and can move freely in gases or stay packed together in solids. On the other hand, when atoms of different kinds combine, they form entirely new substances, often with properties very different from the original atoms. These tiny groups can behave very differently depending on how their atoms are arranged. For example, some are very soft and reactive, while others are hard and very stable. They are everywhere—in the air, in water, in food, and even in your body—helping to build the world around you.
(a) Which of the following is made up of only one type of atom?
(i) Water (ii) Oxygen gas (iii) Carbon dioxide (iv) Ammonia
(b) Which one is a molecule of a compound?
(i) O₂ (ii) N₂ (iii) H₂O (iv) Cl₂
(c) When two or more atoms of different elements combine, they form:
(i) An atom
(ii) A molecule of an element
(iii) A molecule of a compound (iv) A molecule of the same element
(d) A molecule of water contains how many atoms?
(i) 1 (ii) 2 (iii) 3 (iv) 4
Very Short Answer Questions (30–50 words)
17. An element ‘X‘ has a valency 3:
(a) Write the formula of its phosphate.
(b) Write the formula of its carbonate.
(3 × 2 = 6 Marks)
18. How does Dalton’s theory relate to the law of conservation of mass in chemical reactions?
19. Which of the following represents a correct chemical formula? Name it.
(a) CaCl (b) BiPO4 (c) NaSO4 (d) NaS
Short Answer Questions (50–80 words)
(4 × 3 = 12 Marks)
20. Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
21. Calculate the molecular mass of (a) water (H2O) (b) Nitric acid (HNO3).
22. (a) What is the valency of iron in Fe2O3 and Fe3O4?
(b) Calculate the formula unit mass of CaCl2 if atomic mass of Ca is 40.0 u and Cl is 35.5 u.
23. (a) What do you mean by polyatomic cation or polyatomic anion?
(b) Give one example of a polyatomic cation and one example of polyatomic anion and tell their formulae and charges.
Long Answer Questions (80–120 words)
(2 × 5 = 10 Marks)
24. Write the formulae for the following and calculate the molecular mass for each one of them. (a) Baking powder (b) Lime stone (c) Common salt
25. (a) What are the postulates of Dalton’s atomic theory?
(b) What are the limitations of this theory? How was the theory modified?
4 Structure of the Atom
This chapter deals with the structure of an atom by exploring the key subatomic particles—electrons, protons, and neutrons. Various models explaining atomic structure, like Rutherford’s and Bohr’s, are discussed. We learn about particle arrangement, atomic number, mass number, and electronic configuration. Understanding these concepts is crucial for grasping chemical reactions, bonding, and the behaviour of elements. The chapter also introduces isotopes, isobars, valency, and principles governing element reactivity, providing a foundation for chemistry studies.
Bohrʹs Model
Main postulates of this model are as:
• Electrons revolve around the nucleus in circular orbit.
• Only certain orbits called discrete orbits or energy levels of electrons are allowed inside the atom.
• While revolving in discrete orbits, the electrons do not radiate energy.
• Energy levels or shells are designated by K, L, M, N etc or the numbers, n = 1, 2, 3, 4 etc.
Atomic Models
Thomsonʼs Mod
• It considered atom like Plum pudding in which the electrons were embedded in a sphere of positive charge like raisins in a spherical Plum pudding. Here, negative and positive charges balance each other, thus atom as a whole is electrically neutral.
Limitations
It is failed to explain the results of a-particles scattering experiment of Rutherford.
Atomic Number (Z)
• It is equal to the number of protons present in the nucleus of an atom.
• It is denoted by Z. Mass Number (A)
• It is the sum of number of protons and neutrons present in the nucleus of an atom.
A = p + n.
Characteristics of Atom
These are the properties by which an atom can be identified.
ATOM
Rutherfordʼs Model
• On the basis of the a-particles scattering experiment, Rutherford stated that:
• There is a positively charged, highly densed centre in an atom, called the nucleus.
• The nucleus is very small as compared to the size of the atom.
• Nearly the whole mass of an atom resides in it.
• The electrons revolve around the nucleus in circular paths. Limitation
• It is failed to explain the stability of atoms.
Different Type of Atomic Species
Isobars
• Atoms of different elements with different atomic numbers but same mass number.
e.g. 4018Ar and 4020Ca
• Their chemical properties are different because of different atomic number.
Isotopes
• The atoms of the same element, having same atomic number but different mass numbers. e.g., 1H1, 1H2, 1H3
• Their chemical properties are same due to same number of valence electrons.
• Their physicai properties are different be cause of having different atomic masses.
Sub-Atomic Particles
These are the particles from which an atom is made.
Electron
• It was identified by J.J. Thomson.
• It is represented as ʻe–ʼ
• It is the negatively charged particle present in atom.
• Its charge is taken as -1. The mass of an electron is considered to be negligible.
Valence Electrons
Neutrons
• These were discovered by J. Chadwick in 1932.
• In general, a neutron is represented as ʻnʼ.
• These are electrically neutral particles.
• These are present inside the nucleus along with protons.
It is the number of electrons present in outermost or valence shell of an atom.
Valency
It is the combining capacity of an element with the atom(s) of other element(s) in order to complete its octet.
Proton
• It was discovered by E. Goldstein in 1886.
• It is represented as p+ .
• These are the positively charged particles.
• Its mass is approximately 2000 times as that of the electron.
• Its mass is taken as one unit and charge as +1.
Distribution of Electrons in Different Orbits
• Suggested by Bohr and Bury.
• These rules are:
Maximum number of electrons present in a shell = 2n2
(Here, n = 1, 2, 3....)
Outermost orbit can accommodate a maximum of 8 electrons.
The shells are filled in a step-wise manner.
Chapter at a Glance
• Atoms are the basic building blocks of matter, with elements distinguished by different atomic structures.
• Dalton’s atomic theory’s postulate, ‘Atoms are indivisible’ was later on proved wrong because of the discovery of subatomic particles by different scientists.
• The three sub-atomic particles are: (i) electrons, (ii) protons and (iii) neutrons.
• Credit for the discovery of electron, proton and neutron goes to J.J. Thomson, E. Goldstein and J. Chadwick, respectively.
• J.J. Thomson proposed that electrons are embedded in a positive sphere
• Electrons are negatively charged; protons are positively charged and neutrons have no charges.
• The mass of an electron is about 1/2000 times the mass of a hydrogen atom.
• The mass of a proton and a neutron is taken as one unit each.
• Rutherford’s alpha-particle scattering experiment led to the discovery of the atomic nucleus.
• Rutherford’s model of the atom proposed that a very tiny nucleus is present inside the atom and electrons revolve around this nucleus. The stability of the atom could not be explained by this model.
• Neils Bohr’s model of the atom was more successful. He proposed that electrons are distributed in different shells with discrete energy around the nucleus. If the atomic shells are complete, then the atom will be stable and less reactive.
• Electron Configuration is the distribution of electrons in atomic shells.
• Electrons fill the K, L, M, and N shells in order following the 2n² rule for maximum capacity.
• K shell can contain up to 2 electrons, L can have 8, M can hold 18, and N can accommodate 32.
• Atomic Number is the number of protons present in an atom’s nucleus, defining the element’s identity.
• Atomic number (Z) determines the element and its position in the periodic table. It is equal to the number of protons in an atom.
• Mass Number (A) is the total number of protons and neutrons in an atom’s nucleus.
• Atoms with different mass numbers, the atomic mass of the element is the average value which comes out to be fractional. Example: Chlorine has a mass number of 35.5.
• Valency is the combining capacity of an element. It indicates how many bonds an atom can form.
• Isotopes are atoms of the same element that have the same atomic number but different mass numbers due to varying neutrons.
• Isobars are the atoms of different elements that have the same mass number but different atomic numbers.
• Elements are defined by the number of protons they possess.
Important Elements
Important Formulae
• Maximum electrons in a shell: 2n²
• Valency = 8 - (number of valence electrons)
• Mass number (A) = Number of protons (Z) + Number of neutrons (N)
NCERT Zone
Intext Questions
1. What are canal rays?
Ans. Canal rays are streams of positively charged particles, also known as anode rays, discovered by E. Goldstein in 1886. These rays are formed when electricity is passed through gases at low pressure, leading to the emission of positive ions.
2. If an atom contains one electron and one proton, will it carry any charge or not?
Ans. If an atom contains one electron (negative charge) and one proton (positive charge), the charges will cancel each other. Therefore, the atom will be neutral and carry no overall charge.
3. On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Ans. According to Thomson’s model, an atom consists of a positively charged sphere with negatively charged electrons embedded within it. These electrons are scattered throughout the positive sphere in such a way that the negative and positive charges are equal in magnitude. This balance ensures that the atom as a whole remains electrically neutral.
4. On the basis of Rutherford’s model of an atom, which sub-atomic particle is present in the nucleus of an atom?
Ans. In Rutherford’s model of an atom, the nucleus is the central positively charged region, and it contains protons. Rutherford’s experiments showed that the mass of an atom and all its positive charge are concentrated in the nucleus, and therefore, protons are present in the nucleus of the atom.
5. Draw a sketch of Bohr’s model of an atom with three shells.
Ans.
M shell (n = 3)
L shell (n = 2)
K shell (n = 1)
Nucleus
6. What do you think would be the observation if the α-particle scattering experiment is carried out using a foil of a metal other than gold?
Ans. If Rutherford’s a-particle scattering experiment used a metal foil other than gold, such as silver or aluminum, the observations would be similar. But with the foil of a light metal, e.g., that of lithium, the heavy alpha particles may push the nucleus and may not be deflected back.
7. Name the three sub-atomic particles of an atom.
Ans. The three sub-atomic particles of an atom are electrons, protons, and neutrons. Electrons are negatively charged, protons are positively charged, and neutrons have no charge.
8. Helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Ans. The helium atom has an atomic mass of 4 u, with 2 protons in its nucleus.
Mass number (Atomic number) = 4
Atomic number = Number of protons = 2
Number of Neutrons = Mass Number - Atomic Number = 4 - 2 = 2
9. Write the distribution of electrons in carbon and sodium atoms.
Ans. The electron distribution in carbon, which has an atomic number of 6, is 2 electrons in the K-shell and 4 electrons in the L-shell (2, 4).
2 4
Sodium, with an atomic number of 11, has electron distribution as 2 electrons in the K-shell, 8 in the L-shell, and 1 in the M-shell (2, 8, 1). These distributions follow the rule 2n² for the capacity of each shell.
K L M
2 8 1
10. If K and L shells of an atom are full, then what would be the total number of electrons in the atom?
Ans. The maximum number of electrons in a shell can be calculated by the formula 2n2. Maximum number of electrons in K- shell (n = 1) = 2 × 12 = 2
Maximum number of electrons in K- shell (n = 1) = 2 × 22 = 8
If both K and L shells are full, the total number of electrons in the atom would be the sum of the electrons in these two shells: 2 (K-shell) + 8 (L-shell) = 10 electrons.
12. How will you find the valency of chlorine, sulphur and magnesium?
Ans. Atomic number of Chlorine = 17
Electronic configuration of Chlorine =
Since chlorine has 7 valence electrons, it needs 1 more electron for stability. Hence, the valency of chlorine is 1.
Atomic number of Sulphur = 16
Electronic configuration of Sulphur =
Since sulphur has 6 valence electrons, it needs 2 more electron for stability. Hence, the valency of sulphur is 2.
Atomic number of Magnesium = 12
Electronic configuration of Magnesium =
Since Magnesium has 2 valence electrons, it will give 2 electron to complete its octet. Hence, the valency of magnesium is 2.
13. If number of electrons in an atom is 8 and number of protons is also 8, then (a) what is the atomic number of the atom? and (b) what is the charge on the atom?
Ans. (a) Since the number of protons in an atom = atomic number of atom = 8 (b) With 8 electrons and 8 protons, the positive and negative charges cancel out, so the atom is neutral (Charge = 0).
14. Oxygen atom has 8 electrons, 8 protons and 8 neutrons whereas sulphur atom has 16 electrons, 16 protons and 16 neutrons. Find out the mass number of oxygen and sulphur atom.
Ans. Mass number of an atom is the sum of number of protons and neutrons.
Mass number of oxygen atom = No. of protons + No. of neutrons = 8 + 8 = 16
Mass number of sulphur atom = No. of protons + No. of neutrons = 16 + 16 = 32
15. For the symbol H, D and T, tabulate three sub-atomic particles found in each of them.
Ans. Atomic number of H (Hydrogen ) = 1
Mass number of H = 1
Number of protons = No. of electrons = Atomic number = 1
Number of neutrons = Mass number - Atomic number = 1 - 1 = 0
Atomic number of D (Deuterium) = 1
Mass number of D = 2
Number of protons = No. of electrons = Atomic number = 1
Number of neutrons = Mass number - Atomic number = 2 - 1 = 1
Atomic number of T (Tritium) = 1
Mass number of H = 3
Number of protons = No. of electrons = Atomic number = 1
Number of neutrons = Mass number - Atomic number = 3 - 1 = 2
17. Write the electronic configuration of any one pair of isotopes and isobars.
Ans. Example of isotopes of chlorine are 1735Cl and 1737Cl
Electronic configuration of both the isotopes of chlorine is same as their atomic number is same.
Example of isobars are 2040Ca and 1840Ar.
Electronic configuration of both isobars would be different as both have different atomic numbers but have same mass number.
1840Ar K L M
2 8 8
2040Ca K L M N
2 8 8 2
NCERT Exercises
1. Compare the properties of electrons, protons and neutrons.
Ans. Properties of electrons, protons and neutrons are as follows
Location They are present around the nucleus Inside nucleus Inside nucleus
Relative charge -1 unit +1 unit
Mass Nearly 1 2000 of that of H-atom
Nearly equal to that of H-atom
2. What are the limitations of J.J. Thomson’s model of the atom?
Ans. Limitations of J.J. Thomson’s Plum-pudding model of the atom are:
Nil
Nearly equal to that of H-atom
(a) It could not explain the experimental results of other scientists such as Rutherford, as there is no nucleus in the atomic model proposed by Thomson.
(b) It could not explain the stability of an atom, i.e. how positive and negative charges could remain so close together.
3. What are the limitations of Rutherford’s model of the atom?
Ans. Limitations of Rutherford’s model of the atom are as follows
(a) It could not explain why a moving charge (electrons) does not lose energy and fall into the nucleus.
(b) It could not explain the stability of an atom when charged electrons are moving under the attractive force of positively charged nucleus.
(c) The model could not explain as to how the electrons are distributed in the extra nuclear portion of an atom.
4. Describe Bohr’s model of an atom.
Ans. The main postulates of Bohr’s model of an atom are as follows
(a) Atom has nucleus in the centre and electrons revolve around the nucleus.
(b) Certain special orbits known as discrete orbits of electrons, also known as energy shells are allowed inside the atom. While revolving in discrete orbits the electrons do not radiate energy.
(c) These orbits or shells are represented by the letters K, L, M, N or the numbers n = 1, 2, 3, 4
N shell (n = 4)
M shell (n = 3)
L shell (n = 2)
K shell (n = 1)
Nucleus
Bohr’s model
5. Compare all the proposed models of an atom given in this chapter.
Ans.
Feature Thomson’s Model Rutherford’s Model Bohr’s Model
Key Idea
Structure of Atom
Discovery Explained
Electrons
Nucleus
Stability of Atom
Limitation
Atom is a positively charged sphere with electrons embedded (like a watermelon)
Positively charged sphere with scattered electrons.
Atom has a small dense nucleus, electrons move around it.
Nucleus at centre, electrons revolve around it
Electrons move in fixed orbits around the nucleus.
Electrons revolve in fixed paths called shells
Discovery of electron Gold foil experiment Hydrogen line spectrum
Present, embedded inside atom Present, revolving around nucleus
Not present
Couldn’t explain stability
Failed to explain results of scattering experiment
Discovered nucleus
Couldn’t explain stability
Couldn’t explain electron stability
Present, revolving in fixed orbits
Same as Rutherford’s model
Explained stability using fixed energy levels
Couldn’t explain atoms with more than one electron
6. Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Ans. The rules for writing of distribution of electrons in various shells for the first eighteen elements are:
(a) The maximum number of electrons present in a shell is given by the formula - 2n2
∴ n = orbit number i.e., 1, 2, 3
∴ Maximum number of electrons in different shells are:
K shell n = 1, 2n2 = 2(1)2 = 2
L shell n = 2, 2n2 = 2(2)2 = 8
M shell n = 3, 2n2 = 2(3)2 = 18
N shell n = 4, 2n2 = 2(4)2 = 32
(b) The maximum number of electrons that can be accommodated in the outermost orbit is 8.
(c) Electrons are not accommodated in a given shell unless all the inner shells are filled.
7. Define valency by taking examples of silicon and oxygen.
Ans. Valency is the combining capacity of an atom.
Atomic number of oxygen = 8
Electronic configuration of oxygen (O) =
Atomic number of silicon = 14
Electronic configuration of silicon (Si) =
In the atoms of oxygen the valence electrons are 6. To fill the orbit, 2 electrons are required. In the atom of silicon, the valence electrons are 4. To fill this orbit 4 electrons are required. Hence, the combining capacity of oxygen is 2 and of silicon is 4.
i.e., Valency of oxygen = 2
Valency of silicon = 4
8. Explain with examples:
(a) Atomic number (b) Mass number, (c) Isotopes and (d) Isobars.
Give any two uses of isotopes.
Ans. (a) Atomic number: The atomic number of an element is the number of protons present in the nucleus of its atom. E.g., Carbon has 4 protons, hence atomic no. = 4.
(b) Mass number: The mass number of an atom is the number of protons and neutrons present in its nucleus.
Mass number = number of nucleons = number of protons + number of neutrons
Example: For Carbon atom; Mass number = Protons + Neutrons = 6 + 6 = 12
(c) Isotopes: Isotopes are atoms of the same element which have same atomic number but different mass number.
E.g., 11H, 21H, 31H
(d) Isobars: Isobars are atoms having the same mass number but different atomic numbers.
E.g., 2040Ca 1840Ar
Two uses of isotopes are:
(i) An isotope of iodine is used in the treatment of goitre.
(ii) An isotope of uranium is used as a fuel in nuclear reactors.
9. Na+ has completely filled K and L shells. Explain.
Ans. Sodium atom (Na), has atomic number = 11
Number of protons = 11
Number of electrons = 11
Electronic configuration of Na =
Sodium atom (Na) loses 1 electron to become stable and form Na+ ion. Hence, Na+ has completely filled K and L shells.
10. If bromine atom is available in the form of say, two isotopes 7935Br (49.7%) and 8135Br (50.3%), calculate the average atomic mass of bromine atom.
Ans. The average atomic mass of bromine atom
11. The average atomic mass of a sample of an element X is 16.2u. What are the percentages of isotopes 168X and 188X in the sample?
Ans. Let the percentage of 168X be x and the percentage of 168X be 100 - x.
12. If Z = 3, what would be the valency of the element? Also, name the element.
Ans. Z or atomic number = 3
∴ Electronic configuration = 2, 1
Hence, valency of element = 1
From the periodic table, we know that the element with atomic number 3 is lithium.
13. Composition of the nuclei of two atomic species X and Y are given as under X Y
Protons = 6 6
Neutrons = 6 8
Give the mass number of X and Y. What is the relation between the two species?
Ans. Mass number of X = Protons + Neutrons = 6 + 6 = 12
Mass number of Y = Protons + Neutrons = 6 + 8 = 14
Both X and Y are isotopes of same element, because they have same atomic number. i.e., Z = 6. [atomic number = number of protons].
14. For the following statements, write T for True and F for False.
(a) J.J. Thomson proposed that the nucleus of an atom contains only nucleons.
(b) A neutron is formed by an electron and a proton combining together. Therefore, it is neutral.
(c) The mass of an electron is about 1/2000 times that of proton.
(d) An isotope of iodine is used for making tincture iodine, which is used as a medicine.
Ans. (a) F (b) F (c) T (d) F
15. Rutherford’s alpha-particle scattering experiment was responsible for the discovery of (a) Atomic nucleus (b) Proton (c) Electron (d) neutron
Ans. (a) Atomic nucleus
16. Isotopes of an element have
(a) the same physical properties (b) different number of neutrons
(c) different number of neutrons (d) different atomic numbers.
Ans. (c) different number of neutrons
17. Number of valence electrons in Cl ion are: (a) 16 (b) 8 (c) 17 (d) 18
Ans. (b) 8
18. Which one of the following is a correct electronic configuration of sodium?
(a) 2, 8 (b) 8, 2, 1
Ans. (d) 2, 8, 1
19. Complete the following table.
(c) 2, 1, 8 (d) 2, 8, 1
Multiple Choice Questions
1. If an atom has 8 protons and 7 electrons, what is its overall charge? (a) +1 (b) +8 (c) Neutral (d) -1
2. Identify the Mg2+ ion from the figure where, n and p represent the number of neutrons and protons respectively (NCERT Exemplar) (a) (b) (c) (d)
n = 12 p = 12
n = 12 p = 10
n = 10 p = 12
n = 12 p = 12
3. Which of the following correctly represent the electronic distribution in the Al atom?
(a) 4, 8, 1 (b) 1, 8, 4 (c) 2, 8, 3 (d) 8, 2, 3
4. Deuterium, an isotope of hydrogen has (a) No neutrons and one proton (b) One neutrons and two protons
(c) One electron and two neutron (d) One proton and one neutron
5. Which of the following statement is always correct? (NCERT Exemplar)
(a) An atom has equal number of electrons and protons.
(b) An atom has equal number of electrons and neutrons.
(c) An atom has equal number of protons and neutrons.
(d) An atom has equal number of electrons, protons and neutrons.
6. What is the electronic configuration of potassium (atomic number 19)?
7. Isotope of which element is used in the treatment of cancer.
(a) Iodine. (b) Cobalt (c) Uranium (d) Potassium
8. An atom with 3 protons and 4 neutrons will have a valency of (a) 3 (b) 7 (c) 1 (d) 4
9. In Rutherford’s alpha scattering experiment, one out of every particles appear to rebound.
(a) 1200000 (b) 12000 (c) 1200 (d) 120000
10. In which element neutron is not present?
(a) Carbon (b) Hydrogen (c) Nitrogen (d) Helium
11. In a sample of ethyl ethanoate (CH3COOC2H5) the two oxygen atoms have the same number of electrons but different number of neutrons. Which of the following is the correct reason for it?
(NCERT Exemplar)
(a) One of the oxygen atoms has gained electrons.
(b) One of the oxygen atoms has gained two neutrons.
(c) The two oxygen atoms are isotopes.
(d) The two oxygen atoms are isobars.
12. An alpha particle contains (a) 4 positive charge and 2 mass unit (b) 2 positive charge and 4 mass unit (c) 2 positive charge and 2 mass unit (d) 4 positive charge and 4 mass unit
13. In the Thomson’s model of atom, which of the following statements are correct? (NCERT Exemplar)
(i) the mass of the atom is assumed to be uniformly distributed over the atom (ii) the positive charge is assumed to be uniformly distributed over the atom (iii) the electrons are uniformly distributed in the positively charged sphere (iv) the electrons attract each other to stabilise the atom (a) (i), (ii) and (iii) (b) (i) and (iii) (c) (i) and (iv) (d) (i), (iii) and (iv)
14. The subatomic particle that determines the chemical properties of an element is the: (a) Proton (b) Neutron (c) Electron (d) Nucleus
15. Dalton’s atomic theory successfully explained (NCERT Exemplar) (i) Law of conservation of mass (ii) Law of constant composition (iii) Law of radioactivity (iv) Law of multiple proportion (a) (i), (ii) and (iii) (b) (i), (ii) and (iv) (c) (ii), (iii) and (iv) (d) (i), (ii) and (iv)
16. The primary difference between isotopes and isobars is that: (a) Both terms refer to the same concept (b) Isobars have different numbers of electrons (c) Isotopes have different atomic numbers (d) Isotopes have different numbers of neutrons while isobars share the same mass number
17. Which of the following in given figure do not represent Bohr’s model of an atom correctly? (i) (ii) (iii) (iv) (a) (i) and (ii) (b) (ii) and (iii) (c) (ii) and (iv) (d) (i) and (iv)
18. The elements of which of the following pairs represent isobars? (a) 11H and 21H (b) 11H and 31H (c) 126C and 146C (d) 4018Ar and 4020Ca
19. Atomic models have been improved over the years. Arrange the following atomic models in the order of their chronological order. (NCERT Exemplar)
(i) Rutherford’s atomic model
(ii) Thomson’s atomic model
(iii) Bohr’s atomic model
(a) (i), (ii) and (iii) (b) (ii), (iii) and (i) (c) (ii), (i) and (iii) (d) (iii), (ii) and (i)
20. The symbol “X” in the notation (Y X Z) represents the:
(a) Atomic number (b) Atomic mass
(c) Mass number (d) Chemical symbol of the element
Answers
Constructed Response Questions
Very Short Answer Questions (30–50 words)
1. Write down the electron distribution of chlorine atom. How many electrons are there in the L shell? (Atomic number of chlorine is 17). (NCERT Exemplar)
Ans. The electron distribution in a chlorine atom (atomic number 17) is 2 electrons in the K shell, 8 electrons in the L shell, and the remaining 7 electrons in the M shell. Thus, the L shell of the chlorine atom contains 8 electrons.
2. What were the drawbacks of Rutherford’s model of an atom?
Ans. Rutherford’s atomic model failed to explain the stability of electrons in a circular path. He stated that electrons revolve around the nucleus in a circular path, but particles in motion would undergo acceleration and cause energy radiation. Eventually, electrons should lose energy and fall into the nucleus. But it never happens.
3. Calculate the number of neutrons present in the nucleus of an element X which is represented as 1531X.
(NCERT Exemplar)
Ans. The element X is represented as 1531X. The atomic number (Z) is 15, indicating 15 protons. The mass number (A) is 31. The number of neutrons is calculated by subtracting the number of protons from the mass number: 31 - 15 = 16 neutrons.
4. If a neutral atom has filled K and L shells, determine its electron configuration and also tell the ion formed upon removal of the outer electron.
Ans. Maximum electrons that K shell can hold = 2
Maximum electrons that L shell can hold = 8
So, a filled K and L shell means the atom has (2 + 8 = 10 electrons)
So, electronic configuration will be 2, 8
This configuration belongs to neon, which has a fully filled and stable outer shell. Hence, it does not normally form ions. However, hypothetically, if one electron is removed, the atom would have 9 electrons with a configuration of 2, 7, forming a Ne⁺ ion. This requires high energy and is not likely under normal conditions.
5. Will 35Cl and 37Cl have different valencies? Justify your answer. (NCERT Exemplar)
Ans. No, 35Cl and 37Cl will not have different valencies. Valency is determined by the number of electrons in the outer shell, which is the same for both these isotopes as they have the same atomic number. Thus, both 35Cl and 37Cl have a valency of 1.
6. Give any two uses of radioactive isotopes.
Ans. Two uses of radioactive isotopes are:
(a) An isotope of uranium (U-235) is used as a fuel in nuclear reactors.
(b) An isotope of cobalt (Co-60) is used in the treatment of cancer.
7. Which of the following pairs represent isotopes and which represents isobars?
(a) 23He and 13He (b) 1224Mg and 1225Mg (c) 1940K and 2040Ca (d) 2040Ca and 1940K
Ans. (a) ₂³He and ₁³H
Same mass number (3), different atomic numbers (2 and 1). Hence, they are isobars.
(b) ₁₂²⁴Mg and ₁₂²⁵Mg
Same atomic number (12), different mass numbers (24 and 25). Hence, they are isotopes.
(c) ₁₉⁴⁰K and ₂₀⁴⁰Ca
Same mass number (40), different atomic numbers (19 and 20). Hence, they are isobars.
(d) ₂₀⁴⁰Ca and ₁₉⁴⁰K
Same mass number (40), different atomic numbers (20 and 19). Hence, they are isobars.
8. Find out the valency of the atoms represented by the figures (a) and (b). (NCERT Exemplar) (a) (b)
Ans. Atom (a) has zero valency as it has 8 electrons in its valence shell making the configuration stable. Atom (b) has a valency of +1 as it has 7 electrons in its outermost shell. It can accept 1 electron to achieve octet configuration.
9. Explain why isotopes have similar chemical properties even though their mass numbers differ.
Ans. Because isotopes of the same element possess identical numbers of protons and electrons, they have the same chemical properties despite having different numbers of neutrons. The change in mass does not affect the electronic configuration which is responsible for chemical behaviour.
10. An element X has a mass number 4 and atomic number 2. Write the valency of this element?
(NCERT Exemplar)
Ans. The element X with atomic number 2 and mass number 4 is helium (He). Helium is a noble gas with a filled outer shell, and its valency is 0. This indicates that it does not typically form bonds with other elements due to its stability.
11. Show diagrammatically the electron distributions in a sodium atom and a sodium ion and also give their atomic number.
Ans.
Both atoms have same number of protons and hence, they have same Atomic number. Atomic number of Na and Na+ ion = 11.
12. (a) Which observation of Rutherford’s α-ray scattering experiment led to the fact that there is large empty space inside the atom?
(b) Rutherford’s model of atom was similar to that of solar system. Justify the statement.
Ans. (a) Most of a-particles passed through the gold foil without getting deflected and very few particles were deflected from their path, which led to the fact that most of the space is empty inside the atom.
(b) Rutherford’s model was similar to that of solar system. Like the different planets are revolving around the sun in a similar way the electrons are revolving around the nucleus in an atom.
13. Which of the two would be chemically more reactive: element A with atomic number 18 or element D with atomic number 16 and why?
Ans. Electronic configuration of 18A is 2, 8, 8. It would be chemically inert due to its complete octet.
Electronic configuration of 16D is 2, 8, 6. To complete its octet, it needs to gain 2 electrons, therefore it will be more reactive.
14. An atom of element X has an atomic number of 81 and a mass number of 35. Calculate the number of protons, neutrons, and electrons in its neutral state, and provide its electron configuration.
Ans. Given, Atomic number = No. of protons = 35
In a neutral atom, No. of protons = No. of electrons = 35
Mass Number = No. of protons + No. of neutrons
⇒ No. of neutrons = 81 - 35 = 46
Electronic configuration = 2, 8, 18, 7
15. Complete the given table on the basis of information available in the symbols given below
6
Ans.
(a) 126C
16. A student weighs 45.5 kg. Suppose his entire body is made up of electrons. How many electrons are there in his body?
Ans. Since, mass of 1 electron = 9.1 × 10–31 kg ∴ 1 kg is the mass of
45.5 kg is the mass of 45.51031 9.1 electrons = 5 × 1031 electrons.
Short Answer Questions (50–80
words)
1. Elements P, Q, R, S and T have atomic numbers 4, 9, 13,17 and 20 respectively.
(a) Write their electronic distribution.
(b) Determine their valency.
Ans. (a) The electronic distribution of elements P, Q, R, S and T are given below:
2 2
2 7
2 8 3
2 8 7
2 8 8 2
(b) As valency of an element depends on the valence shell electrons thus, for the given elements valencies are P = 2; Q = 1; R = 3; S = 1; T = 2.
2. (a) Answer the following questions.
(i) Name the scientist who proposed the model for the structure of an atom for the first time.
(ii) Where is proton located in an atom?
(b) An atom of an element has mass number 24 and its atomic number is 12. How many neutrons does it have? Also, name the element.
Ans. (a) (i) J.J. Thomson was the first scientist to propose a model for the structure of an atom.
(ii) Proton is located within the nucleus of an atom.
(b) Number of neutrons = Mass number - Atomic number = 24 - 12 = 12
The element is magnesium 2412Mg
3. Write the conclusions drawn by Rutherford when he observed the following.
(a) Most of the α-particles passing straight through the gold foil.
(b) Some α-particles getting deflected from their path.
(c) Very small fraction of α-particles getting deflected by 180°.
Ans. (a) Most of the space inside the atom is empty.
(b) It indicates that the positive charge of the atom occupies a very little space.
(c) All the positive charge and mass of the atom were concentrated in a very small volume called nucleus.
4. You are given the atom of an element 147X. Find out the (a) number of protons, electrons and neutrons in X.
(b) valency of X.
(c) chemical formula of the compound formed when X reacts with (i) hydrogen
(ii) carbon
Ans. (a) Number of protons = 7
Number of electrons = 7
Number of neutrons = 14 - 7 = 7
(b) Electronic configuration = 2, 5
Hence, valency = 8 - 5 = 3
(c) (i) XH3
(ii) X4C3
5. What information do you get from the given figure about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form.
6. (a) What will be the maximum number of electrons which can be filled in Y-shell of an imaginary atomic model?
(b) Why is it almost impossible to find such an atom in nature?
Ans. (a) According to Bohr and Bury rule, the maximum number of electrons in Y-shell (15th shell)
= 2n2 = 2 × (15)2 = 2 × 225 = 450
(b) Atom with greater number of electrons have greater number of protons and neutrons. Therefore, they have unstable nucleus which radiates energy or divide itself into smaller stable nucleus. Hence, elements with atomic number greater than 100 are rarely found in nature.
7. An atom of an element has 6 electrons in L-shell.
(a) What is the atomic number of the element?
(b) State its valency.
(c) Identify the element and write its name.
Ans. (a) Since the atom has 6 electrons in L shell, that means it K shell is fully filled with 2 electrons. Therefore, the total number of electrons in atom = atomic number = 8
(b) Valency = 8 - 6 = 2
(c) The element with atomic number 8 is oxygen (O).
8. The two isotopes of chlorine have mass number 35 and 37 and number of neutrons 18 and 20 respectively. Which one will have higher valency? Do they have same physical or chemical properties?
Ans. Number of protons in 35Cl = 35 - 18 = 17
Number of electrons = Number of protons = 17
Similarly, number of protons in 37Cl = 37 - 20 = 17
Number of electrons = Number of protons = 17
Since, both have same number of electrons (i.e: 17), with 7 electrons in M shell.
Hence, they will have same valency of 8 - 7 = 1.
Isotopes having same number of electrons show same chemical properties but may show some different physical properties.
9. (a) Identify which of the following pairs are isotopes and which are isobars? Give reasons for your choice.
(b) Do isobars also have identical chemical properties like isotopes? State reason.
Ans. (a) Since samples A, C, and D have the same number of protons, they are isotopes.
Sample B and C have same mass number (Mass number = Number of protons + number of neutrons = 37) but different atomic numbers (i.e. 18 and 17 respectively). Hence, they are a pair of isobars.
(b) Isobars are atoms of different elements that have the same mass number but different atomic numbers. Since chemical properties depend on the atomic number (i.e., the number of electrons and their arrangement), isobars have different chemical properties. In contrast, isotopes have the same atomic number and electron configuration, so they exhibit similar chemical behaviour.
10. Match the names of the Scientists given in column A with their contributions towards the understanding of the atomic structure as given in column B (NCERT Exemplar)
Column A
Column B
(a) Ernest Rutherford (i) Indivisibility of atoms
(b) J. J. Thomson (ii) Stationary orbits
(c) Dalton (iii) Concept of the nucleus
(d) Neils Bohr (iv) Discovery of electrons
(e) James Chadwick (v) Atomic number
(f) E. Goldstein (vi) Neutron
(g) Mosley (vii) Canal rays
Ans.
Column A
Column B
(a) Ernest Rutherford (iii) Concept of the nucleus
(b) J. J. Thomson (iv) Discovery of electrons
(c) Dalton (i) Indivisibility of atoms
(d) Neils Bohr (ii) Stationary orbits
(e) James Chadwick (vi) Neutron
(f) E. Goldstein (vii) Canal rays
(g) Mosley (v) Atomic number
11. An element X forms the compound XH3 when it reacts with hydrogen.
(a) Find the valency of element X.
(b) Write the electronic configuration and valency of a neutral sulphur atom.
(c) What will be chemical formula for product of reaction between X and sulphur.
Ans. (a) Valency of X is 3 because it combines with three atoms of hydrogen.
(b) Electronic configuration of sulphur is 2, 8, 6. Hence, valency of sulphur, is 8 - 6 = 2.
(c) X with valency of 3 and sulphur with valency of two will react to form X2S3
X 3 S 2
12. In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer. (NCERT Exemplar)
Ans. The given statement is incorrect.
According to this statement: p > n > e
But actually, number of protons can never be greater than the number of neutrons (except Protium 11H).
Number of neutrons can be equal to or greater than the number of protons because mass number is equal to double the atomic number or greater than double the atomic number.
Of course, number of neutrons can be greater than the number of electrons because number of electrons are equal to number of protons in the neutral atom.
13. (a) What is the mass number of an element whose atomic number is 17 and number of neutrons are 18?
(b) What is the number of electrons in M-shell of atom of an element whose atomic number is 8?
(c) Find the number of neutrons in an atom with atomic number and mass number are 9 and 19 respectively.
Ans. (a) Atomic number = Number of protons = 17
Mass number = Number of protons + Number of neutrons = 17 + 18 = 35
(b) Electronic configuration of the element having atomic number 8 is
Hence, there are zero electrons present in M-shell.
(c) Atomic number = Number of protons = 9;
Mass number = Number of neutrons + Number of protons
Number of neutrons = Mass number - Number of protons
Number of neutrons = 19 - 9 = 10
14. Enlist the conclusions drawn by Rutherford from his α-ray scattering experiment. (NCERT Exemplar) Ans. Rutherford concluded from his a-ray scattering experiment that:
1. Most a-particles passed through the gold foil undeflected, suggesting that atoms are mostly empty space.
2. Some a-particles were deflected at small angles, indicating the presence of a positive charge in the atom.
3. A few particles rebounded back, signifying a dense, positively charged core or nucleus.
4. The nucleus is small compared to the size of the atom but contains most of its mass. These findings contradicted the “plum pudding model” and led to the nuclear model of the atom, with electrons orbiting a central nucleus.
15. In the following table the mass numbers and the atomic numbers of certain elements are given.
(a) Select a pair of isobars from the above table.
(b) What would be the valency of element C listed in the above table?
(c) Which two sub-atomic particles are always equal in number in a neutral atom?
Ans. (a) D and E have same mass number but different atomic numbers. Hence, they are a pair of isobars.
(b) Electronic configuration of C is 2,5. Hence, its valency is three because it will gain three electrons to attain stable electronic configuration.
(c) For a neutral atom, Number of electrons = Number of protons
16. Two ions, one carrying a 3⁻ charge and the other a 3⁺ charge, are found to have 7 and 14 neutrons respectively. If the electronic configuration of both ions is the same as that of neon (Ne), determine the mass number of each ion and identify the elements to which they belong.
Ans. Electronic configuration of Neon (Ne) is 2, 8, i.e., it has 10 electrons. Both ions also have 10 electrons.
For the 3⁻ ion:
Number of protons = Number of electrons + charge on atom = 10 + (-3) = 7
Number of neutrons = 7
Mass number = Number of proton + Number of neutron = 7 + 7 = 14
For the 3⁺ ion:
Number of protons = Number of electrons + charge on atom = 10 + 3 = 13
Number of neutrons = 14
Mass number = Number of proton + Number of neutron = 13 + 14 = 27
Therefore,
The 3⁻ ion belongs to nitrogen (N) and its mass number is 14.
The 3⁺ ion belongs to aluminium (Al) and its mass number is 27.
17. If one atom of the element 24 22 X loses 2 electrons and gains 2 neutrons, determine its new atomic number and mass number.
Ans. Given,
Atomic number of element X = 12
Mass number of element X = 24
Number of neutrons in X = Mass number - Atomic number = 24 - 12 = 12
When the atom loses 2 electrons, the atomic number remains the same because electrons are not part of the nucleus.
So, new atomic number = 12
When the atom gains 2 neutrons,
New number of neutrons = 12 + 2 = 14
New mass number = Protons + Neutrons = 12 + 14 = 26
Therefore,
New atomic number = 12
New mass number = 26
The new representation of the ion will be ²⁶₁₂X²⁺. (Since it lost 2 electrons, it becomes a 2⁺ ion)
Long Answer Questions (80–120 words)
1. Compare the atomic models proposed by Thomson, Rutherford and Bohr.
Ans.
Feature
Structure
Model
Atom is a positively charged sphere with electrons in it
Nucleus No nucleus
Electron Arrangement
Stability of Atom
Electrons are scattered inside the sphere like plums
Could not explain stability
Model
Atom has a small, dense, positive nucleus with electrons around it
Model
Electrons revolve in fixed orbits (shells) around the nucleus
Nucleus present at the center Nucleus present at the center
Electrons move around the nucleus randomly
Could not explain electron stability in orbits
Electrons move in fixed circular paths called energy levels
Explained stability by placing electrons in specific energy levels
Nicknamed As Plum pudding model / Watermelon model Nuclear model Planetary model
2. (a) An element X has an atomic number 20 and mass number 40. Draw a diagram showing the distribution of electrons in the orbits and the nuclear composition of the neutral atom of the element. What is the valency of the element and why?
(b) If this element X combines with another element Y whose electronic configuration is 2,8,7. What will be the formula of the compound thus formed? State how did you arrive at this formula.
Ans. (a) Element X:
Atomic number = number of protons = number of electrons = 20 (neutral atom)
Mass number = number of neutrons + number of protons
Number of neutrons = Mass number - number of protons = 40 - 20 = 20
So, electronic configuration of X is: 2, 8, 8, 2 and the valency is 2.
(b) Since valency of X is 2 and Y is 1, the formula of the compound will be XY2.
3. What is Rutherford’s gold foil experiment? Write the conclusions and drawbacks of Rutherford’s model of atom.
Ans. Rutherford bombarded a stream of a-particles on a thin sheet of gold foil. An alpha-particle is a positively charged helium ion (He2+).
The following observations were made.
• Most of the a-particles passed straight through the foil without any deflection. This concluded that most of the space inside of an atom is empty.
• A few a-particles were deflected through small angle and few through larger angles. This happened due to positive charge on a-particles and core (nucleus) of the atom. The heavy positively charged ‘core’ was named as nucleus.
• The number of a-particles which bounced back was very small. This concluded that the volume of the nucleus is very small in comparison to the total volume of the atom.
On the basis of gold foil experiment, Rutherford concluded that an atom consists of nucleus which has positive charge and it is surrounded with electrons which are moving around the nucleus.
Drawbacks in the Rutherford’s model
According to classical electro-magnetic theory, a moving charged particle, such as an electron under the influence of attractive force loses energy continuously in the form of radiations. As a result of this, electron should lose energy and therefore, should move in even smaller orbits ultimately falling into the nucleus. But the collapse does not occur. There is no explanation for this behaviour.
Paths of several alpha particles in the scattering experiment
4. (a) Differentiate between isotopes and isobars.
(b) Name two isotopes and their real world application.
Ans. (a)
Feature Isotopes Isobars
Definition Atoms of the same element having the same atomic number but different mass numbers.
Element
Type
Chemical Properties
Same element
Identical chemical properties
Atoms of different elements having the same mass number but different atomic numbers.
Different elements
Different chemical properties
Examples 11H, 21H, 31H (Hydrogen isotopes) 4018Ar and 4020Ca
(b) Two Isotopes and Their Applications are
Cobalt-60 (Co-60): Used in the treatment of cancer through radiation therapy.
Uranium-235 (U-235): Used as a fuel in nuclear reactors for generating electricity.
5. Number of electrons, protons and neutrons in chemical species A, B, C and D is given below.
Now answer the following questions using given data in the table.
(a) What is the mass number of A and B?
(b) What is the atomic number of B?
(c) Which two chemical species represent a pair of isotopes and why?
(d) What is the valency of element C?
(e) Which of these species carry a charge?
Also justify your answers.
Ans. (a) Mass number = Number of protons + Number of neutrons
Mass number of A = 3 + 4 = 7
Mass number of B = 9 + 8 = 17
(b) Atomic number of B = Number of protons = 9
(c) C and D are isotopes as they have same atomic numbers but different mass numbers.
(d) Electronic configuration of C: 2, 6
Its valency is 8 - 6 = 2. Hence, it needs two electrons to complete its octet.
(e) In species A and B, number of electron ≠ number of protons. Hence, both A and B carry a charge. A has +1 charge and B has -1 charge.
Competency Based Questions
Multiple Choice Questions (Choose the most appropriate option/options)
1. According to the Bohr-Bury rules, what is the maximum number of electrons that can be accommodated in the M-shell?
(a) 6 (b) 8 (c) 18 (d) 32
2. Which of the following statements about Rutherford’s model of atom are correct? (NCERT Exemplar) (i) considered the nucleus as positively charged (ii) established that the α–particles are four times as heavy as a hydrogen atom (iii) can be compared to solar system (iv) was in agreement with Thomson’s model (a) (i) and (iii) (b) (ii) and (iii) (c) (i) and (iv) (d) only (i)
3. Which among the following ions have 10 electrons?
(a) Mg+ (b) Na+ (c) Ar (d) C–3
4. The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element? (NCERT Exemplar) (a) 3115X (b) 3116X (c) 1615X (d) 1516X
5. Which of the following formula is used to find the maximum number of electrons that can be accommodated in an orbit?
(c) A large part of an atom’s volume is empty space.
(d) Atoms are solid and impenetrable spheres.
7. Which of the following are true for an element? (NCERT Exemplar)
(i) Atomic number = number of protons + number of electrons
(ii) Mass number = number of protons + number of neutrons
(iii) Atomic mass = number of protons = number of neutrons
(iv) Atomic number = number of protons = number of electrons
(a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) (d) (ii) and (iv)
8. Rutherford’s α–particle scattering experiment showed that (NCERT Exemplar) (i) electrons have negative charge (ii) the mass and positive charge of the atom is concentrated in the nucleus (iii) neutron exists in the nucleus (iv) most of the space in atom is empty Which of the above statements are correct?
(a) (i) and (iii) (b) (ii) and (iv) (c) (i) and (iv) (d) (iii) and (iv)
9. Rutherford’s alpha (α) particles scattering experiment resulted in to discovery of (a) Electron (b) Proton (c) Nucleus (d) Atomic mass
10. Elements with valency 1 are (NCERT Exemplar) (a) always metals (b) always non-metals (c) either metals or non-metals (d) always metalloids
11. The sub-atomic particle found in canal rays is known as: (a) Neutron (b) Proton (c) Electron (d) Positron
12. The ion of an element has 3 positive charges. Mass number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion? (NCERT Exemplar) (a) 13 (b) 10 (c) 14 (d) 16
13. Which of the following have the minimum number of electrons? (a) O2– (b) N2– (c) Mg2+ (d) Na+
Assertion–Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): The mass of the total number of protons and neutrons is a measure of the approximate mass of an atom.
Reason (R): The mass of an electron is negligible.
2. Assertion (A): In Bohr’s model, electron orbits are quantized and electrons do not radiate energy while in stationary states.
Reason (R): Discrete energy levels prevent the continuous loss of energy by orbiting electrons.
3. Assertion (A): Atom is electrically neutral.
Reason (R): A neutral particle, i.e. neutron, is present in the nucleus of atom.
4. Assertion (A): The atomic masses of some elements are in fractions and not a whole numbers.
Reason (R): The fractional atomic masses of some elements are due to the existence of their isotopes having different masses.
5. Assertion (A): Electronic configuration of neon is 2, 8.
Reason (R): Atomic number of neon is 8.
6. Assertion (A): Isobars are identical in chemical properties.
Reason (R): Isobars have same mass number.
7. Assertion (A): The atoms of different elements having same mass number are known as isotopes.
Reason (R): The sum of number of protons and neutrons in the isotopes is always different.
8. Assertion (A): In Rutherford’s gold foil experiment, very few alpha-particles are deflected back.
Reason (R): Nucleus present inside the gold atom is heavy.
Case-Based/Source-Based/Passage-Based Questions
1. Read the given passage and answer the questions that follow:
In 1897, J.J. Thomson discovered the electron, a negatively charged particle present within the atom. This marked a major step in understanding atomic structure. Following this discovery, it became clear that atoms contain internal particles and are not indivisible as previously thought. Scientists understood that atoms must contain both positive and negative charges to remain neutral overall. This raised new questions about how these charges are distributed within the atom and how such a system remains stable. Although early ideas tried to address this, they were later refined by further experimental evidence and models proposed by scientists like Rutherford and Bohr.
(a) Which statement is true according to Thomson’s model?
(i) Atom has a central nucleus
(ii) Electrons are uniformly scattered within the atom
(iii) Protons and neutrons are in the centre
(iv) Atom is mostly empty space
(b) Which discovery after Thomson’s model proved its major assumption wrong?
(i) Electron’s mass
(ii) Proton’s charge
(iii) Nucleus in an atom
(iv) Neutron’s neutrality
(c) Why is Thomson’s model often compared to a “plum pudding” or “watermelon”?
(i) Because atoms are round like a pudding
(ii) Electrons are like seeds embedded in a soft positive sphere
(iii) Protons stick out like plums
(iv) Neutrons orbits around the nucleus like grapes
(d) If an atom has 6 electrons according to Thomson’s model, what must be the total positive charge in it?
(i) +3 (ii) +6 (iii) +12 (iv) 0
2. Read the given passage and answer the questions that follow:
Atoms have a central region called the nucleus, which contains certain subatomic particles responsible for many of the atom’s properties. This core houses positively charged particles and others with no charge, both of which contribute significantly to the atom’s overall mass. These particles are tightly packed in the nucleus, making it the densest part of the atom. Interestingly, all atoms of a particular element share a common identity based on the number of these specific particles found in their nuclei. This shared feature allows us to distinguish one element from another. While the outer region of the atom contains lighter, negatively charged particles, the inner core plays the main role in defining the atom’s structure and behaviour.
(a) Atomic number of an atom is equal to the total number of ____________ within the atom.
(i) electrons (ii) protons (iii) neutrons (iv) nucleons
(b) Mass number of an atom is independent of the number of _____________ within the atom.
(i) electrons (ii) protons (iii) neutrons (iv) nucleons
(c) Which of the following is the correct representation of an atom? (i) ZAX (ii) AZX (iii) XAZ (iv) AXZ
(d) Identify the correct statement
Statement 1 – Protons are present in the nucleus of an atom.
Statement 2 – Mass number is the number of nucleons within an atom.
Statement 3 – Atomic number is denoted by ‘A’.
Statement 4 – The mass of electron in an atom can be ignored.
(i) Only 2 (ii) Both 3 & 4 (iii) Both 1 & 2 (iv) All of the above
3. Read the given passage and answer the questions that follow:
Atoms are all made from the same basic ingredients, the sub-atomic particles. The only difference is the recipe-how many of each of these sub atomic particles are present in the atoms of different elements. All the elements are given a number, which describes their relative position in the periodic table. It tells us something about the structure of the atoms of the element. It is defined as the number of protons in the atom. As an atom has no overall charge, the positive charge is equal to the negative charge of the electrons. The atomic number is also equal to the number of electrons. When an atom loses electrons, it forms a positive ion or cation as now number of protons is greater than the number of electrons. Negative ions or anions are formed when atoms gain electrons. How can an element have a fractional relative atomic mass if both protons and neutrons have a relative atomic mass of 1? The reason is that atoms of the same element with different mass numbers exist. These are called isotopes.
The table given below gives the number of protons, neutrons and electrons in atoms or ions.
(a) Are isotopes and isobars identical? Explain with reasons.
(b) Which atom in the table is an isotope of the atom with the composition 11p, 11e and 14n? Give a reason for your choice.
(c) Write the correct symbol or formula for B, C, D and E.
4. Read the given passage and answer the questions that follow:
In the early 20th century, scientists were deeply interested in uncovering the mysteries of atomic structure, especially the way electrons are arranged within an atom. To explore this, they designed a groundbreaking experiment using a source of alpha (a) particles—positively charged, fast-moving particles emitted from certain radioactive materials. These particles were directed at an extremely thin sheet of gold foil, just a few atoms thick. The purpose of this experiment was to observe how the alpha particles interacted with the atoms in the gold foil. This experiment was a turning point in atomic research. It provided valuable clues that atoms are not solid spheres, as once believed, but have a much more complex internal structure. The findings laid the groundwork for future developments in atomic theory and nuclear physics, eventually leading to more accurate models that explained the behaviour of atoms and subatomic particles in greater detail.
(a) Which of the following scientist was known as the ‘Father of nuclearʹ physics?
(i) J.J. Thomson (ii) John Dalton (iii) Ernest Rutherford (iv) Neils Bohr
(b) Positively charged centre in an atom is termed as
(i) Nucleus (ii) Molecule (iii) Neutron (iv) Proton
(c) Identify the correct statement
Statement 1 – Positively charged particles are present inside the nucleus.
Statement 2 – The electrons revolve around the nucleus in circular paths.
Statement 3 – Nearly all the mass of an atom resides in the nucleus.
Statement 4 – The size of the nucleus is very small as compared to the size of the atom.
(i) Only 2 (ii) Both 3 & 4 (iii) Both 1 & 2 (iv) All of the above (d) Why were the results of the alpha scattering experiment surprising?
(i) Most of the alpha particles were deflected at large angles, suggesting atoms are solid and impenetrable.
(ii) The majority of alpha particles passed through the gold foil, indicating that atoms are mostly empty space with a dense central region.
(iii) All of the alpha particles bounced back, showing that the gold foil was impenetrable.
(iv) The alpha particles lost their energy upon striking the foil and did not pass through it.
Isotopes are atoms of the same element that have the same atomic number but different mass numbers. They have the same number of protons and electrons but a different number of neutrons. Since they belong to the same element, they have similar chemical properties but different physical properties.
Example: 11H, 12H, 13H (isotopes of hydrogen)
Isobars are atoms of different elements that have the same mass number but different atomic numbers. Since they belong to different elements, both their chemical and physical properties are different.
Example: 1840Ar and 2040Ca
Thus, isotopes and isobars differ in their atomic structure, chemical behavior, and physical properties.
(b) B. Isotopes have same number of protons.
(c) 1123B, 1840C, 1531D, 1327E
4. (a) (iii) (b) (i) (c) (iv) (d) (ii)
Practice Questions
Multiple Choice Questions
1. In an atom with a fixed number of protons, an increase in the number of neutrons leads to a change in which of the following?
(a) Both atomic number and mass number remain the same.
(b) Both atomic number and mass number increase.
(c) Atomic number remains unchanged but mass number increases.
(d) Atomic number decreases while mass number remains unchanged.
2. An element exhibits an electronic configuration of 2, 8, 6. What is its valency?
(a) 0 (b) 8 (c) 2 (d) 6
3. In Rutherford’s α-particle scattering experiment, the occurrence of a few large-angle deflections is best attributed to:
(a) Encounters with a tiny, dense, positively charged nucleus
(b) Random scattering due to thermal motion
(c) Collisions with electrons in the outer orbits
(d) Diffuse positive charge spread throughout the atom
4. Which statements are correct in Thomson’s model for atoms?
(i) The mass of an atom is assumed uniformly distributed across the atom
(ii) The positive charge is assumed uniformly distributed throughout the atom
(iii) The electrons are evenly distributed in the positively charged space
(iv) The electrons attract one another to stabilise the atom
(a) (i), (ii) and (iii) (b) (i) and (iii) (c) (i) and (iv) (d) (i), (iii) and (iv)
5. A neutral sodium atom has an electron configuration of 2, 8, 1. When it forms a positive ion with completely filled K and L shells, which ion is formed?
(a) Na2+ (b) Na– (c) Na+ (d) Na
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
6. Assertion(A): Isobars are identical in chemical properties.
Reason (R): Isobars have same mass number.
7. Assertion (A): Anions are larger in size than the parent atom.
Reason (R): In an anion, the number of protons in the nucleus is less than the number of electrons moving around it.
Very Short Answer Questions (30-50
words)
8. Why did Rutherford select a gold foil in his a-ray scattering experiment? (NCERT Exemplar)
9. Consider an element X with an atomic number of 11 and a mass number of 24. If one atom of X loses 2 electrons and gains 2 neutrons, determine its new atomic number and mass number.
10. Write the distribution of electrons in argon and nitrogen atoms? Determine their valencies.
11. An element with atomic number 9 forms an ion by gaining one electron. Calculate the number of electrons in the ion. Identify the ion and provide its electronic configuration.
12. Compare the number of protons, neutrons and electrons in 1840Ar and 2040Ca? Are they isotopes or isobars? Justify your answer.
Short Answer Questions (50-80 words)
13. A metal (A = 24) having same number of protons and neutrons, combines with two atoms of a nonmetal (Z = 17). Identify the elements and write the electronic configuration of these elements in combined state.
14. Explain why chlorine, whether as the element or its compounds, always has relative atomic mass of about 35.5.
15. The given figure depicts the atomic structure of an atom of an element ‘X’. Write the following information about the element ‘X’.
(a) Atomic number of ‘X’
(c) Valence electrons
(e) ‘X’ should be metal or non-metal.
16. Differentiate between isotopes and isobars.
(b) Atomic mass of ‘X’
(d) Valency of ‘X’
17. (a) An atom of an element has three electrons in the outermost M-shell. State its (i) electronic configuration. (ii) valency.
(b) What is the number of protons in a neutral atom whose L-shell is half-filled?
Long Answer Questions (80-120 words)
18. (a) Consider an ion of element 18692X²⁺.
(i) What is the number of electrons in this atom?
(ii) What is the mass number?
(iii) Calculate the number of neutrons in this atom.
(b) A certain isotope of calcium has 20 protons and 20 neutrons. If another isotope of calcium has 24 neutrons, then
(i) Compare atomic numbers and mass numbers
(ii) Are chemical properties same? Justify your answer.
19. An element A has atomic number 16 and mass number 32. When it reacts with hydrogen, it gains 2 electrons to become an ion.
(a) Identify the element.
(b) What is the name of the ion formed by element A? Also, Write its symbol.
(c) Calculate the number of protons, neutrons, and electrons in the ion.
20. Consider the following pairs,
(i) 2658A, 2858B
(ii) 3579Y, 3580Y
(a) Which of the above pairs are isotopes and isobars?
(b) What factors are responsible for the change in superscripts, 79, 80 (in case ii), though the element is the same?
(c) Write the nuclear composition of all four elements.
Brain Charge
1. Crossword puzzle ACROSS
2. Neutral subatomic particle (7)
6. Discovered nucleus using gold foil experiment (10)
7. Model of atom with fixed orbits (4)
8. Maximum number of electrons in K shell (3)
9. Total number of protons and neutrons in an atom (8)
10. Central part of an atom (7) DOWN
1. Negatively charged particle (8)
3. Name of scientist who discovered the electron (7)
4. Positively charged particle in an atom (7)
5. Atomic number is equal to number of __________ (8)
2. Word Jumble – Unscramble the words
(a) ORNPTO
(b) SNLUECUI
(c) EROTUNN
(d) CTREOELN
(e) SMAS RUEBMN
3. Who Am I?
(a) I am negatively charged and revolve around the nucleus. Who am I?
(b) I have no charge but I add to the mass of the atom. Who am I?
(c) I discovered the nucleus with my gold foil experiment. Who am I?
(d) I proposed the model where electrons revolve in fixed orbits. Who am I?
(e) I always equal the number of electrons in a neutral atom. Who am I?
Challenge Yourself
1. Calculate the total number of protons in 1 g of H+ ion. (given that mass of proton is 1.67262192 × 10–27 kg)
2. What would happen to the identity of an atom if:
(a) You change the number of protons?
(b) You change the number of neutrons?
(c) You change the number of electrons?
3. If an atom has equal number of protons and neutrons, but twice the number of electrons as protons, what can you say about this species?
4. An atom has atomic number 8 and mass number 16. During a nuclear reaction, it converts two of its protons into neutrons. What is the new element formed and what will be its relation with original atom?
5. Atom A has 18 protons and 18 electrons. Atom B has 20 protons and 20 electrons. Which atom is more chemically reactive and why?
6. An atom has only 1 neutron and no protons. Is such an atom possible? Justify your answer.
Answers
Practice Questions
9. New atomic number = 11 (protons unchanged) New mass number = 11 + 15 = 26
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10. Argon (Z = 18), Electron distribution: 2, 8, 8
Valency: 0
Nitrogen (Z = 7), Electron distribution: 2, 5
Valency: 3
11. Atomic number = 9, Fluorine
Gains 1 electron → Electrons = 10
Electronic distribution: 2, 8
12. 4018Ar (Argon): Protons = 18, Neutrons = 22, Electrons = 18, Atomic number = 18, Mass number = 40
4020Ca (Calcium): Protons = 20, Neutrons = 20, Electrons = 20, Atomic number = 20, Mass number = 40
13. Metal: Magnesium (Mg)
Non-metal: Chlorine (Cl)
Electronic configuration of Mg²⁺ = 2, 8
Electronic configuration of Cl– = 2, 8, 8
15. (a) 8 (b) 18 (c) 6 (d) 2 (e) Non-metal
17. (a) 2, 8, 3 (b) 3 (c) 6
18. (a) (i) 90 (ii) 186 (iii) 96 (b) (i) Atomic number: both 20; Mass number: 40 and 44
Brain Charge
2. (a) PROTON
(b) NUCLEUS
(c) NEUTRON
(d) ELECTRON
(e) MASS NUMBER
Challenge Yourself
1. 6 × 1023
3. (a) Electron
(b) Neutron
(c) Rutherford
(d) Bohr
(e) Proton
4. Original: Z = 8 (Oxygen); New: Carbon with mass number 16; relation: Isobar
5. Atom B
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1.
SELF-ASSESSMENT
Time: 1.5 Hour
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Max. Marks: 50
Multiple Choice Questions (10 × 1 = 10 Marks)
1. Which isotope is commonly used as a fuel in nuclear reactors?
2. Iron forms two oxide FeO (A) and Fe2O3 (B). The number of electrons present in Fe of A and Fe of B and O of A and O of B will be respectively. (Given atomic numbers: Fe = 26, O = 8)
7. The hydrogen isotope with one proton and two neutrons is known as:
(a) Protium (b) Deuterium (c) Tritium (d) None of these
8. The first model of an atom was given by (NCERT Exemplar) (a) N. Bohr (b) E. Goldstein (c) Rutherford (d) J.J. Thomson
9. An atom has 7 neutrons and 9 electrons. If the atom has a charge of -2, what is its atomic number? (a) 11 (b) 16 (c) 9 (d) 7
10. Which experiment provided the first clear evidence for the existence of a small, dense, positively charged nucleus within the atom?
(a) Thomson’s cathode ray experiment
(b) Millikan’s oil drop experiment
(c) Faraday’s electrolysis experiment
(d) Rutherford’s gold foil experiment
Assertion–Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false. (d) A is false, but R is true.
11. Assertion (A): Isotopes of an element exhibit nearly identical chemical behaviour.
Reason (R): They have the same number of protons and, consequently, the same electron configuration.
12. Assertion (A): For non-metals, valency can be calculated by subtracting the number of valence electrons from eight.
Reason (R): Non-metals achieve stability by completing their octet through the gain of electrons..
13. Assertion (A): The gold foil experiment demonstrated that an atom’s positive charge is concentrated in a very small nucleus.
Reason (R): Most alpha particles passed through the foil un-deflected because atoms are mostly empty space.
14. Assertion (A): The chemical properties of an element are determined by its innermost electrons.
Reason (R): Valence electrons participate in chemical bonding, thus influencing an element’s reactivity.
15. Read the given passage and answer the questions that follow:
Some atoms of the same element can have the same number of protons but a different number of neutrons, which gives them different mass numbers. These different forms of the same element are naturally occurring and are found in varying proportions. Because of the difference in the number of neutrons, they have different masses, even though they behave the same in chemical reactions. The physical properties, such as density or boiling point, may vary slightly due to the difference in mass. When calculating the atomic mass of such an element, scientists take the average of the masses of all its naturally existing forms, weighted by how abundant each form is in nature. This is why the atomic mass of many elements is not a whole number.
(a) The atoms of the same element, having the same atomic number but different mass numbers are termed as _________
(i) Isotopes (ii) Isobars (iii) Isotones (iv) Isomers
(b) Which of the following are the isotopes of hydrogen atom.
(i) Protium (ii) Deuterium (iii) Tritium (iv) All of the above
(c) Identify the correct statement for isotopes
Statement 1 – Chemical properties are the same.
Statement 2 – Physical properties are different.
Statement 3 – Chemical properties are different.
Statement 4 – Physical properties are same.
(i) Only 2 (ii) Both 3 & 4 (iii) Both 1 & 2 (iv) All of the above
(d) If a proton inside the nucleus of an oxygen decays into a neutron, it will become (i) isotope of oxygen (ii) isotope of nitrogen (iii) isotope of fluorine (iv) isotope of neon
16. Read the given passage and answer the questions that follow:
Niels Bohr received the Nobel Prize in 1922 for his research on the structure of the atom. He also wrote several important books, including The Theory of Spectra and Atomic Constitution, Atomic Theory,
and The Description of Nature. To solve the issues found in Rutherford’s model of the atom, Bohr proposed a new model. According to Bohr, electrons move only in certain fixed paths inside the atom. These fixed paths are called discrete orbits or energy levels. While moving in these energy levels, electrons do not lose any energy. These orbits or energy levels are named using letters such as K, L, M, N or by numbers like n = 1, 2, 3, 4, and so on.
(a) The third shell of the atom is represented by alphabet (i) K (ii) L (iii) M (iv) N
(b) These orbits or shells are called (i) energy levels. (ii) nuclear orbit. (iii) atomic levels. (iv) none of the these.
(c) The maximum number of electrons that can be accommodated in M shell: (i) 2 (ii) 32 (iii) 18 (iv) 8
(d) An atom has seven electrons in M shell. What is its atomic number? (i) 7 (ii) 9 (iii) 17 (iv) 18
Very Short Answer Questions (30–50 words) (3 × 2 = 6 Marks)
17. In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed? (NCERT Exemplar)
18. A neutral magnesium atom has an atomic mass of 24 u. When it loses two electron to form Mg²⁺, compute the number of protons, neutrons, and electrons in Mg²⁺ and provide its electron configuration.
19. Explain why electrons do not emit radiation when in their stationary orbits according to Bohr’s postulates.
Short Answer Questions (50–80 words)
(4 × 3 = 12 Marks)
20. Describe the significance of the Bohr-Bury rules in determining the electronic configuration of elements,
21. Determine the number of electron, proton and neutron in sodium atom and sodium ion. Also show diagrammatically the electron distributions in both of them.
22. The atomic number of Sulphur is 16. Its mass number is 32.
(a) How many protons and neutrons are present in a sulphur atom?
(b) Draw the diagram of a sulphur atom.
(c) Write electronic configuration of sulphur ion (S2-).
23. One electron is present in the outermost shell of the atom of an element AZX. (Given that A > Z > 1)
(a) Is it a metal or non-metal?
(b) What will be the value of charge of the ion formed, if this electron is removed from the outermost shell?
(c) Can we remove one more atom from this ion? Justify your answer.
Long Answer Questions (80–120 words)
(2 × 5 = 10 Marks)
24. (a) The average atomic mass of an element is 16.2u. If its two isotopes have mass numbers 16 and 18, determine the percentage abundance of the 16u isotope.
(b) Explain why isotopes exhibit nearly identical chemical behaviour.
25. (a) 22286Rn is an isotope of noble gas, radon. How many protons, neutrons and electrons are there in one atom of this radon isotope?
(b) An atom ‘M’ of an element reacts with oxygen to form M2O3. Calculate the valency of the element ‘M’.
(c) Calculate the total number of protons, neutrons and electrons in Na2O.
5 The Fundamental Unit of Life
This chapter deals with the cell which is the basic structural and functional unit of life. It carries out essential life processes such as growth, metabolism and reproduction. Cells form the building blocks of all living organisms and can exist as independent, single-celled organisms or as part of complex multicellular organisms. Cells are structured and various organelles and inclusions, which are characteristics of all living forms, are present in the cells.
Cell Division: Formation of new cell from pre existing cell
Cell
Structural and functional unit of living organisms was discovered by Robert Hooke in the year 1665 on his own designed microscope.
Cell Number
The count of cells varies in organisms.
Types of Cells
On the basis of number
Unicellular (Amoeba, Paramecium, Bacteria)
Multicellular (Some fungi, plants, animals)
On the basis of the kind of nucleus and presence of membrane bound organelles
Prokaryotic (Bacteria, blue green algae)
Eukaryotic (All plants, animals and fungi)
Mitochondria
Power houses of the cell, energy production
Plastids
Photosynthesis, food production, storage
Morphology of a Cell
Cell Shape
The shape and size of cells are related to the specific function they perform.
Mitosis Meiosis
Cell Size
The size of a cell can vary from 0.0001 mm to 18 cm.
Components of the Cell
Cell wall
In bacterial cell and some plant and fungal cells, it lies outside the plasma membrane.
Cell membrane
Outermost covering of the cell
Cytoplasm
Cell Organelles
Golgi complex
Transport, packaging, lysosome formation
Chloroplast (Green coloured)
Chromoplast (Coloured)
Leucoplast (Colourless)
Fluid content inside the plasma membrane and outside the nucleus (Living cell organelles + Non-living cell inclusions)
Nucleus
Control centre of the cell
Nuclear membrane
Lysosome
Suicidal bags, digestive bag
Nucleoplasm Chromosome
Vacuoles
Storage, water balance
Endoplasmic Reticulum (E.R)-Intracellular transport, lipid and protein synthesis
RER (With ribosomes)
SER (Without ribosomes)
Chapter at a Glance
• A cell is the basic structural and functional unit of life.
• In the year 1665, the scientist Robert Hooke discovered the cell while examining a thin slice of cork.
• The cell theory, that all plants and animals are composed of cells and that the cell is the basic unit of life, was presented by two biologists, Schleiden (1838 and Schwann (1839).
• Virchow (1855 suggested that all cells arise from pre-existing cells.
• The shape and size of cells are related to the specific function they perform.
• Amoeba, Paramecium and bacteria are examples of some unicellular organisms while fungi, plants and animals are examples of multicellular organisms.
• The plasma membrane, nucleus and cytoplasm are the main components of every cell.
• Cells are enclosed by a plasma membrane composed of lipids and proteins.
• The cell membrane is an active part of the cell. It regulates the movement of materials between the ordered interior of the cell and the outer environment.
• The movement of water molecules through a selectively permeable plasma membrane is called osmosis. Osmosis is a type of diffusion that occurs when water molecules move from a region of high concentration to a region of low concentration.
• If the medium surrounding the cell has a higher water concentration than the cell, meaning that the outside solution is very diluted, the cell will gain water by osmosis. Such a solution is known as a hypotonic solution.
• If the medium has exactly the same water concentration as the cell, there will be no net movement of water across the cell membrane. Such a solution is known as an isotonic solution.
• If the medium has a lower concentration of water than the cell, meaning that it is a very concentrated solution, the cell will lose water by osmosis. Such a solution is known as a hypertonic solution.
• In plant cells, a cell wall composed mainly of cellulose is located outside the cell membrane. Cellulose is a complex substance that provides structural strength to plants.
• The presence of the cell wall enables the cells of plants, fungi and bacteria to exist in hypotonic media without bursting.
• The nucleus in eukaryotes is separated from the cytoplasm by a double-layered membrane and it directs the life processes of the cell.
• The nucleus contains chromosomes that contain information for the inheritance of characteristics from parents to the next generation in the form of DNA (Deoxyribo Nucleic Acid molecules.
• Cells are prokaryotic and eukaryotic based on the presence of true nucleus and some membrane-bound organelles.
• The cytoplasm is the fluid content inside the plasma membrane. It contains many specialised living cell organelles and non-living cell inclusions.
Fig. 5.1 A plant cell in different solutions
• Cell organelles are enclosed by membranes. Some important cell organelles are the endoplasmic reticulum, Golgi apparatus, lysosomes, mitochondria and plastids.
• The ER functions both as a passageway for intracellular transport and as a manufacturing surface.
• There are two types of ER—rough endoplasmic reticulum (RER and smooth endoplasmic reticulum (SER) RER is involved in protein synthesis while SER is involved in lipid synthesis. SER plays a crucial role in detoxifying many poisons and drugs.
• Ribosomes are particles that are found attached to the rough endoplasmic reticulum. Protein synthesis occurs in ribosomes. Thus, ribosomes are called protein factories.
• The Golgi apparatus was first described by Camillo Golgi
• The Golgi apparatus consists of stacks of membrane-bound vesicles that function in the storage, modification and packaging of substances manufactured in the cell. The Golgi apparatus is also involved in the formation of lysosomes.
• Lysosomes are membrane-bound sacs filled with digestive enzymes. These enzymes are made by RER. Lysosomes are a kind of waste disposal system of the cell. These organelles are also known as the ‘suicide bags’ of a cell.
• Mitochondria are known as the powerhouses of the cell. The energy required for various chemical activities needed for life is released by mitochondria in the form of ATP (Adenosine triphosphate molecules.
• Most plant cells have large membranous organelles called plastids, which are of two types— chromoplasts and leucoplasts.
• Chromoplasts contain coloured pigments that give colour to fruits, flowers and vegetables. Chromoplasts containing the pigment chlorophyll are known as chloroplasts. Chloroplasts are important for photosynthesis in plants.The primary function of leucoplasts is storage.
• Most mature plant cells have a large central vacuole that helps to maintain the turgidity of the cell and stores important substances including wastes. Vacuoles are small in animal cells while plant cells have very large vacuoles.
• The division of cells occurs in organisms for the body to grow, for replacing dead cells and for forming gametes for reproduction.
NCERT Zone
Intext Questions
1. Who discovered cells, and how?
Ans. In the year 1665, Robert Hooke discovered cells while examining a thin slice of cork. He observed the cells through a self-designed microscope. He observed that the cork resembled the structure of a honeycomb consisting of numerous tiny compartments. The small box-like structures are referred to as cells.
2. Why is the cell called the structural and functional unit of life?
Ans. Cells form the structure of an entity. A group of cells forms a tissue, further an organ and ultimately an organ system. They perform fundamental functions and life processes such as respiration, digestion and excretion, in both unicellular and multicellular entities. Because cells perform all the activities independently, they are referred to as the structural and functional units of life.
3. How do substances like CO2 and water move in and out of the cell? Discuss.
Ans. Carbon dioxide and water move across the cell by the process of diffusion and osmosis, respectively. CO2 moves by diffusion. The cellular waste accumulates in high concentrations in the cell, whereas the concentration of CO2 in the external surroundings is comparatively lower. This difference in the concentration level inside and outside of the cell causes the CO2 to diffuse from a region of higher (within the cell to a lower concentration (outside the cell).
H2O diffuses by osmosis through the cell membrane. It moves from a region of higher concentration to a lower concentration region through a selectively permeable membrane until the concentrations on both sides of the membrane are equal.
4. Why is the plasma membrane called a selectively permeable membrane?
Ans. The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. Not all molecules are free to diffuse.
5. Fill in the gaps in the following table illustrating differences between prokaryotic and eukaryotic cells.
Prokaryotic Cell
1. Size: generally small (1–10 m) 1 m = 10–6 m
2. Nuclear region: and is known as
3. Chromosome: Single
4. Membrane-bound cell organelles absent
Eukaryotic Cell
1. Size: generally large (5–100 m)
2. Nuclear region: well defined and surrounded by a nuclear membrane
3. More than one chromosome
4.
Ans. 2. The nuclear region is poorly defined due to the absence of a nuclear membrane and is known as the nucleoid.
4. Membrane-bound cell organelles present.
6. Can you name the two organelles we have studied that contain their own genetic material?
Ans. The two organelles which have their own genetic material are
• Mitochondria
• Plastids
7. If the organisation of a cell is destroyed due to some physical or chemical influence, what will happen?
Ans. If the organisation of a cell is destroyed due to physical or chemical influences, the cell will lose its ability to perform essential life processes such as respiration, growth and reproduction. This disruption may lead to the leakage of cellular contents, malfunction of organelles, and eventually, the death of the cell.
8. Why are lysosomes known as suicide bags?
Ans. When cells are damaged during a disturbance in cellular metabolism and their revival is not possible, lysosomes may burst, and the enzymes digest their own cell. This is the reason why lysosomes are known as suicide bags.
9. Where are proteins synthesised inside the cell?
Ans. Protein synthesis in cells takes place in the ribosomes. Hence, they are also referred to as protein factories. Ribosomes are particles that are found attached to the rough endoplasmic reticulum (RER).
NCERT Exercises
1. Make a comparison and write down ways in which plant cells are different from animal cells.
Ans.
Characteristics
Cell Wall Present
Shape of the Cell
Plant Cell
With distinct edges, the shape is either rectangular or square
Animal Cell
Round and irregular
Nucleus
Present. It lies on one side of the cell. Present. It lies in the centre of the cell.
Lysosomes Rarely present Always present
Plastids Present Absent
Structure of Vacuoles
Single or a few large vacuoles that are centrally located Presence of numerous small vacuoles
2. How is a prokaryotic cell different from a eukaryotic cell?
Ans.
Prokaryotic Cell
1. Size: generally small (1–10 m), 1 m = 10–6 m
2. The nuclear region is not well defined as the nuclear membrane is absent and is referred to as the nucleoid.
3. There is a single chromosome.
4. Membrane-bound cell organelles are absent.
Eukaryotic Cell
1. Size: generally large (5–100 m)
2. Nuclear region: well-defined and encircled by a nuclear membrane.
3. There is more than one chromosome.
4. Membrane-bound cell organelles are present.
3. What would happen if the plasma membrane ruptures or breaks down?
Ans. If the plasma membrane ruptures or breaks down, molecules of some substances will freely move in and out of the cells. As the plasma membrane acts as a mechanical barrier, the exchange of material from its surroundings through osmosis or diffusion in a cell won’t take place. Consequently, the cell would die due to the disappearance of the protoplasmic material.
4. What would happen to the life of a cell if there was no Golgi apparatus?
Ans. The Golgi apparatus consists of stacks of membrane-bound vesicles whose functions are as follows: • Storage of substances • Packaging of substances • Manufacturing of substances Without the Golgi apparatus, the cells will not be able to package and dispatch materials produced by the cells. Since the Golgi apparatus is also involved in the formation of cells, new cells will not be produced.
5. Which organelle is known as the powerhouse of the cell? Why?
Ans. Mitochondria are known as the power houses of the cell. It is because they release the energy required for different activities of life. Mitochondria release energy in the form of ATP (Adenosine triphosphate molecules, essential for the numerous chemical activities of life. Hence, ATP is often referred to as the ‘energy currency of the cell’.
6. Where do the lipids and proteins constituting the cell membrane get synthesised?
Ans. Lipids and proteins are synthesised in the ER (Endoplasmic Reticulum).
7. How does an Amoeba obtain its food?
Ans. An Amoeba obtains its food through the process of endocytosis. Its flexible cell membrane, assisted by the pseudopodia, engulfs the food particles by forming a food vacuole around them. The Amoeba then secretes digestive enzymes to bring about the digestion of the trapped food. Next, the digested food passes into the cytoplasm and the undigested matter is expelled from the cell (egestion).
Fig. 5.2 The process of digestion in Amoeba
8. What is osmosis?
Ans. The process of movement of a water molecule from a region of higher concentration to a region of lower concentration through a semipermeable membrane is known as osmosis.
9. Carry out the following osmosis experiment:
Take four peeled potato halves and scoop each one out to make potato cups. One of these potato cups should be made from a boiled potato. Put each potato cup in a trough containing water. Now, follow these steps.
(I) Keep cup A empty.
(II) Put one teaspoon of sugar in cup B.
(III) Put one teaspoon of salt in cup C.
(IV) Put one teaspoon of sugar in the boiled potato cup D.
Set these aside for two hours. Then, observe the four potato cups and answer the following:
(a) Explain why water gathers in the hollowed portion of B and C.
(b) Why is potato A necessary for this experiment?
(c) Explain why water does not gather in the hollowed-out portions of A and D.
Ans.
(a) Water accumulates in the hollowed portions of B and C as a difference in the water concentration is observed. Thereby, endosmosis occurs as the cells act as a semipermeable membranes.
(b) Potato A is essential in this experiment as it is necessary to compare different scenarios seen in potato cups B, C and D. Potato A in this experiment clearly shows that the potato cavity on its own cannot bring about water movement.
(c) The cavity in A does not have water in the hollowed-out portion because we did not add any salt or sugar to cause osmosis.
The cells in cup D are dead, since the semipermeable membrane doesn’t exist for the water to flow—osmosis does not occur.
10. Which type of cell division is required for the growth and repair of the body and which type is involved in the formation of gametes?
Ans. There are two ways in which a cell divides:
• Mitosis
• Meiosis
Mitosis is a type of cell division that is involved in the growth and repair of the body, whereas meiosis is a type of cell division which results in the formation of gametes.
Multiple Choice Questions
1. Name the organelle which plays a crucial role in detoxification of poisons and drugs in a cell.
(a) The Golgi apparatus
(c) SER
(b) Lysosomes
(d) Vacuoles and RER
Fig. 5.3 An experiment of osmosis using potato halves
Empty potato cavity
Empty potato
Potato with sugar
Boiled potato
Potato with sugar
Potato with salt
Water- lled cavity
2. Amoeba acquires its food through a process, termed (a) exocytosis. (b) endocytosis. (c) plasmolysis.
(d) exocytosis and endocytosis both.
3. Cell wall of which one of these is not made up of cellulose
(a) Bacteria
(c) Mango tree
4. 1 μm is (a) 10–6 m
m
(b) Hydrilla
(d) Cactus
m
5. Lysosome contains digestive enzymes. It arises from which of the following organelles?
(a) endoplasmic reticulum.
(b) Golgi apparatus.
(c) nucleus.
(d) mitochondria.
m
6. A cell will swell up if: (NCERT Exemplar)
(a) the concentration of water molecules in the cell is higher than the concentration of water molecules in the surrounding medium.
(b) the concentration of water molecules in the surrounding medium is higher than the concentration of water molecules in the cell.
(c) the concentration of water molecules is the same in the cell and in the surrounding medium.
(d) the concentration of water molecules does not matter.
7. Which of these options are not a function of ribosomes? (NCERT Exemplar)
(I) It helps in the manufacture of protein molecules.
(II) It helps in the manufacture of enzymes.
(III) It helps in the manufacture of hormones.
(IV) It helps in the manufacture of starch molecules.
(a) (I) and (II)
(c) (III) and (IV)
(b) (II) and (III)
(d) (IV) and (I)
8. The following are a few definitions of osmosis. Read carefully and select the correct definition. (NCERT Exemplar)
(a) Movement of water molecules from a region of higher concentration to a region of lower concentration through a semipermeable membrane
(b) Movement of solvent molecules from its higher concentration to lower concentration
(c) Movement of solvent molecules from higher concentration to lower concentration of solution through a permeable membrane
(d) Movement of solute molecules from a lower concentration to a higher concentration of solution through a semipermeable membrane
9. Find out the false sentence. (NCERT Exemplar)
(a) Golgi apparatus is involved in the formation of lysosomes.
(b) Nucleus, mitochondria and plastids have DNA; hence they are able to make their own structural proteins.
(c) Mitochondria are said to be the power houses of the cell as ATP is generated in them.
(d) The cytoplasm is called the protoplasm.
10. Find out the correct sentence. (NCERT Exemplar)
(a) Enzymes packed in lysosomes are made by the RER (rough endoplasmic reticulum).
(b) The rough endoplasmic reticulum and the smooth endoplasmic reticulum produce lipid and protein, respectively.
(c) The endoplasmic reticulum is related with the destruction of the plasma membrane.
(d) The nucleoid is present inside the nucleoplasm of a eukaryotic nucleus.
11. Lipid molecules in the cell are synthesised by
(a) The movement of water across a semi permeable membrane is affected by the amount of substances dissolved in it.
(b) Membranes are made of organic molecules like proteins and lipids.
(c) Molecules soluble in organic solvents can easily pass through the membrane.
(d) Plasma membranes contain chitin sugar in plants.
14. All are functions of the endoplasmic reticulum except
(a) It transports proteins and carbohydrates to other organelles.
(b) It helps in the biogenesis of cellular membranes.
(c) It involves the formation of lysosomes.
(d) It provides a large area for cellular reactions.
15. Select the odd one out.
(a) Mitochondria – ATP
(c) Vacuoles – Turgidity
16. The main function of the cytoplasm in a cell
(a) Storage of genetic material
(b) Site of photosynthesis
(b) Lysosomes – Storage
(d) RER – Digestive enzymes
(c) Location of cellular metabolism and organelle support
(d) Regulation of cell functions
17. Which structure helps in the movement of chromosomes during cell division?
(a) Centrosome
(c) Nucleolus
(b) Golgi apparatus
(d) Ribosomes
18. The membrane of the Golgi apparatus has connections with the (a) membranes of mitochondria.
(c) plasma membrane.
19. Which of the following functions is performed by SER?
(b) endoplasmic reticulum.
(d) membrane of the nucleus
(a) It helps to expel excess water and waste out of the cell.
(b) It helps to produce energy in the form of ATP.
(c) It helps to detoxify drugs.
(d) It helps to digest some foreign particles.
20. The endoplasmic reticulum helps in transporting proteins between various regions of the cytoplasm. How?
(a) By directing cell organelles to perform some biochemical activity
(b) By forming a network of membrane-bound tubes in the cytoplasm
(c) By occupying most of the space in the cytoplasm
(d) By generating small transport vesicles throughout the cell
21. How can you differentiate a prokaryotic cell from a eukaryotic cell?
(a) Presence or absence of genetic material
(b) Presence or absence of the nucleus and vacuole
(c) Presence or absence of the cytoplasm
(d) Presence or absence of membrane-bound organelles
22. Which of the following organelle is involved in regulating cellular processes such as mitosis, DNA repair, stress responses, autophagy, apoptosis and inflammation?
(a) RER (b) Golgi bodies (c) Ribosomes (d) SER
23. An unripe green fruit changes colour when it ripens because (a) chromoplasts change to chlorophyll. (b) chromoplasts change to chromosomes. (c) chromosomes change to chromoplasts. (d) chloroplasts change to chromoplasts.
24. Choose the location in plant where the meiotic cell division takes place. (a) Leaves and stem (b) Stem and branches (c) In anther and ovary (d) All of these
Answers
Constructed Response Questions
Very Short Answer Questions
(30-50 words)
1. Discuss the microscopic structure of a cell.
Ans. Cork cells were the first cells to be observed. They were composed of box-like compartments, forming a honeycomb structure. Cell organelles are found embedded in the cytoplasm. These are smaller and bounded by the plasma membrane.
2. How can you calculate the magnification of a microscope?
Ans. The magnification of a microscope is calculated by multiplying the powers of the eyepiece and objective lenses.
Mathematically, M = P1 × P2, where P1 is the power of the eyepiece and P2 is the power of the objective lens.
3. Are chromatids the same as chromosomes? Explain.
Ans. No, chromatids are not the same as chromosomes, but they are related. A chromosome is a single DNA molecule before replication. A chromatid is one of the two identical halves of a chromosome.
4. Answer the following questions.
(a) A cell of fungi can withstand very dilute external media without bursting but an animal cell cannot. Justify.
(b) What is the main complex substance found in the cell wall?
Ans. (a) A cell of fungi is covered by a cell wall which can withstand greater changes without bursting. But, in an animal cell, there is no cell wall, so normally bursting occurs in the changes.
(b) The cell wall is made up of cellulose.
5. Answer the following questions.
(a) How is a hypotonic solution different from an isotonic solution?
(b) The plasma membrane is made up of which two components?
Ans. (a) A solution having solute concentration lower than that of the cell sap is called a hypotonic solution. A solution having solute concentration same as that of the cell sap is called a isotonic solution.
(b) The plasma membrane is mostly made up of lipids and proteins.
6. What is a cell wall and how is it formed?
Ans. The cell wall is non-living and freely permeable rigid structure that binds the plant cell. It is secreted by the cell itself for the protection of its plasma membrane and cytoplasm.
7. Name the organelle
(a) which is involved in the formation of lysosomes.
(b) which is the storage sac of solid and liquid materials.
Ans. (a) Golgi apparatus
(b) Vacuoles
8. Answer the following questions.
(a) Name two structures found in animal cells but not in plant cells.
(b) What is the intracellular source of digestive enzymes?
Ans. (a) Lysosomes and centrioles
(b) Lysosome
9. Why were scientists not able to observe most of the cell organelles before 1940?
Ans. Before 1940, scientists could view the cell only under a light microscope. The invention of the electron microscope in 1940 enabled the scientists to observe the cell in greater detail.
10. There would be no plant life if chloroplasts did not exist. Justify.
Ans. Chloroplasts contain the pigment chlorophyll which is responsible for food preparation in plants through the process of photosynthesis. Hence, if there were no chloroplasts then there would not have been any plant life.
11. What are the functional regions of a cell?
Ans. There are three major functional regions of cells (i) the cell membrane or plasma membrane (ii) nucleus and (iii) cytoplasm.
12. What are cytosol and cytoskeleton?
Ans. Cytosol is the semifluid part of the cell cytoplasm which is embedded in between cell organelles. Cytoskeleton is a network of protein fibres present in the cell which provides a supporting framework for the organelles.
13. What are secretory proteins? Give an example of secretory protein.
Ans. Proteins which are synthesised by the cell and then released into the outer medium of the cell are called secretory proteins.
Examples of secretory proteins include mucus, digestive enzymes and hormones.
14. What do you mean by plasmodesmata?
Ans. The presence of a cell wall in plant cells prevents direct exchange of materials between them. To facilitate this exchange, plant cells are connected by cytoplasmic channels called plasmodesmata, which pass through the cell walls. These channels allow the transfer of materials between adjacent cells.
15. Why do animals not have a cell wall?
Ans. Animal cells do not have rigid walls because cell walls are incompatible with the way in which an animal moves and grows. The flaccid cell membrane provides the animal cell freedom of mobility and formation of different tissues which is not present in plants.
16. What is the significance of pores present on the nuclear membrane?
Ans. The pores present on the nuclear membrane allow transport of water-soluble molecules across the nuclear envelope. RNA and ribosomes move out of the nucleus, whereas carbohydrates, lipids and proteins move into the nucleus.
17. A cell is called a building unit of an organism. Justify.
Ans. An organism is made up of various organ systems like the digestive system and nervous system. These organ systems in turn are made up of various organs which are made up of tissues. Also, tissues are a group of cells performing the same function. Hence, a cell is the building unit of an organism. Cell → tissue → organ → organ system → organism.
18. Why does the skin of your finger wrinkle when you wash clothes for a long time?
Ans. While washing clothes, the skin of your fingers shrink and appear wrinkled because the soap solution acts as a hypertonic solution, meaning it has a higher concentration of dissolved substances than the cells in the skin. So, it causes the removal of water from the skin cells through a process called osmosis. It leads to shrinkage of the skin on the fingers.
19. Why is endocytosis found in animals? (NCERT Exemplar)
Ans. Endocytosis is a process where a cell engulfs a substance from the external environment and brings it into the cell. It requires a flexible plasma membrane that can undergo extensive invaginations. The external membrane in animal cells is the plasma membrane. Thus, endocytosis is seen only in animal cells.
Plant cells have rigid cell walls that prevent the plasma membrane from doing this.
20. A person takes concentrated solution of salt and after some time, he starts vomiting. What is the phenomenon responsible for such a situation? Explain. (NCERT Exemplar)
Ans. A concentrated salt solution is a hypertonic solution. This causes exosmosis in the cells of the alimentary tract which further results in dehydration and irritation thus resulting in vomiting.
21. We eat food composed of all the nutrients such as carbohydrates, proteins, fats, vitamins, minerals and water. After digestion, these are absorbed in the form of glucose, amino acids, fatty acids, glycerol, etc.
What are the mechanisms involved in the absorption of digested food and water? (NCERT Exemplar)
Ans. The mechanisms that are involved in the total process in the absorption of digested food are simple diffusion, active transport and passive transport.
The mechanism involved in the absorption of water is osmosis.
22. Bacteria do not have chloroplast but some bacteria are photoautotrophic in nature and perform photosynthesis. Which part of the bacterial cell performs this? (NCERT Exemplar)
Ans. Bacteria are prokaryotes with no specific organelles for photosynthesis. The photosynthetic pigments bacteriochlorophyll are found in the folds of the plasma membrane. So, photosynthesis occurs in the small vesicles associated with the plasma membrane.
23. How is a bacterial cell different from an onion peel cell? (NCERT Exemplar) Ans.
Bacterial Cell
1. Size is small (1–10 mm).
2. Nucleus is absent.
3. It is a prokaryotic cell.
4. Cell division takes place by fission or budding.
Onion-peel Cell
1. Size is larger (5–100 mm).
2. Nucleus is present.
3. It is a eukaryotic cell.
4. Cell division takes place by mitosis.
24. Why are lysosomes also known as ʻscavengers of the cellsʼ?
Ans. Lysosomes contain hydrolytic digestive enzymes that break down biomolecules like proteins, nucleic acids, carbohydrates and lipids. They digest worn-out, dead or poorly functioning cell organelles, as well as foreign materials like bacteria. Since lysosomes help in the removal of unnecessary particles and organelles, they are also known as ʻscavengers of the cellsʼ .
25. Which cell organelle is responsible for controlling most of the activities of the cell?
Ans. The nucleus is the organelle that controls most of the cell’s activities by storing genetic information, directing protein synthesis and regulating cell division. It acts as the central command that ensures the cell functions properly to maintain life. Without a nucleus, cells (except prokaryotes like bacteria) would not be able to grow, divide or produce proteins.
Short Answer Questions (50-80 words)
1. Differentiate between diffusion and osmosis.
Ans.
Diffusion
1. It occurs in any medium.
2. Diffusing molecules may be solid, liquid or gaseous solutes.
3. Semipermeable membrane is not required.
2. Answer the following questions.
Osmosis
1. It occurs only in a liquid medium.
2. It involves the movement of solvent molecules only.
3. Semipermeable membrane is required.
(a) What is the main function of cellulose in plant cells?
(b) What do you mean by plasmolysis?
(c) What is active transport?
Ans. (a) Cellulose gives structural strength to the plant cells.
(b) The shrinkage or contraction of the contents of the cell away from the cell wall, when kept in a hypertonic medium is known as plasmolysis.
(c) The movement of molecules across a membrane in cells against a concentration gradient with the help of ATP units is called active transport.
3. What is membrane biogenesis? How is plasma membrane formed during this process?
Ans. The process of plasma membrane formation is called membrane biogenesis. The following organelles are involved in this process:
The proteins and lipids are first synthesised in the rough endoplasmic reticulum and in the smooth endoplasmic reticulum, respectively. These are then transported to the Golgi complex for their modification. After modification, these are transported to the cell surface through vesicles which bud off from the Golgi bodies to fuse with cell membrane and form a part of the membrane.
4. What are the functional differences between a plasma membrane and cell wall?
Ans.
1. It holds cellular contents and controls passage of materials in and out of the cell.
2. It is semipermeable in nature and allows entry of selected molecules into the cell.
3. It is not elastic.
1. It gives protection, strength and rigidity to the cell.
2. It is completely permeable in nature.
3. It is elastic and controls the cell’s turgidity preventing it from bursting.
5. What are the functions of the nuclear membrane?
Ans. Functions of the nuclear membrane:
• A nuclear envelope separates the environment of the nucleus from that of the rest of the cell.
• It protects the genetic material from damage.
• It facilitates and regulates the exchange of materials in and out of the nucleus.
6. Write a note on the Golgi apparatus and the functions it performs.
Ans. The Golgi apparatus or Golgi bodies or Golgi complex is composed of membrane-bound fluid-filled vesicles, vacuoles and cisternae. In animal cells, they are larger and only one or two in number, while in plants they are smaller and more in number. Also, in plant cells, they are distributed throughout the cytoplasm and are called dictyosomes.
Functions:
• It is involved in the transport and modification of protein, lipids as well as carbohydrates.
• It helps in the formation of the cell plate during cell division.
• It is also involved in the formation of the cell wall, plasma membrane, lysosomes and peroxisomes.
• The material synthesised near the endoplasmic reticulum is packaged and transported to various targets as well as outside the cell through the Golgi apparatus.
7. How would the absence of a single cell organelle impact the overall functioning of the cell?
Ans. Functions of all the organelles are inter-linked and ultimately to help in the working of the cell. So, if even a single link is missing, the cell ultimately suffers and dies. For example, DNA from the nucleus passes the information for protein formation to the ribosomes which send the proteins to the Golgi complex and ER for modification and transport.
8. Draw a neat diagram of a plant cell and label any four parts which are common in both plant and animal cells.
Ans. (You can label any four parts.)
Cell membrane
Golgi apparatus
9. Answer the following questions.
Ribosomes
Nucleus
Nucleolus
Endoplasmic reticulum
Mitochondrion
Cytoplasm
(a) Which organelle is associated with the formation of ribosomes?
(b) Name an infectious particle that has no cellular structure, no membranes and does not show any characteristics of life.
(c) Define diffusion.
Ans. (a) Nucleolus
(b) Virus
(c) The movement of molecules from a region of higher concentration to a region of lower concentration is called diffusion.
Fig. 5.4 A plant cell
10. Answer the following questions.
(a) Name the organelle in an animal cell that can make their own proteins.
(b) Name the cellular structure that plays a crucial role in cellular reproduction.
(c) Name the only cell organelle seen in prokaryotic cells.
(d) Which organelle serves as a channel for the transport of materials between the cytoplasm and the nucleus?
(e) What is a nucleoid?
Ans. (a) Mitochondria
(b) Nucleus
(c) Ribosomes
(d) Endoplasmic reticulum
(e) The undefined nuclear region in the cytoplasm of prokaryotes is known as the nucleoid.
11. Answer the questions.
(a) What is endocytosis? How is endocytosis related to exocytosis?
(b) Why is the light microscope called a compound microscope?
Ans. (a) Endocytosis is the ingestion or engulfment of food and other material by folding of the plasma membrane as seen in Amoeba.
Endocytosis and exocytosis are both membrane transport processes. Exocytosis is the opposite process, where the vesicle fuses with the plasma membrane and expels its contents to the outside of the cell.
(b) The light microscope is called a compound microscope because it consists of two or more lens systems.
12. Illustrate an animal cell and label those organelles which are absent in the plant cell.
Ans.
13. How does nourishment occur in Amoeba? (NCERT Exemplar)
Fig. 5.6 Different stages of digestion in Amoeba
Ans. Amoeba uses pseudopodia, which are arm-like projections on the cell surface, to take in food. This is followed by formation of food vacuole surrounding the food by joining of pseudopodia of both sides and engulfment of food vacuole in the body. These food vacuoles serve as the site of food digestion. The digested food is then diffused into the cytoplasm. The expulsion of undigested matter from the cell occurs through the plasma membrane, and is called egestion.
Fig. 5.5 An animal cell
Lysosome
Centriole
14. Why do plant cells possess a large vacuole?
Ans. Plant cells have a large central vacuole because it plays several critical roles in maintaining cell structure, growth and function. Here are the key reasons for its large size:
• The vacuole stores large amounts of water, minerals, sugars, amino acids and waste products. This storage allows the plant to maintain a constant supply of water and nutrients, even during dry conditions.
• The large vacuole maintains turgidity, which keeps the plant rigid and upright. If water is scarce, the vacuole shrinks, causing the plant to wilt.
• As plant cells grow, the vacuole enlarges, allowing the cell to increase in size without needing to produce more cytoplasm (which would require more energy and nutrients).
• The vacuole acts as a storage site for waste materials and harmful by-products.
15. How are chromatin, chromatid and chromosomes related to each other?
Ans. Chromatin is a mass of thread-like structures made of DNA and protein. In a non-dividing nucleus, chromatin appears as a diffuse network of fine filaments.
Chromatids are short thread-like structures that make up chromosomes. Two chromatids are attached at the centromere to form a chromosome. Chromatids are seen in dividing cells.
Chromosomes are rod-like structures that are formed when chromatin condenses during cell division. Each chromosome is made up of two chromatids.
Thus, chromosomes are made up of chromatids and chromatids are made up of chromatin.
16. What are the consequences of the following conditions?
(a) A cell containing a higher water concentration than the surrounding medium
(b) A cell with a lower water concentration than the surrounding medium
(c) A cell with an equal water concentration as its surrounding medium
Ans. (a) Exosmosis
(b) Endosmosis
(c) No effect
17. Identify A, B and C in the given diagram. Differentiate between B and C.
Ans. A- Centriole, B- Asters, C- Spindle
Asters are two star-like structures with fibres radiating from the centrosome. They help in locating the spindle, and trigger the cleavage of cytoplasm.
Spindle fibres are formed from microtubules during cell division. They pull the chromosomes apart and bring them towards the poles. These are broader in the middle and narrower at the poles.
18. Illustrate only a plant cell as seen under an electron microscope. How is it different from an animal cell? Ans.
Fig. 5.8 Illustration of a plant cell
Fig. 5.7
A plant cell differs from an animal cell as it has
• a cell wall outside the plasma membrane.
• a large central vacuole.
• Absence of centrosomes and lysosomes
• Presence of plastids (chloroplasts, chromoplasts, leucoplasts)
Long Answer Questions (80–120 words)
1. Make a pencil sketch of a plant cell as seen under an electron microscope. Label all the parts of the cell clearly. How can you differentiate the parts from an animal cell?
Ans. The major differences are:
• Plant cells have chloroplasts
• Plant cells have large vacuoles.
• Plant cells have cell walls.
• Plant cells do not have centrioles.
• Plant cells do not have lysosomes.
Cell membrane
Ribosomes
Nucleus
Nucleolus
Endoplasmic reticulum
Mitochondrion
Cytoplasm
2. Draw a labelled diagram of a mitochondrion. Write the functions of this organelle.
Ans. Outer membrane DNA
Inner membrane
Functions of mitochondria:
The mitochondria are the main sites for cellular respiration, the process in which the cell converts the food and oxygen into energy in the form of ATP molecules. ATP is used by various bodies as a source of energy to perform functions. This is the reason, mitochondria are called the power houses of the cell.
3. What do you mean by active transport? Differentiate between active and passive transport.
Ans. The process in which the molecules are moved uphill against the concentration gradient is called active transport. Active transport always involves the expenditure of energy because the materials are pumped against the concentration gradient.
Active Transport
1. It involves the movement of molecules against the concentration gradient.
2. It requires energy in the form of ATP molecules.
3. It is a rapid movement.
4. Large molecules are moved through active transport.
Passive Transport
1. It involves the movement of molecules along the concentration gradient.
2. No energy is required.
3. It is a slow movement.
4. Only small molecules or water molecules are transported passively.
Matrix
Ribosome
Cristae
Granule
Fig. 5.10 Mitochondria
Golgi apparatus
Vacuole Cell wall
Chloroplast
Fig. 5.9 A plant cell
4. Write the similarities and differences between mitochondria and chloroplast.
Ans. Similarities
1. Both have their own DNA.
2. Both have their own ribosomes.
3. They are surrounded by a double membrane.
4. They are semi-autonomous organelles.
Differences
1. Mitochondria are the power houses of the cell while the chloroplast is called the kitchen of the cell.
2. Mitochondria are responsible for energy production while the chloroplast is responsible for photosynthesis in plants.
3. Mitochondria are present in all cells while chloroplast is seen in plants and algal cells.
5. Look at the Fig. 5.11. Name it. Write five important features about it.
Ans. The Golgi complex
Important features of the Golgi complex
• The Golgi complex is a stack of membrane-bound sacs called cisternae.
• The Golgi complex receives substances from the endoplasmic reticulum.
• Substances enter the Golgi complex from the cis face, also known as the forming face or entry face. They exit through the trans face, also known as the releasing face.
• The Golgi complex processes and packages cellular metabolites, forms lysosomes and synthesises the cell wall material. It also modifies and synthesises the carbohydrate portions of glycoproteins.
• The Golgi complex plays a crucial role in lysosome biogenesis, which is essential for maintaining cellular functions and the survival of the cell.
6. Draw a plant cell and label the parts which (NCERT Exemplar)
(a) determines the function and development of the cell.
(b) packages materials coming from the endoplasmic reticulum.
(c) provides resistance to microbes to withstand hypotonic external media without bursting.
(d) is the site for many biochemical reactions necessary to sustain life.
7. Draw a well-labelled diagram of a eukaryotic nucleus. How is it different from a nucleoid?
Ans.
Nuclear pore
Nucleolus
Nucleoplasm
Chromatin network
Nuclear membrane
Fig. 5.13
Nucleus
1. A nucleus is found in eukaryotes and stores genetic material.
2. It is enclosed by a nuclear membrane.
3. It is generally spherical.
Nucleoid
1. A nucleoid is found in prokaryotes and stores genetic material.
2. It is not enclosed by a nuclear membrane.
3. It is generally irregular.
8. Differentiate between the rough and the smooth endoplasmic reticulum. How is the endoplasmic reticulum important for membrane biogenesis?
Ans. RER (Rough Endoplasmic Reticulum) SER (Smooth Endoplasmic Reticulum)
1. It has ribosomes attached to its membrane.
2. Its main function is protein synthesis.
3. It is found mostly deep inside the cytoplasm.
4. It may develop from the nuclear envelope.
1. It has no ribosomes attached to its membrane.
2. Its main function is lipid synthesis.
3. It is found in the peripheral region.
4. It may develop from the RER.
The plasma membrane in cells is formed of proteins and lipids through the process of membrane biogenesis. The endoplasmic reticulum performs a vital role in this process by providing the essential lipids and proteins. The SER helps in lipid synthesis while the RER helps in protein synthesis as it contains ribosomes.
9. In brief, state what happens when
(a) dry apricots are left for some time in pure water and later transferred to a sugar solution.
(b) a Red Blood Cell is kept in a concentrated saline solution.
(c) the plasma membrane of a cell breaks down.
(d) Rhoeo leaves are boiled in water first and then a drop of sugar syrup is put on them.
(e) the Golgi apparatus is removed from the cell.
Ans. (a) First, it swells due to endosmosis and then exosmosis occurs when it is put in to sugar solution.
(b) It will lose water and shrink.
(c) The cell will die.
(d) The cells are killed on boiling, so plasmolysis does not occur when the sugar syrup solution is put on it.
(e) Vesicle formation stops on the removal of the Golgi apparatus.
10. Draw a neat diagram of plant cell and label any three parts which differentiate it from an animal cell.
Ans.
Competency Based Questions
Multiple Choice Questions (Choose the most appropriate option/options)
1. Amoeba obtains its food by endocytosis. Which of the following properties of the plasma membrane helps during the above process?
(a) Plasma membrane is selectively permeable.
(b) Plasma membrane is flexible in nature.
(c) Plasma membrane allows diffusion of substances.
(d) Plasma membrane is composed of lipids, proteins and carbohydrates.
2. The following are some observations about a cell in a living organism under the microscope. On the basis of the features, identify the kind of cell.
Description: The nucleus is clearly visible in one side of the cell. The vacuole is large and occupies most of the space. The cell wall is clearly visible. Centrioles are invisible.
(a) It is a fungal cell.
(c) It is a plant cell.
(b) It is a bacterial cell.
(d) It is a cell of a human being.
3. Which functions are carried out by all unicellular organisms?
(a) growth homeostasis photosynthesis
(b) growth homeostasis metabolism
(c) metabolism photosynthesis reproduction
(d) growth nutrition reproduction
4. Look at the Venn diagram. Choose the correct option about P.
The organelle is enclosed by a double membrane and its has own DNA.
The organelle is found in both plant cells and animal cells. P
(a) It is the smallest organelle involved in protein synthesis.
(b) It helps in lipid metabolism along with transportation of substances.
(c) It is semiautonomous in nature and acts as the power house of the cell.
(d) It is involved in the secretion of substances and acts as scavenger of the cell.
Fig. 5.14
Big vacuole
Chloroplast
Cell wall
5. Select the incorrect statement regarding the vacuole.
(a) It provides turgidity and rigidity to the cell.
(b) It acts as the storage sac containing amino acids, sugars, various organic acids and some proteins of the cell.
(c) It gets rid of harmful toxins and waste products.
(d) It generates ATP for various cellular activities in case of an emergency.
6. Look at the Fig. 5.15. What is the incorrect function of part B?
(a) It is the site of lipid and protein synthesis.
(b) It is the site where drugs and poisons are detoxified.
(c) It is semiautonomous in nature and acts as the power house of the cell.
(d) It is involved in the secretion of substances and acts as scavenger of the cell.
7. Choose the correct statement.
(a) Cytoplasm – Nucleus = Protoplasm
(c) Protoplasm + Cytoplasm = Nucleus
(b) Nucleus – cytoplasm = Protoplasm
(d) Protoplasm – Nucleus = Cytoplasm
8. Look at the Fig. 5.16. Identify the kind of solution in A, B and C.
(a) A. Hypertonic solution, B. Hypotonic solution, C. Isotonic solution
(b) A. Hypertonic solution, B. Isotonic solution, C. Hypotonic solution
(c) A. Hypotonic solution, B. Hypertonic solution, C. Isotonic solution
(d) A. Hypotonic solution, B. Isotonic solution, C. Hypertonic solution
9. Look at the Fig. 5.17. It shows how food is captured by the Amoeba. Nucleus
Food particle
Plasma membrane
Fig. 5.17 Food capturing process in Amoeba
Which of the following statements about Amoeba states that it is a eukaryote?
(a) It is a unicellular organism having a single cell.
(b) It needs food to get energy.
(c) The single cell controls all the activities of the body.
(d) It has a membrane-bound nucleus.
Food vacuole
Swollen red blood cell
Normal red blood cell
Shrunken (crenated) red blood cell
Fig. 5.16
B A
Fig. 5.15
10. What distinguishes prokaryotic cells from eukaryotic cells?
Prokaryotic cells Eukaryotic cells
(a) no plasma membrane plasma membrane
(b) ribosomes DNA
(c) Golgi apparatus mitochondria
(d) no internal membrane compartments internal membrane compartments
11. Which of the following statements does not pertain to the ER?
(a) It serves as a conduit for protein transportation between the nucleus and cytoplasm.
(b) It facilitates the movement of materials within the cytoplasm.
(c) It may function as a site for certain cellular biochemical processes.
(d) It can serve as the site for energy production and mechanical support to the cell.
12. An unknown cell is observed using a microscope. A cell wall, ribosomes and DNA are identified. What can be concluded from these observations?
(a) It can only be a prokaryotic cell.
(b) It can only be a eukaryotic cell.
(c) It could be a prokaryotic or eukaryotic cell.
(d) It can only be a plant cell.
13. Where are proteins synthesised by free ribosomes used?
(a) Outside the cell after secretion
(c) Within the lysosomes
14. Look at the diagram of a mitochondrion.
Analyse the following statements and choose the correct option.
(a) Cristae are present in the outer membrane.
(b) The inner membrane is longer than the outer membrane.
(c) The outer membrane is folded to increase the surface area.
(d) Matrix is the main site of ATP formation.
(b) Within the nucleus
(d) Within the cytoplasm
15. Most of the proteins are synthesised near the endoplasmic reticulum. Where are these proteins transported further in the cell?
(a) To the mitochondria
(c) To the nucleus
(b) To the Golgi body
(d) To the cell membrane
16. Which of the following is the role of P in a cell in the given Fig. 5.19?
(a) It helps the cell to divide into two daughter cells.
(b) It helps the cell to enlarge for the division of the cell.
(c) It helps separate sister chromatids.
(d) It organises microtubules and helps join chromosomes together.
17. Mita has a bacterial infection. Which of the following organelles is responsible to remove the infection from her body?
(a) Vacuoles can expel the foreign substance.
(b) Lysosomes break down the material as they have digestive enzymes.
(c) Golgi bodies help in destroying the foreign bodies.
(d) Rough endoplasmic reticulum removes the unwanted particles through the plasma membrane.
Fig. 5.19
Fig. 5.18
18. During summer, when the soil becomes dry, leaves of a potted plant droops. Which of the following cell organelles is involved in making the leaves to droop?
(a) Nucleus, as it is the control centre of the cell. (b) Vacuole, as it loses water content.
(c) Cell wall, as it starts to shrink. (d) Lysosome as it releases digestive enzymes.
19. To carry out cellular processes, a large amount of energy is needed. Which part of the mitochondrion helps to generate maximum energy required for the various chemical activities?
(a) The folds present in the outer membrane increase the surface area for more ATP production.
(b) The folds present in the inner membrane increase the surface area for more ATP production.
(c) The folds present in the outer membrane decrease the surface area for more ATP production.
(d) The folds present in the inner membrane decrease the surface area for more ATP production.
20. What is a function of the plant cell wall?
(a) Formation of vesicles for transport of large molecules
(b) Prevention of excessive water uptake
(c) Communication with other cells by means of glycoproteins
(d) Active transport of ions
21. Choose the name of the scientist who coined the term ‘protoplasm’ in a cell.
(a) Purkinje
(c) Virchow
22. Select the incorrect statement regarding mitosis.
(a) It helps in growth and repairing of tissues
(b) It helps to replace old, dead and injured cells
(c) The mother cell forms two identical daughter cells
(d) It occurs in two stages
Assertion-Reason Based Questions
(b) Robert Hooke
(d) Robert Brown
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): Golgi bodies help in storing, modifying and packaging.
Reason (R): They are involved in the formation of lysosomes.
2. Assertion (A): All unicellular organisms take in their food by the process of endocytosis.
Reason (R): The taking in of matter by a living cell by invagination of its membrane to form a vacuole is called endocytosis.
3. Assertion (A): Chromosomes are thread-like structures found in the nucleus.
Reason (R): They are composed of lipids, proteins and DNA.
4. Assertion (A): Prokaryotes lack cytoplasmic organelles and a true nucleus.
Reason (R): They are primitive, single-celled small organisms.
5. Assertion (A): Osmosis is a special type of diffusion.
Reason (R): Osmosis involves the movement of water molecules from a region of a high concentration to one of a low concentration through a semipermeable membrane.
6. Assertion (A): Mitochondria are able to make their own proteins. Reason (R): They are the power houses of the cells.
Case-Based/Source-Based/Passage-Based Questions
1. Read the paragraph and answer the following questions.
Cell membranes are composed primarily of fatty-acid-based lipids and proteins. Membrane lipids are principally of two types—phospholipids and sterols. Both types share the defining characteristic of lipids. They dissolve readily in organic solvents but in addition they both have a region that is attracted to and soluble in water. This ʻamphiphilicʼ property is basic to the role of lipids as building blocks of cellular membranes. Membrane proteins are also of two general types. One type, called the extrinsic proteins, is loosely attached by ionic bonds or calcium bridges to the electrically charged phosphoryl surface of the bilayer. They can also attach to the second type of protein, called the intrinsic proteins. The intrinsic proteins, as their name implies, are firmly embedded within the phospholipid bilayer.
(a) What are sterols? Is cholesterol the same as sterols?
(b) Differentiate between extrinsic and intrinsic proteins.
(c) Explain the importance of the plasma membrane on the basis of flexibility.
(d) What do you mean by the amphiphilic property of lipids? What is its basic role in context of the plasma membrane?
2. Look at the Fig. 5.20 and read the following paragraph. Answer the following questions.
Inner membrane
Intermembrane space
(b) Which of the following contains chlorophyll?
A chloroplast is a type of plastid (a sac-like organelle with a double membrane) that serves as the site of photosynthesis, the process by which energy from the Sun is converted into chemical energy for growth. Chloroplasts contain the pigment chlorophyll to absorb light energy. The flat structures called thylakoids are arranged in groups called grana. Stroma is the semi-fluid substance that fills the interior of the chloroplast and contains water, dissolved ions, enzymes and other molecules.
(a) Chloroplast is found in (i) fungal cells. (ii) plant cells. (iii) bacterial cells. (iv) plant and algal cells.
(i) The inner membrane of the chloroplast (ii) Thylakoids (iii) Stroma (iv) Grana and stroma lamella
(c) Differentiate between grana and stroma.
(d) Write the role of stroma in the process of photosynthesis.
3. Look at the diagram of a kind of cell division (Fig. 5.21) and answer the following questions.
(a) What is the kind of division shown in the picture?
(b) Define the kind of cell division and explain the result of this division.
(c) What are the merits of this kind of division?
(d) What would happen if division of this kind did not occur in the cell?
Fig. 5.20 Structure of a chloroplast
Fig. 5.21
Answers
Multiple Choice Questions
Assertion-Reason Based Questions
Case-Based/Source-Based/Passage-Based Questions
1. (a) Sterols are a type of lipids found in the membrane of eukaryotic cells. No, cholesterol is a type of sterol found in animal cells. All sterols are not cholesterol.
(b) Extrinsic proteins are located on the surface of the lipid bilayer. They are not embedded in the membrane but are typically bound to the membrane.
Intrinsic proteins are embedded within the lipid bilayer and span the membrane (either partially or completely).
(c) The flexibility of the plasma membrane is important because for reasons such as cell movement, endocytosis and transportation of molecules.
(d) The amphiphilic property of lipids refers to the characteristic of having two distinct regions. One part of the lipid molecule is hydrophilic (water loving), while the other part is hydrophobic (water hating). The amphiphilic property of lipids is crucial in the formation and function of the plasma membrane.
2. (a) (iv) (b) (ii)
(c) Grana are stacks of flattened membranes called thylakoids, while stroma is the fluid that fills the chloroplast's inner space.
(d) Stroma contains several enzymes that are needed for the process of photosynthesis.
3. (a) The picture shows the mitotic division of the cell.
(b) Mitosis is the division of the nucleus of a cell into two nuclei, each with the same number of chromosomes as the original cell. It ensures that the two new cells (daughter cells) are identical to the original (parent) cell.
(c) Merits:
• Growth: It allows organisms to grow by producing more cells.
• Repair: It helps in replacing damaged or dead cells.
• Asexual reproduction: Some organisms reproduce asexually through mitosis, producing offspring identical to the parent.
(d) Without mitotic division, organisms would not be able to grow, repair damaged tissue or reproduce and life would cease to exist.
Practice Questions
Multiple Choice Questions
1. Which pair of cells keeps changing their shapes?
(a) Sperm cells and ovum
(c) Fat and bone cells
(b) Amoeba and WBCs
(d) RBCS and WBCs
2. Which of the following is covered by a single membrane?
(a) Mitochondria
(c) Lysosome
(b) Nucleus
(d) Plastid
3. Which cell organelle plays the role of a scavenger in a cell?
(a) The Golgi apparatus (b) Lysosomes
(c) The smooth endoplasmic reticulum (d) Vacuoles
4. Silver nitrate solution is used to study
(a) the endoplasmic reticulum. (b) the Golgi apparatus. (c) the nucleus. (d) mitochondria.
5. Select the incorrect statement about the vacuole.
(a) It provides turgidity and rigidity to the cell.
(b) It acts as the stored sac containing amino acids, sugars, various organic acids and some proteins of the cell.
(c) It gets rid of harmful toxins and waste products.
(d) It generates ATP for various cellular activities in case of an emergency.
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false.
(d) A is false but R is true.
6. Assertion (A): Mitochondria are called the power houses of the cell.
Reason (R): Mitochondria their own DNA and ribosomes.
7. Assertion (A): Gametes are formed by meiotic cell division.
Reason (R): Mitotic cell division is absent in sexually reproducing organisms.
Very Short Answer Questions (30–50 word)
8. (a) State two important functions of the nucleus.
(b) Name the organelle which is called as the protein factories of the cell.
9. (a) Name the animal cell that lacks lysosomes.
(b) Name the membrane that surrounds the vacuole.
10. State the major function of the cell wall in plants.
11. Differentiate between mitosis and meiosis.
12. State two functions of the vacuole.
Short Answer Questions (50–80 word)
13. How do the cells of fungi withstand very dilute external media without bursting?
14. Differentiate between the cell wall and plasma membrane.
15. Sate three functions of Golgi apparatus.
16. State the reason for the following.
(a) Chloroplasts are able to make their own protein.
(b) The nucleus is the control room of the cell.
17. Compare chloroplasts, chromoplasts and leucoplasts.
Long Answer Questions (80–120 word)
18. Answer the following questions.
(a) Differentiate between mitosis and meiosis.
(b) Design an activity to show endosmosis and exosmosis.
19. Explain about a eukaryotic nucleus with the help of a well-labelled diagram.
20. How are the following related to each other?
(a) Chromatids and chromosomes
(b) Chloroplast and chlorophyll
Brain Charge
1. Solve the crossword puzzle below.
Across Down
2. Synthesises protein
6. Performs photosynthesis
9. Provides storage
10. Controls the cell
11. Processes and packages micromolecules
1. Transports things around the cell
3. ATP production
4. Provides structure and support
5. Filters what comes into the cell
7. Contains digestive enzymes
8. Holds the organelles and inclusions
2. Fill in the gaps in the following table.
Smooth Endoplasmic Reticulum (SER)
Rough Endoplasmic Reticulum (RER) Ribosomes are _______. Ribosomes are _______. It is involved in _________ synthesis. It is involved in protein synthesis. It is present near the __________. It is present near the nucleus. It may develop from the RER. It may develop from _________.
3. Complete the following flow chart.
Cytoplasmic Organelles
ER
Mitochondria
Golgi complex
Lysosome
Plastids
Non-cytoplasmic Inclusions
Challenge Yourself
1. The image shows a stage of cell division. On the basis of the image, what can be the function of the centrosome in the cell division?
(a) It releases spindle fibres and holds the chromosomes during cell division.
(b) It organises spindle fibres and allows equal distribution of chromosomes in the daughter cells.
Crystals and oildrops
Other inclusions
Spindle fibres
(c) It forms spindle fibres before cell division starts.
(d) It degrades the chromosomes and spindle fibres when division is completed.
5.22 A stage of cell division
2. A solution of 3% glucose and a solution of 8% glucose are kept in a trough separated by a semipermeable membrane. What will you observe after 1 hour? Justify your answer.
3. If the cells of a Rhoeo plant and RBC are separately kept in a hypotonic solution, what will happen to each of them? Explain the reason for your answer.
(a) Both the cells will swell.
(b) The RBC will burst easily while cells of Rhoeo will resist the bursting to some extent.
(c) Only the RBC will swell and the Rhoeo cells will show no change.
(d) The RBC and onion peel cells will behave similarly.
Scan me
Chromosomes
Centrosome
Fig.
4. Match Column I with Column II. Select the correct option.
Column I
i. Lysosome
ii. Cell wall
iii. Ribosome
iv. Nucleus
(a) i-p,ii-o,iii-n,iv-m
(c) i-p,ii-q,iii-o,iv-n
Column II
m. Robert Hooke
n. Robert Brown
o. Palade
p. C. de Duve
(b) i-q,ii-o,iii-n,iv-p
(d) i-n,ii-m,iii-p,iv-q
5. Look at the diagram. Which of the following is not the function of the organelle shown in the diagram?
2. SER – Ribosonmes are absent. It is involved in lipid synthesis. It is located near the periphery. RER – Ribosonmes are present. It is involved in protein synthesis. It is located near the nucleus. It may develop from the nucleolus.
Challenge Yourself
1. (b) The centrosome, a major microtubule-organising center (MTOC), plays a crucial role in cell division by organising microtubules to form the mitotic spindle.
2. After 1 hour, the solutions on both the sides of the semipermeable membrane will become isotonic because of the process of osmosis.
3. (b) When kept in a hypotonic solution, the onion cells will become turgid because the water will enter the cell due to osmosis. But the cell wall present outside the cell provides rigidity and does not let any harm occur to the cell. Whereas, in the RBC the movement of water into the cell due to osmosis will lead to it bursting because of the absence of a rigid cell wall. In this scenario, the RBC will burst easily while onion peel cells will resist bursting to some extent.
4. (c) Christian de Duve, a Belgian cytologist and biochemist, discovered the lysosome.
5. (b) The picture is showing the vacuole of a plant cell.
Scan me for Exemplar Solutions
Tonoplast
Cell sap
Fig. 5.23
SELF-ASSESSMENT
Time: 1.5 Hour
Scan me for Solutions
Max. Marks: 50
Multiple Choice Questions (10 × 1 = 10 Marks)
1. Which cell organelle plays the role of packaging and distribution in a cell?
(a) The Golgi apparatus (b) Lysosomes
(c) The smooth endoplasmic reticulum (d) Vacuoles
2. Which of the following is a cell inclusion?
(a) Lysosomes (b) Centrioles (c) Lipid droplets (d) All of the options
3. Which of these options are correct for the function of ribosomes?
(I) They help in the manufacture of protein molecules.
(II) They help in the manufacture of enzymes.
(III) They help in the manufacture of hormones.
(IV) They help in the manufacture of starch molecules.
(a) (I) and (II) (b) (II) and (III) (c) (III) and (iv) (d) (IV) and (I)
4. Select the false statement regarding the vacuole.
(a) It provides turgidity and rigidity to the cell.
(b) It acts as the stored sac containing amino acids, sugars, various organic acids and some proteins of the cell.
(c) It gets rid of harmful toxins and waste products.
(d) It generates ATP for various cellular activities in case of an emergency.
5. Which of these options are incorrect about lysosomes?
(I) They contain about 50 degradative enzymes.
(II) They are known as the waste disposal system of cells.
(III) They are known as the ribosome factory of the cell.
(IV) They are involved in cellular transportation.
(a) (I) and (II) (b) (II) and (III) (c) (III) and (IV) (d) (iv) and (I)
6. A diagram of an animal cell is given in Fig. 5.24. Which part acts as the suicide bag of the cell? Justify your answer.
(a) A (b) B
(c) E (d) C
7. Which of these options are correct?
(I) Small granular shaped mitochondria are found in bacterial cells.
(II) Mitochondria are double-membranous organelles.
(III) ATP is synthesised in mitochondria.
(IV) Cellular transportation is one of the functions of mitochondria.
(a) (I) and (II) (b) (II) and (III)
(c) (III) and (IV) (d) (IV) and (I)
5.24
Fig.
8. The table lists functions performed by some plant cell organelles.
Organelles
Functions
Q Allows transportation of materials in and out of the cell
R Stores products of the ER in the vesicle
S Acts as scavengers of the cell
T Helps in production of food in the cell
Which of the options shows the correct labelling of organelles in the plant cell based on the given tabulated functions?
9. Look at the Fig. 5.25. Analyse the statements and choose the correct option.
(a) Cristae are present in the outer membrane.
(b) The inner membrane is longer than the outer membrane.
(c) The outer membrane is folded to increase the surface area.
(d) Matrix is the main site of ATP formation.
5.25
10. Living cells were discovered by (a) Robert Hooke (b) Purkinje (c) Leeuwenhoek (d) Robert Brown
Assertion-Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): The cell wall is the dead part of the cell.
Reason (R): It is semipermeable in nature as found in the plasma membrane.
12. Assertion (A): The endoplasmic reticulum acts as an intracellular transport system of the cell.
Reason (R): It transports proteins and other materials within the cell.
13. Assertion (A): Protein synthesis occurs in both eukaryotes and prokaryotes.
Reason (R): Ribosomes are found in both kinds of cells.
14. Assertion (A): There is no energy requirement during passive transport.
Reason (R): Passive transport occurs due to the concentration gradient of materials.
Fig.
Case-Based/Source-Based/Passage-Based Questions
15. Read the following paragraph and answer the following questions.
(2 × 4 = 8 Marks)
The nucleus has a membrane around it that keeps all the chromosomes inside and makes the distinction between keeping the chromosomes inside the nucleus and the other organelles and components of the cell outside. There are pores in this nuclear membrane too.
(a) Look at the Fig. 5.26. The nuclear pores are (i) T and S
(iii) Q
(ii) S only
(iv) R and S
(b) Which of the following is/are called vehicles of heredity?
(i) R and P
(iii) R only
(ii) R and T
(iv) T only
(c) What is the role of "S" in the diagram?
(d) What is T and what is its role?
16. Read the following paragraph and answer the following questions.
The cytoplasm is a thick solution that fills each cell and is enclosed by the cell membrane. It is mainly composed of water salts and proteins. In eukaryotic cells, the cytoplasm includes all of the material inside the cell and outside of the nucleus. All of the organelles in eukaryotic cells, such as the nucleus, endoplasmic reticulum, and mitochondria, are located in the cytoplasm. The portion of the cytoplasm that is not contained in the organelles is called the cytosol. Although cytoplasm may appear to have no form or structure, it is actually highly organised. A framework of protein scaffolds called the cytoskeleton provides the cytoplasm and the cell with their structure.
(a) How you define cytosol?
(b) How is the cytoplasm of prokaryotes different from the cytoplasm of eukaryotes?
(c) Is the nucleoplasm a part of the cytoplasm? Justify.
(d) How is the protoplasm different from the cytoplasm?
Very Short Answer Questions (30-50 words)
17. State important functions of Golgi apparatus.
18. Where will you find more ribosomes—in cancer cells or in fat cells? Give reason.
19. Define plasmolysis. Give an example of plasmolysis you observe in your daily life.
Short Answer Questions (50-80 words)
(3 × 2 = 6 Marks)
(4 × 3 = 12 Marks)
20. How will the absence of any one of the cell organelles affect the overall function of the cell?
21. Differentiate between a plant cell and an animal cell.
22. What is endocytosis? Why does endocytosis occur in animal cells?
23. (a) All multicellular organisms come from a single cell. Justify.
(b) What is the intracellular source of digestive enzymes?
Long Answer Questions (80-120 words)
(2 × 5 = 10 Marks)
24. Explain the structure and function of mitochondria with the help of a labelled diagram.
25. Answer the following questions.
(a) What is active transport?
(b) Write three differences between active and passive transport.
Fig. 5.26
6 The Tissues
This chapter discusses the various types of plant and animal tissues, and their structure and functions. A group of cells that are similar in structure and/or work together to achieve a particular function forms a tissue. The students will be able to understand how different organs and organ systems, in plant and animal bodies, are made up of different tissues and how these tissues perform together in a coordinated manner.
Tissues
A group of cells that are similar in structure and work together to achieve a particular function forms a tissue.
Meristematic Tissues (Dividing tissues - Cells of meristematic tissue are very active, they have dense cytoplasm, thin cellulose walls, prominent nuclei and they lack vacuoles.)
Apical Meristem (Growing tips of stem and root)
Intercalary Meristem (Nodal part of the plant)
Lateral Meristem (Lateral part of the plant)
Plant
Tissues
The tissues have a large quantity of supportive tissues having dead cells
Permanent Tissues (Non-dividing tissuesThey have lost their ability to divide and are specialised for specific functions.)
Simple Permanent Tissues (Made up of one type of cells)
Parenchyma (Most common living cells with intercellular spaces, and generally stores food.)
Collenchyma (The cells of this tissue are living, elongated and irregularly thickened at the corners.)
Sclerenchyma (They are dead cells and are long and narrow as the walls are thickened.)
Complex Permanent Tissues (Conducting tissues, made up of more than one type of cells)
(On the basis of the functions they perform there are different types of animal tissues, such as epithelial tissue, connective tissue, muscular tissues and nervous tissues.)
Epithelial Tissues
(Protective, simple)
Squamous
Simple flat kind of epithelium forms a delicate lining of skin. They are two types, simple squamous and stratified squamous.
Ciliated Cells have cilia, hair-like projections found on the outer surface of epithelial cells.
Cuboidal Cube shaped cells have rounded nuclei, providing mechanical support.
Glandular
The cells have gland cells to secrete substances.
Columnar Tall, pillar-like cells have elongated nuclei.
Muscular Tissues (Elongated, help in body movement)
Striated having striations and voluntary in function.
Smooth Unstriated and are involuntary in functions.
Cardiac having striations but involuntary in function.
Connective Tissues (Connect and bind different tissues)
Areolar (Loose connective tissue)
Adipose (Made up of Adipocytes that store fats)
Fibrous connective (Supports and protects bones and muscles)
Skeletal (Strong and flexible)
Fluid connective (Helps in transportation of materials)
Nervous Tissues (Made up of nerve cells)
Lymph (Strawcoloured fluid with inorganic salts)
Plasma (Fluid part in the blood in which blood cells are suspended.)
Red Blood cells (Erythrocytes)
Chapter at a Glance
Tendons (Connect muscles to bones)
Ligaments (Connect bones to bones)
Cartilage (Semi-rigid, flexible)
Bones (Strong, not flexible, hard)
Blood (Alkaline fluid that transports gases and nutrients)
Blood cells (Three types of blood cells are found in the plasma.)
White Blood cells (Leukocytes)
Blood platelets (Thrombocytes)
• A group of cells that are similar in structure and/or work together to achieve a particular function forms a tissue.
Plant Tissues
• Based on the dividing capacity of the tissues, various plant tissues can be classified as growing or meristematic tissue and permanent tissue.
• There is no such demarcation of dividing and non-dividing regions in animals.
• Meristematic tissue is the dividing tissue present in the growing regions of plants.
• Depending on the location of meristematic tissues, they are classified as apical, intercalary and lateral.
• Apical meristem is present at the growing tips of stems and roots. It increases the length of the stem and the root.
• Intercalary meristem is found near the nodal region of the plant and helps in the growth and elongation of the plant.
• Lateral meristem helps in increasing the diameter or girth of the stem and root.
• Meristematic tissues are active, with dense cytoplasm and prominent nuclei with thin cellulose walls. They lack vacuoles.
• Permanent tissues are derived from meristematic tissue once they lose the ability to divide.
• Parenchyma, collenchyma and sclerenchyma are three types of simple tissues.
• Parenchyma is the most common simple permanent tissue. They are living cells and are usually loosely arranged, with large spaces between cells (intercellular spaces).
• When parenchyma contains chlorophyll and performs photosynthesis, it is called chlorenchyma.
• Similarly, when parenchyma contains large air cavities, they are called aerenchyma
• Collenchyma cells are living, elongated and irregularly thickened at the corners, and they provide mechanical support and flexibility to plants.
• Sclerenchymatous tissues are hard and their cells are dead. Sclereids and fibres are both types of sclerenchyma cells that provide structural support in plants.
• Epidermis is the outermost layer of cells and is usually made of a single layer of cells. It protects all the parts of the plant.
• Stomata are some small pores found in the epidermis of the leaf. These pores are enclosed by two kidney-shaped cells called guard cells. They are necessary for the exchange of gases with the atmosphere.
• Xylem and phloem are types of complex tissues. They are also called as conducting or vascular tissues.
• Xylem conducts water and minerals; consists of tracheids, vessels, xylem fibres and xylem parenchyma.
• Xylem components are mostly dead, except xylem parenchyma.
• Phloem conducts food; consists of sieve tubes, companion cells, phloem fibres and phloem parenchyma.
• Phloem components are mostly living, except phloem fibres.
Animal Tissues
• Animal tissues are classified into four types: Epithelial, Connective, Muscular and Nervous.
• Depending on the shape and function, epithelial tissue is classified as squamous, cuboidal, columnar, ciliated and glandular.
o Squamous epithelium: Thin and flat; found in blood vessels and alveoli.
o Cuboidal epithelium: Cube-shaped; found in kidney tubules.
Fig. 6.2 Sclerenchyma Thick lignified walls
(a)
(b)
Narrow lumen Lignified thick wall
Wall thickenings Nucleus
Cell wall
Vacuole
Fig. 6.1 Collenchyma
o Columnar epithelium: Tall cells; lines intestines.
o Ciliated epithelium: Has hair-like cilia; found in respiratory tract.
o Glandular epithelium: Secretes substances like mucous and enzymes.
• The different types of connective tissues in our body include areolar tissue, adipose tissue, skeletal (bone, cartilage), fibrous connective tissues (tendon, ligament) and fluid connective (blood).
o Bone: Rigid and strong; supports the body.
o Cartilage: Flexible; found in nose, ear and joints.
o Ligaments: Connects bones to bones; elastic.
o Tendons: Connects muscles to bones; tough and non-elastic.
o Blood: Fluid tissue; transports nutrients, gases and waste.
o Adipose: Stores fat and insulates the body.
• Striated, unstriated and cardiac are three types of muscle tissues.
o Striated (Skeletal): Voluntary, striated and multinucleated: found in limbs, neck and face
o Smooth: Involuntary, non-striated and uninucleated: walls of the hollow organs and blood vessels throughout the body
o Cardiac: Involuntary, striated and branched; found in the heart
SKELETAL MUSCLE
Usually attached to skeleton
MUSCLE
Usually covering wall of internal organs
Usually covering wall of the heart
• Nervous tissue is made of neurons that receive and conduct impulses.
• The cell body (cyton), dendrites and axon are parts of a neuron.
NCERT Zone
Intext Questions
1. What is a tissue?
Ans. A tissue is a group of similar cells that work together to perform a specific function. In multicellular organisms, cells do not work independently; instead, they are organised into tissues that carry out specialised tasks. For example:
• Plant tissues include xylem (conducts water) and phloem (transports food).
• Animal tissues include muscle tissue (helps in movement) and nervous tissue (transmits signals).
2. What is the utility of tissues in multi-cellular organisms?
Ans. In multicellular organisms, tissues play a crucial role in ensuring proper functioning and efficiency. Their utility can be understood as follows:
• Division of labour – Different tissues perform specific functions, allowing specialisation and efficiency. For example, muscle tissue enables movement, while nervous tissue transmits signals.
• Structural support – Some tissues, such as connective tissues (bones, cartilage), provide structural strength and support to the body.
• Coordination and communication – Nervous tissue helps in transmitting signals and coordinating body functions.
SMOOTH
CARDIAC MUSCLE
Fig. 6.3
• Transport of substances – Vascular tissues like xylem and phloem in plants help in the transport of water, minerals and food. Similarly, blood in animals transports oxygen, nutrients and waste products.
• Protection – Epithelial tissue acts as a barrier, protecting internal organs from damage, infections and dehydration.
• Energy storage and repair – Some tissues, like adipose tissue, store energy, while others, like areolar connective tissue, help in repair and healing.
Ans. The apical meristem is found at the tips of roots and shoots in plants.
• In the root tip, it helps in root elongation, enabling the plant to grow deeper into the soil.
• In the shoot tip, it facilitates the growth of stems and branches, increasing the height of the plant.
5. Which tissue makes up the husk of coconut?
Ans. The husk of a coconut is made up of sclerenchyma tissue.
• It consists of thick-walled, dead cells that provide strength and rigidity.
• The fibres in the husk are called sclerenchymatous fibres, which make the coconut husk tough and protective.
6. What are the constituents of phloem?
Ans. Phloem is made up of four types of elements:
• sieve tubes, • companion cells
• phloem fibres and • phloem parenchyma.
7. Name the tissue responsible for movement in our body.
Ans. 1. Muscular tissue; 2. Nervous tissue
The combination of both tissues is responsible for movement in our body.
8. What does a neuron look like?
Ans. A neuron consists of a cell body with a nucleus and cytoplasm, from which long, thin hair-like parts arise. Each neuron has a single, long part called the axon, and many small, short-branched parts called dendrite. An individual nerve cell is called neuron; it may be up to a metre long.
Axon
Myelin
Node of Ranvier Nucleus
Cell body
Dendrites
Synaptic knob
Axon terminals
Fig. 6.4 Structure of a neuron Shutter id is provided to keep the image
9. Give three features of cardiac muscles.
Ans. Features of cardiac muscles:
(a) Heart muscles (cardiac muscles) are cylindrical, branched and uninucleated.
(b) They are striated muscle fibres.
(c) They are involuntary muscles and cannot be controlled by us.
10. What are the functions of areolar tissue?
Ans. Areolar tissues are connective tissues found in animals. It is found between skin and muscles, around blood vessels and nerves and in the bone marrow. It fills the space inside the organs, supports internal organs and helps in the repair of tissues.
NCERT Exercises
1. Define the term “tissue”.
Ans. A group of cells that are similar in structure and perform the same function is called a tissue.
2. How many types of elements together make up the xylem tissue? Name them.
Ans. The xylem is made up of tracheids, vessels, xylem fibres and xylem parenchyma.
3. How are simple tissues different from complex tissues in plants?
Ans. Differences between simple and complex tissues in plants:
Features
Simple Tissues
Complex Tissues
Definition Made up of only one type of cells. Made up of more than one type of cells working together.
Cell Nature
Occurrence
Cells are similar in structure and function.
Found in all parts of the plant, mainly in soft tissues.
Function Provide support, storage, and photosynthesis.
Types
Parenchyma, collenchyma, sclerenchyma.
Cells have different structures and functions.
Found in vascular bundles, helping in conduction of water and food.
Help in the transport of water, minerals and food.
Xylem (water transport) and Phloem (food transport).
4. Differentiate between parenchyma, collenchyma and sclerenchyma on the basis of their cell wall.
Ans.
Parenchyma
The cells have thin cell walls made up of cellulose.
Collenchyma
Sclerenchyma
The cells have cell walls thickened at the corners due to pectin deposition. Their walls are thickened due to lignin deposition.
5. What are the functions of the stomata?
Ans. Stomata are the tiny pores found on the surface of plant leaves, stems and other green parts of the plant. They help in transpiration and exchange of gases.
6. Diagrammatically show the difference between the three types of muscle fibres.
Ans.
Striated Muscles
They are connected to bones (skeletal muscles).
Smooth Muscles
Cardiac Muscles
They are found in the alimentary canal and lungs. They are found in the heart.
They are voluntary muscles. They are involuntary muscles. They are involuntary in action.
Striated Muscles
Smooth Muscles
Cardiac Muscles
The cells are long, cylindrical with many nuclei and are unbranched. They are spindle-shaped and have a single nucleus. They are branched and have one nucleus.
• They are involuntary muscles.
Ans.
7. What is the specific function of the cardiac muscle?
Ans. • Cardiac muscle cells are cylindrical, branched and uninucleated.
• They show rhythmic contraction and relaxation throughout life.
Muscles
Structure and shape
Striated Cells are long, cylindrical, non-tapering and are unbranched.
Unstriated Cells are long with tapering ends and are unbranched.
Cardiac Cells are non-tapering and cylindrical in shape and are branched.
9. Draw a labelled diagram of a neuron.
Ans. Nucleus Dendrites Scwann cell
hillock (trigger area)
10. Name the following.
In hands, legs and skeletal muscles.
• Their rhythmic contraction and relaxation help in the pumping action of the heart.
nerve endings
The wall of stomach, intestine, ureter, bronchi, etc.
Present
Absent
8. Differentiate between striated, unstriated and cardiac muscles on the basis of their structure and site/location in the body.
Site/Location Light and dark bands
In the heart. Present but less prominent.
(a) Tissue that forms the inner lining of our mouth. (b) Tissue that connects muscle to bone in humans.
(c) Tissue that transports food in plants. (d) Tissue that stores fat in our body.
(e) Connective tissue with a fluid matrix. (f) Tissue present in the brain.
Ans. (a) Tissue that forms the inner lining of our mouth: Squamous epithelium
(b) Tissue that connects muscle to bone in humans: Tendons
(c) Tissue that transports food in plants: Phloem
(d) Tissue that stores fat in our body: Adipose tissue
(e) Connective tissue with a fluid matrix: Blood
(f) Tissue present in the brain: Nervous tissue
Fig. 6.5 Striated muscles
Fig. 6.6 Smooth muscles
Fig. 6.7 Cardiac muscles
Axonal
Effector
Axon
Node of Ranvier
Myelin sheath
Soma
Fig. 6.8 Structure of a neuron
11. Identify the type of tissue in the following: skin, bark of tree, bone, lining of kidney tubule, vascular bundle.
Ans. (a) Skin—Striated squamous epithelium
(b) Bark of tree—Cork, protective tissue
(c) Bone—Connective tissue
(d) Lining of kidney tubule—Cuboidal epithelium tissue
(e) Vascular bundle—Conducting tissue
12. Name the regions in which parenchyma tissue is present.
Ans. Parenchyma tissue is found in the cortex, or outer layers, and pith, or innermost layers of stems and roots. It is found in the mesophyll, or internal layers, of leaves.
In the pith of the roots and stems, the parenchymatous tissue contains chlorophyll, and is called chlorenchyma, found in green leaves. In aquatic plants, parenchyma contains large air cavities, which help the plants to float. Such a type of parenchyma is called aerenchyma.
13. What is the role of epidermis in plants?
Ans. Cells of epidermis form a continuous layer without intercellular spaces. It protects all the parts of plants.
14. How does the cork act as a protective tissue?
Ans. Cork acts as a protective tissue because
• Its cells are dead and compactly arranged without intercellular spaces.
• They have deposition of suberin on the walls that make them impervious to gases and water.
• Cork cells are dead at maturity, meaning they lack living cytoplasm and organelles. This lack of living components contributes to their structural rigidity and impermeability.
15. Complete the following chart:
Ans.
Parenchyma
Permanen Tissue
Simple Complex
Collenchyma
Permanen Tissue
Simple Complex
Collenchyma Phloem Sclerenchyma
Xylem
Xylem
Multiple Choice Questions
1. Which meristem is responsible for the growth of plants in diameter?
(a) Both apical and lateral meristem (b) Apical meristem only
(c) Lateral meristem only
2. Find out the false sentence.
(a) Parenchymatous tissues have intercellular spaces.
(d) Both intercalary and lateral meristem
(b) Collenchymatous tissues are irregularly thickened at corners.
(c) Apical and intercalary meristems are permanent tissues.
(d) Meristematic tissues, in its early stage, lack vacuoles.
(a) (i) and (ii) (b) (ii) and (iii) (c) (iii) and (iv) (d) (i) and (iv)
5. Meristematic tissues in plants are (a) localised and permanent (b) not limited to certain regions (c) localised and dividing cells (d) growing in volume
6. Cartilage is not found in (a) nose (b) ear (c) kidney (d) larynx
7. Bone matrix is rich in (NCERT Exemplar) (a) fluoride and calcium (b) calcium and phosphorus (c) calcium and potassium (d) phosphorus and potassium
8. Contractile proteins are found in (a) bones (b) blood (c) muscles (d) cartilage
9. Voluntary muscles are found in (a) alimentary canal (b) limbs (c) iris of the eye (d) bronchi of lungs
10. What material comprises the husk of a coconut?
(a) Xylem and phloem both
(b) Phloem only (c) Sclerenchyma only (d) Collenchyma and parenchyma
11. Which cell type primarily protects against water loss? (a) Epidermis (b) Guard cell (c) Cutin (d) Cortex
12. Which of the following helps in repair of tissue and fills up the space inside the organ?
13. The muscular tissue which function throughout the life continuously without fatigue is (a) skeletal muscle (b) cardiac muscle (c) smooth muscle (d) voluntary muscle
14. Which of the following cells is found in the cartilaginous tissue of the body?
16. In desert plants, rate of water loss gets reduced due to the presence of (a) cuticle (b) stomata (c) lignin (d) suberin
17. Which of the following is NOT a type of plant tissue?
(a) Epidermis (b) Xylem (c) Stroma (d) Parenchyma
18. Which of the following is incorrect?
(a) Vessels–Xylem
(c) Companion cell–phloem
(b) Sieve cells–Xylem
(d) Sieve tubes–Phloem
19. Which epithelial tissue allows exchange of gases in the lungs?
(a) (b) (c)
(d)
20. Parenchyma cells are (NCERT Exemplar)
(a) relatively unspecified and thin walled
(b) thick walled and specialised (c) lignified (c) none of these
21. Cork cells are made impervious to water and gases by the presence of (NCERT Exemplar) (a) cellulose (b) lipids (c) suberin (d) lignin
22. Survival of plants in terrestrial environment has been made possible by the presence of (NCERT Exemplar)
(a) intercalary meristem
(c) apical meristem
(b) conducting tissue
(d) parenchymatous tissue
23. The water conducting tissue generally present in gymnosperm is (NCERT Exemplar)
(a) vessels
(c) tracheids
(b) sieve tube
(d) xylem fibres
24. Which of the following functions is not performed by the nervous tissue
(a) Sensory input
(c) Secretion of enzymes
(b) Muscle contraction
(d) Transmission of electrical impulses
25. Which structure of neuron is responsible for transfer of impulse to muscles?
(a) Cyton
(c) Nerve endings
(b) Nucleus
(d) Dendrites
26. Which should be the likely structure of a connective tissue whose framework supports the organs of the body?
(a) Multiple layers like the skin
(c) Strong and non-flexible like bones
(b) Relaxing and contracting like muscles
(d) Bundle of neurons like nerve
27. Look at the diagram. Identify the tissue type.
Nucleus
Cytoplasm
Basement membrane
Connective tissue
Fig. 6.9
(a) Stratified squamous epithelium
(c) Simple squamous epithelium
28. What is the function of meristematic tissue in plants?
(a) Photosynthesis and growth
(c) Water and nutrient transport
29. Find out the false sentence.
(a) Parenchymatous tissues have intercellular spaces.
(b) Simple cuboidal epithelium
(d) Stratified columnar epithelium
(b) Growth and repair of the body parts
(d) Storage of carbohydrates
(b) Collenchymatous tissues are irregularly thickened at corners.
(c) Intercalary meristems help in lateral growth of the plants.
(d) Meristematic tissues, in its early stage, lack vacuoles.
30. Find out the false statement regarding connective tissue.
(a) Cells are loosely packed.
(b) Cells are living.
(c) Cells help in the absorption of water and minerals.
(d) Cells connect various tissues and organs with each other.
Answers
Constructed Response Questions
Very Short Answer Questions (30-50
words)
1. (a) Which tissue in plants provides them flexibility?
(b) In desert plants, how does the rate of loss of water get reduced?
Ans. (a) Collenchyma
(b) The presence of cuticle on the surface of desert plants reduces the rate of loss of water.
2. What happens to the cells formed by meristematic tissue?
Ans. The cells formed by meristematic tissue take up a specific role and lose their ability to divide. As a result, they form a permanent tissue. This process of taking up a permanent shape, size and function is called differentiation.
3. Answer the questions.
(a) What is cutin?
(b) What is the composition of the cartilage matrix?
Ans. (a) Cutin is a chemical substance, with waterproof quality, covering the aerial parts of plants.
(b) Proteins and sugars
4. Why is the epidermis present as a thick waxy coating of cutin in desert plants?
Ans. A thick waxy coating of cutin is present in desert plants to prevent excessive loss of water during transpiration. Due to this, plants can survive despite the scarcity of water in deserts.
5. What do you mean by ‘phellogen’?
Ans. As plants grow older, the outer protective tissue undergoes certain changes. A layer of secondary meristem develops, which is called phellogen. It is also known as cork cambium. It replaces the epidermis of stem and roots.
6. Answer the questions.
(a) What minerals is the bone matrix rich in?
(b) Name the water conducting tissue generally present in gymnosperms.
(c) Which chemical in cork cells makes them impervious to water and gases?
(d) Which animal tissue helps in repair of tissue and fills the space inside the organ?
Ans. (a) Calcium and potassium
(c) Suberin
7. How are glandular epithelium formed?
(b) Tracheids
(d) Areolar tissue
Ans: An epithelial cell often acquires additional specialisation as gland cells, which can secrete substances at the epithelial surface. Sometimes, a portion of the epithelial tissue folds inward and a multicellular gland is formed. This is glandular epithelium.
8. Describe the function of bones.
Ans. Bones form the framework that supports the body. They also anchor the muscles and serve as storage sites of calcium and phosphate. They provide shape to the body and protect vital body organs such as brain, lungs, tissue, etc.
9. How are messages conveyed from one place to another within the body?
Ans. Nervous tissue is made up of neurons that receive and conduct impulses. Neurons are highly specialised for being stimulated and then transmitting the stimulus very rapidly from one place to another within the body. Impulses are the passage of electrical activity along the axon of a nerve cell.
10. Answer the questions.
(a) What is the lining of blood vessels made up of?
(b) What is the lining of small intestine made up of?
(c) What is the lining of kidney tubules made up of?
(d) Where are the epithelial cells with cilia found?
Ans. (a) Squamous epithelium
(c) Cuboidal epithelium
11. What is a goblet cell? What is the role of those cells?
(b) Columnar epithelium
(d) Respiratory tract
Ans. A goblet cell is a unicellular mucous-secreting gland. It protects the lining of the respiratory and gastrointestinal tract. It also plays a key role in defending against pathogens, allergens and toxins.
12. Differentiate between intercalated discs in cardiac muscles and striations in skeletal muscles.
Ans. Intercalated discs: Found in cardiac muscles, they allow synchronised contraction of the heart.
Striations: Found in skeletal muscles, they are alternating bands for voluntary movement.
13. How does the structure of areolar tissue make it multifunctional?
Ans. Areolar tissue contains loosely arranged fibres (collagen and elastin) and cells. They provide flexibility and support, and connect different tissues. They also serve as a packing material in organs.
14. How is cartilage different from bone in context to hardness and weight? What is the role of cartilage in the human body?
Ans. Bones are hard and rigid, while cartilage is soft, flexible and elastic. Cartilage is lighter than bones. Cartilage provides support and flexibility to body parts. It reduces friction between bones and acts as a cushion at the joints.
15. Look at the diagram. Write the characteristic features of this tissues.
Ans. The diagram shows skeletal muscle.
Features of skeletal muscle:
• They have a banded appearance.
• The cells have many nuclei.
• They are cylindrical with blunt ends.
• They are voluntary in nature.
16. What is differentiation?
6.10
Ans. The cells formed by meristematic tissue take up a specific role and lose their ability to divide. As a result, they form a permanent tissue. This process of taking up a permanent shape, size and function is called differentiation.
17. Name the elements of phloem that are living.
Ans. The living elements of phloem are:
1. Sieve tube elements
2. Companion cells
3. Phloem parenchyma (Note: These are living in most plants, though absent in monocots)
18. What are the functions of adipose tissues?
Ans. • They serve as a fat reservoir.
• They provide shape to the limbs and the body.
• They keep the visceral organs in position.
• They form shock-absorbing cushions around the kidneys and eyeballs.
19. Differentiate the following activities on the basis of voluntary (V) or involuntary (IV) muscles.
(a) Jumping of frog
(b) Pumping of the heart
(c) Writing with hand
(d) Movement of chocolate in your intestine
Ans. (a) V (b) IV (c) V (d) IV
Short Answer Questions (50-80 words)
1. Describe the structure and function of stomata.
Ans. Stomata are small pores present in the epidermis of leaves, which are enclosed by two kidney-shaped cells called guard cells.
Function of stomata:
• Necessary for exchanging gases with the atmosphere during photosynthesis and respiration.
• Transpiration, i.e., loss of water, takes place through them.
Fig.
2. Answer the questions.
(a) What is responsible for increase in girth of the stem or root?
(b) What is lignin?
(c) Which tissue forms a barrier to keep different body systems separate?
Ans. (a) The girth of the stem or root increases due to lateral meristem (cambium).
(b) Lignin is a chemical substance present in the cell wall of plants, which acts as a cement and hardens it.
(c) Epithelial tissue
3. How are xylem and phloem different from simple tissues? Differentiate between xylem and phloem. Ans. Xylem and phloem are called complex tissues as they are made up of more than one type of cells. Simple tissues are made up of one type of cells. Following are the differences between xylem and phloem:
Xylem
1. Xylem mainly consists of dead cells (except xylem parenchyma).
2. It conducts water and minerals from the roots to the aerial parts of the plant.
3. Xylem is generally located towards the inner side of plant parts.
4. Write a short note on xylem.
Phloem
1. Phloem mainly consists of living cells (except phloem fibre).
2. It translocates prepared food from the leaves to the storage organs and growing parts of the body.
3. Phloem is generally located towards the outer side of plant parts.
Ans. Xylem is a complex permanent tissue and is also known as conduction tissue. It consists of tracheids, vessels, xylem parenchyma and xylem fibres. The cells have thick walls, and many of them are dead. Tracheids and vessels are tubular structures. This allows them to transport water and mineral vertically upwards. The parenchyma stores food and helps in the lateral conduction of water. Fibres are mainly supportive in function.
5. Discuss about phloem tissue.
Ans. Phloem is a complex permanent tissue. It is made up of four types of elements: sieve tubes, companion cells, phloem fibres and phloem parenchyma. Sieve tubes are tubular cells with perforated walls. Phloem transports food from the leaves to the other parts of the plant. Except for phloem fibres, all phloem cells are living.
6. Answer the questions.
(a) Give the name of a connective tissue lacking fibres.
(b) Why does an organism—plant or animal, require different types of cells in the body?
(c) Why are voluntary muscles also called skeletal muscles?
Ans. (a) Blood
(b) Any organism will have a wide range of cell types. This is because each cell type specialises in one particular function. For the proper working of an organism, many functions, like food transport, immunity, strength, etc., must be performed properly.
(c) Voluntary muscles are also called skeletal muscles because they are mostly attached to the bones and help in body movement.
7. What kind of tissue is blood? What are the components of blood?
Ans. Blood is a type of connective tissue. It has a liquid matrix called plasma, in which the red blood cells (RBCs), white blood cells (WBCs) and platelets are suspended. The plasma contains proteins, salts and hormones. Blood flows and transports gases, digested food, hormones to tissues and waste materials from the tissues to the liver and kidney.
8. What are involuntary muscles? Where are they found?
Ans. The muscles which do not move on our will are called involuntary muscles. The movement of food in the alimentary canal or the contraction and relaxation of blood vessels are involuntary movements. These muscles are also called smooth muscles. They are also found in the iris of the eye, in ureters and in the bronchi of the lungs.
9. Look at the diagram. Write three important features of this kind of tissues.
Ans. The diagram shows cardiac muscles.
Features of cardiac muscles:
• The cells of cardiac muscles are cylindrical, branched, and uninucleated.
• Cardiac muscles are involuntary muscles that contract rapidly but do not get fatigued.
• They control the contraction and relaxation of the heart.
10. Draw the diagrams of a living component and a non-living component of xylem. Discuss their role.
Ans. Xylem parenchyma is a living tissue made up of parenchyma cells with thin cell walls made of cellulose. It is the only living tissue in the xylem. The main function of xylem parenchyma is storage but it also helps in water conduction through ray parenchymatous cells. Xylem fibre is the non-living component of xylem. It provides mechanical strength to the plants.
11. Show the location of meristematic tissues in a plant with the help of a diagram. Write the function of each meristem.
Apical Meristem
Intercalary Meristem
Fig. 6.13
Meristem
12. Differentiate between cartilage and bone.
Ans. Cartilage Bone
1. Matrix is composed of a firm, but flexible material, called chondrion.
2. Cartilage is surrounded by a sheath called perichondrium.
3. Blood vessels are absent.
4. Cartilage cells lie singly or in groups of two or four.
5. Chondrocytes are oval and devoid of processes.
Ans. Apical meristem: These meristems are responsible for primary growth, or the increase in the length of the plant.
Intercalary meristem: These meristems are responsible for increasing the length of internodes. Lateral meristem: These meristems are responsible for secondary growth, or the increase in the width of the stem and root.
1. Matrix is composed of a tough inflexible material, called ossein.
2. Bone is surrounded by a tough sheath, called periosteum.
3. Blood vessels are present.
4. Bone cells lie in lacunae, singly.
5. Osteocytes are irregular and give off branching processes.
Fig. 6.11
Lateral
Stem
Location of meristematic tissues
Fig. 6.12 Xylem parenchyma and xylem fibre
13. What happens when
(a) formation of older stem does not occur?
(b) apical meristem is damaged?
(c) blood platelets are removed from the blood?
Ans. (a) If the formation of cork does not occur in older stems, the outer tissue ruptures due to the increase in girth and the plant is infected by diseases.
(b) If the apical meristem is damaged, there is no growth in length of the plant.
(c) If the blood platelets are removed, there is no blood clotting at the site of an injury. Bleeding continues with a significant loss of blood.
14. Answer the following questions.
(a) The root tips of a plant were cut and then the plant was replanted. What will happen to the plant and why?
(b) Where are fats stored in our body? How is it useful for the organism?
Ans. (a) The plant will die a few days after replanting because its root tips are cut and the meristematic tissues, required for the plant’s growth, is absent. So, without roots, there will be no absorption of water and nutrients.
(b) Fat-storing adipose tissue is found below the skin and between internal organs.The cells of this tissue are filled with fat globules. Storage of fats allows it act as an insulator.
15. How are tendons different from ligaments?
Ans. Tendons Ligaments
1. It is a thick, tough and non-elastic connective tissue. 1. It is a strong but elastic bond of connective tissue.
2. It connects muscles to a bone.
3. It is formed of white fibrous tissues.
4. Fibres are white and formed of collagen protein.
5. It helps in the movement of bone.
2. It connects two bones.
3. It is formed of yellow elastic, fibrous tissues.
4. Fibres are yellow and formed of protein elastin.
5. It restricts or limits the movement of bones.
16. Give three differences between epithelial tissues and connective tissues.
Ans. Differences between epithelial and connective tissues are as follows:
Epithelial tissue
1. It is the covering tissue and protects the body from infections, injury, etc.
2. Cells are tightly packed.
Connective tissue
1. It is a connecting tissue as it connects various body organs.
2. Cells are loosely packed.
3. Cells contain very little or no intercellular matrix. 3. The space between cells is filled with matrix (solid or fluid).
17. What are different types of blood vessels? Write their functions. Draw diagrams of each type.
Ans. Three different types of blood cells, with their functions, are as follows:
(a) Red Blood Cells (RBCs): They contain haemoglobin, help in the transportation of gases, digested food, hormones, etc.
(b) White Blood Cells (WBCs): They are an integral part of the immune system, help in fighting diseases by producing antibodies and engulfing the germs and pathogens. They are of five types: Monocyte, lymphocyte, Eosinophil, basophil and neutrophil.
(c) Platelets: They help in the clotting of blood.
Red Blood Cells
Fig. 6.14
18. Animals of colder regions and fishes of cold water have thicker layer of subcutaneous fat. Describe why? (NCERT Exemplar)
Ans. A thick layer of subcutaneous fat functions as an insulating coat that prevents heat loss from the body and helps to keep the body of the animal warm in a colder environment. Moreover, fat also functions as reserve food during periods of scarcity. Thus, animals in colder regions and fish in cold water have a thicker layer of subcutaneous fat.
19. If a potted plant is covered with a glass jar, water vapours appear on the wall of glass jar. Explain why. (NCERT Exemplar)
Ans. In a potted plant, transpiration occurs through the stomata present on the surface of leaves. When a potted plant is covered with a glass jar, water vapours released by transpiration condense and appear as water droplets on the inner walls of the glass jar.
20. Name the different components of xylem and draw a living component. (NCERT Exemplar)
Ans. The different components of xylem are tracheids, vessels, xylem parenchyma and xylem fibres. Xylem parenchyma is the only living component of xylem.
21. Draw and identify different elements of phloem. (NCERT Exemplar)
Ans. The different elements of phloem are sieve tubes, companion cells, phloem fibres and phloem parenchyma.
22. Water hyacinth floats on water surface. Explain.
Ans. Water hyacinth (Eichhornia crassipes) is a free-floating aquatic plant. The plant possesses aerenchyma in its spongy petiole. Aerenchyma consists of a network, which encloses very large air cavities. These air cavities store gases, making the plant light and helping it float on the surface of water.
23. Which structure protects the plant body against the invasion of parasites? (NCERT Exemplar)
Ans. Epidermis is a layer of parenchymatous cells that forms the outermost covering of plant body. Epidermis consists of compactly arranged cells without any intercellular spaces. On the aerial plant
Fig. 6.15 Xylem parenchyma
Sieve Plate
Phloem fibre
Phloem Parenchyma
Companion Cell
Sieve Tube Cell
Sieve Pores
Fig. 6.16 Elements of phloem
parts, it secretes a thick, waxy, water-resistant layer called cuticle on its outer surface. These features protect the epidermis against loss of water, mechanical injury and the invasion of parasites.
Long Answer Questions (80–120 words)
1. Why are plants and animals made of different types of tissue?
Ans. Plants and animals are two different types of organisms. Plants are autotrophic organisms, so they prepare their own food by photosynthesis. However, plants are stationary or fixed organisms. Since they do not consume or need much energy, most plant tissues are supportive. Tissues, such as xylem, phloem, sclerenchyma and cork, are dead tissues, i.e., they do not contain living protoplasm.
Animals, on the other hand, are heterotrophic organisms. They have to move in search of food, mate and shelter; so they need more energy as compared to plants. Most of these tissues contain living protoplasm.
There are some tissues in plants which divide throughout life. They divide for the growth and reproduction of the plants. In contrast to plants, growth in animals is uniform.
2. Differentiate between parenchyma and collenchyma.
Ans.
Parenchyma
Collenchyma
1. This tissue consists of thin-walled cells. 1. This tissue consists of localised thickenings in the corner.
2. It is distributed in almost all parts of the plant. 2. It occurs mostly in the aerial parts of the plant after epidermis.
3. The cells store food and waste products. 3. They are the chief mechanical tissue in younger plants.
4. They are loosely packed with intercellular spaces. 4. They are compactly packed without intercellular spaces.
3. Differentiate between collenchyma and sclerenchyma.
Ans.
Collenchyma
1. This tissue consists of living cells.
2. The cells contain cytoplasm.
3. Cell wall is cellulosic in nature.
Sclerenchyma
1. This tissue consists of dead cells.
2. The cells do not contain cytoplasm as they are dead.
3. Cell wall is lignified in nature.
4. They are loosely packed with intercellular spaces. 4. They are compactly packed without intercellular spaces.
5. Lumen of cell is wide.
5. Lumen of cell is narrow.
6. It provides elasticity and mechanical support. 6. It is chiefly a mechanical tissue.
4. What is a neuron? Write the structure and functions of a neuron.
Ans. Nervous tissue contains highly specialised unit cells called nerve cells or neurons. Each neuron has the following three parts:
• The cyton or cell body: It contains a central nucleus and cytoplasm with characteristic deeply stained particles, called Nissl granules.
• Dendrites are short processes arising from the cyton.
• The axon is a single, long, cylindrical process of uniform diameter, which carries impulses away from the cell body.
Functions:
Neurons have the ability to receive stimuli from within or outside the body and conduct impulses to different parts of the body. The impulses travel from one neuron to another neuron and, finally, to the brain or spinal cord.
5. Differentiate between meristematic tissue and permanent tissue.
Ans.
Meristematic Tissues
1. The cells divide repeatedly.
2. The cells are undifferentiated.
3. The cells are small and isodiametric in nature.
4. Intercellular spaces are absent.
5. Vacuoles are absent.
6. Metabolism occurs at a high rate because they are dividing in nature.
Permanent Tissues
1. The cells do not divide.
2. The cells are differentiated.
3. The cells are variable in shape and size.
4. Intercellular spaces are present.
5. Vacuoles are present.
6. Metabolism occurs at a low rate because they are non-dividing in nature.
6. Discuss striated and smooth muscles with the help of diagram.
Ans. Striated muscle fibres are long or elongated, non-tapering, cylindrical and unbranched. These cells have a number of nuclei called sarcolemma. These muscle fibres shows alternate dark and light stripes or striations, so they are called striated muscles. These muscles occur in the muscles of limbs, body wall, face, neck, etc.
Functions of striated muscles:
• Striated muscles are powerful and undergo rapid contraction and expansion.
• They provide the force for locomotion and all other voluntary movements of the body.
Smooth muscles are also known as unstriated or involuntary muscles. They are held together by loose connective tissue. These muscle fibres do not bear any bands, stripes or striation across them.Smooth muscles are found in the walls of the alimentary canal and internal organs, ducts of glands and blood vessels. They are also found in the stomach, intestine, ureters, bronchi, iris of the eye, etc.
Functions of smooth muscles:
Smooth muscles do not work according to our will, so they are also called involuntary muscles. The movement of food in the alimentary canal or the contraction and relaxation of blood vessels are involuntary movements.
Axon Myelin sheath
Schwann cells
Node of ranvier Axon terminals
Fig. 6.17
Fig. 6.18
Skeletal muscle
Smooth muscle
7. Identify different elements (A, B, C, D) of phloem. What are the dead parts and what are the living parts?
Fig. 6.19 Components of phloem
Ans. A–Sieve plate, B–Sieve cells, C–Companion cell, D–Phloem parenchyma. Sieve tubes, companion cells and phloem parenchyma are living cells, while phloem fibres are dead cells.
8. How are permanent tissues formed? Classify permanent tissues and describe them.
Ans. Permanent tissues are derived from meristematic tissue but their cells have lost the power of division. Permanent tissues are classified into the following two types:
(i) Simple permanent tissue
(ii) Complex permanent tissue
Simple permanent tissues: These tissues are composed of cells which are structurally and functionally similar.They are classified into the following three types:
• Parenchyma: Parenchyma forms the bulk of the plant body. Parenchyma cells are living and possess the power of division.
• Collenchyma: Collenchyma tissue is also living. It is characterised by the deposition of pectin and cellulose at the corners of the cells.
• Sclerenchyma: Sclerenchyma cells are dead cells and they are devoid of protoplasm. The cell walls of sclerenchyma are largely thickened with the deposition of lignin.
Complex permanent tissues: The complex tissues consist of more than one type of cells with a common origin. All these cells coordinate to perform a common function.Complex tissues are of the following two types:
• Xylem is composed of four different types of cells: (i) Tracheids; (ii) Vessels; (iii) Xylem parenchyma; and (iv) Xylem fibre. Except for xylem parenchyma, all other xylem elements are dead and bounded by thick lignified walls.
• Phloem is composed of the following four elements: (i) Sieve tubes; (ii) Companion cells; (iii) Phloem parenchyma; and (iv) Phloem fibres. Except for phloem fibres, all other phloem elements are living. Xylem and phloem are both conducting tissues and are also known as vascular tissues. Together, both of them constitute vascular bundle.
9. Explain any three types of connective tissues along with their functions.
Ans. There are five types of connective tissues:
(i) Areolar connective tissue: It is a loose and cellular connective tissue.
Functions:
• It acts as a supporting and packing tissue between organs in the body cavity.
• It helps in the repair of tissues after an injury.
(ii) Dense regular connective tissue: It is a fibrous connective tissue, characterised by an ordered and densely packed collection of fibres and cells.
Functions:
• Tendons: Tendons are cord-like, strong, inelastic structures that join skeletal muscles to bones.
• Ligament: They are an elastic structure which connects bones to bones.
(iii) Adipose tissue: Adipose tissue is essentially an aggregation of fat cells.
Functions:
• It serves as a fat reservoir.
• It provides shape to the limbs and the body.
• It forms shock-absorbing cushions around the kidneys and eyeballs.
• It acts as an insulator. It reduces heat loss from the body, i.e., it regulates body temperature.
(iv) Skeletal tissue: Skeletal or supporting tissue includes bone and cartilage
Cartilage is a specialised connective tissue, which is compact and less vascular. Cartilage can be found in the ear pinna, nose tip, and rings of trachea. Bone is a strong and non-flexible tissue. Like cartilage, bone is also a specialised connective tissue.
Functions:
• Cartilage provides support and flexibility to the body parts. It smoothens the surface at the joints.
• Bone provides shape and skeletal support to the body.
• Bone protects vital body organs such as brain, lungs, etc.
• Bone anchors the muscles.
(v) Fluid connective tissue: Fluid connective tissue links the different parts of the body and maintains continuity in the body. It includes blood and lymph.
• Blood: In this tissue, cells move in a fluid or liquid matrix, or medium, called blood plasma. Blood occurs in blood vessels called arteries, veins and capillaries, which are connected to form the circulatory system.
• Lymph: Lymph is a colourless fluid that is filtered out of the blood capillaries.
Functions:
• Blood transports nutrients, hormones and vitamins to the tissues, and transports excretory products from the tissues to the liver and kidney.
• Lymph transports the nutrients (oxygen, glucose) that may have filtered out of the blood capillaries back into the heart to be recirculated in the body.
• Lymph brings C02 and nitrogenous wastes from the tissues to the blood.
Note: Students can write any three types of connective tissues.
10. How is sclerenchyma different from parenchyma? Draw longitudinal section of them.
Ans. The cells of sclerenchyma are made up of dead cells and are thick walled and lignified. While parenchyma cells are thin walled, living having inter cellular spaces.
Narrow lumen
Lignified thick wall
Cytoplasm
Nucleus
Middle lamella
Chloroplast
Vacuole
Intercellular space
Primary cell wall
Fig. 6.20 L.S of sclerenchyma and parenchyma
11. Write a short note on epithelial tissue. Describe the functions of epithelium tissue.
Ans. The covering or protective tissue in the animal body are epithelial tissues. Epithelial tissue cells are tightly packed and form a continuous sheet. They only have a small amount of cementing material between them and almost no intercellular spaces. Epithelium covers most organs and cavities within the body. It forms a barrier to keep different body systems separate. The skin, lining of the mouth, lining of blood vessels, lung alveoli and kidney tubules are all made up of epithelial tissue.
Functions of epithelial tissue:
• Epithelial cells protect the underlying cells from drying, injury and chemical effects. They also protect the body from viral or bacterial infections.
• They help in the absorption of water and nutrients.
• They perform secretary function by secreting useful chemicals like sweat, saliva, enzymes from the food, etc., in the body.
12. Differentiate between sclerenchyma and parenchyma tissues on the basis of some distinguishing features.
Ans. Features
Sclerenchyma
Nature It is a dead, supportive tissue that provides mechanical strength.
Parenchyma
It is a living tissue involved in various functions like storage, photosynthesis and secretion.
Cell wall Thick, lignified secondary walls, making the cells rigid. Thin, non-lignified primary walls, allowing flexibility.
Cell shape Cells are long, narrow and pointed at both ends (fibres) and irregular (sclereids).
Location Found in stems, vascular bundles, veins of leaves, hard seed coats and nutshells.
Function Provides strength, rigidity and support to plant parts like stems and veins of leaves.
Cells are usually oval, round or polygonal with intercellular spaces.
Present in soft plant parts like cortex, pith, mesophyll of leaves and fruits.
13. Describe the structure and function of different types of epithelial tissues. Draw diagram of each type of epithelial tissue.
Ans. Depending upon the shape and function of the constituent cells, epithelial tissues are of the following types:
Squamous epithelium: The cells in this epithelium are extremely thin and flat and are arranged edge to edge. It forms the lining of cavities of ducts and blood vessels, lines the chambers of the heart and covers the skin and lining of the mouth.
It provides protection to the underlying parts against abrasion (mechanical injury).It also helps in excretion, gas exchange and secretion of coelomic fluid.
Columnar epithelium: This epithelium consists of cells and look like a column. It forms the lining of the stomach and intestines; also found in salivary glands in the mouth, sweat glands and oil glands of the skin. It helps in protection, absorption and secretion.
Cuboidal epithelium: Cells are as long as they are broad and appear cube-like. The cuboidal epithelium lines the small salivary ducts, pancreatic ducts, sweat glands, salivary glands and thyroid glands. It helps in protection, secretion, absorption, excretion and gamete formation.
Ciliated epithelium: This epithelium, usually consisting of cuboidal or columnar cells, has numerous, thin, hair-like projections called cilia arising from the outer free surface of the cells. It is found lining the wind-pipe (trachea), kidney tubules and oviducts (fallopian tubes). This epithelium helps in the movement of mucous, eggs, sperms.
Gandular epithelium: This epithelium consists of columnar cells modified to secrete chemicals. It lines the glands such as gastric glands, pancreatic lobules, etc.
Stratified epithelium: This is a compound epithelium in which cells are arranged in many layers, one above the other. Simple squamos epithelium Simple cuboidal epithelium
Stratified squamos epithelium
Stratified cuboidal epithelium
14. Give reasons for
(a) Meristematic cells have a prominent nucleus and dense cytoplasm but they lack vacuole.
(b) Intercellular spaces are absent in sclerenchymatous tissues.
(c) We get a crunchy and granular feeling, when we chew pear fruit.
(d) Branches of a tree move and bend freely in high wind velocity.
(e) It is difficult to pull out the husk of a coconut tree.
Ans. (a) Meristematic cells have a prominent nucleus and dense cytoplasm because they are metabolically highly active and are in a continuous state of division. They lack vacuole because they do not store food material, waste material, sap, etc.
(b) Sclerenchyma cells have lignified cell walls, which makes them compact and leaves no intercellular spaces.
(c) Pear fruit contains sclerenchymatous stone cells or sclereids, which provide a gritty texture to the fruit. Thus, when we chew pear fruit, we get a crunchy and granular feeling.
(d) Collenchyma tissue present in the branches of a tree provides flexibility to them and allows their easy bending without breaking. Thus, the branches move and bend freely in high wind velocity.
(e) The husk of a coconut is made up of sclerenchymatous fibres, which consist of compactly arranged cells with thick lignified cell walls. So, they are tightly joined together. Thus, it is difficult to pull out the husk of a coconut.
15. List the characteristics of cork. How are they formed? Mention their role.
Ans. Cork covers the old stems of woody trees. The characteristics of cork are as follows:
• The cells of cork are dead at maturity.
• These cells are compactly arranged.
• The cells do not possess intercellular spaces.
• They possess a chemical substance, suberin, in their walls.
• They are several layers thick.
• Cork is impervious to gases and water.
As plants grow older, a strip of secondary lateral meristem (called cork cambium) develops in the cortical region. It cuts cells towards both the outer and inner sides. Gradually, this secondary tissue replaces the epidermal layer of the stem. This forms the several-layer-thick cork.
Fig. 6.21 Epithelial tissue
The role of cork is as follows:
• It protects the internal tissues from mechanical injury and from parasitic attack.
• It contains small pores (called lenticels) for gaseous exchange.
• It provides mechanical strength.
16. Why are xylem and phloem called complex tissues? How are they different from one other?
Ans. Both xylem and phloem consist of more than one type of cells, which coordinate to perform a common function. Therefore, they are called complex tissues.
Differences between xylem and phloem are given in the table.
Xylem Phloem
1. Xylem mainly consists of dead cells (except xylem parenchyma).
2. It conducts water and minerals from the roots to the aerial parts of the plant.
3. Xylem is generally located towards the inner side of plant parts.
Competency Based Questions
1. Phloem mainly consists of living cells (except phloem fibre).
2. It translocates prepared food from the leaves to the storage organs and growing parts of the body.
3. Phloem is generally located towards the outer side of plant parts.
Multiple Choice Questions (Choose the most appropriate option/options)
1. Intestine absorbs the digested food materials. What type of epithelial cells are responsible for that? (NCERT Exemplar)
(a) Stratified squamous epithelium
(b) Columnar epithelium
(c) Spindle fibres
(d) Cuboidal epithelium
2. While doing work and running, you move your organs like hands, legs etc. Which among the following is correct? (NCERT Exemplar)
(a) Smooth muscles contract and pull the ligament to move the bones.
(b) Smooth muscles contract and pull the tendons to move the bones.
(c) Skeletal muscles contract and pull the ligament to move the bones.
(d) Skeletal muscles contract and pull the tendon to move the bones.
3. Select the incorrect sentence. (NCERT Exemplar)
(a) Blood has matrix containing proteins, salts and hormones.
(b) Two bones are connected with ligament.
(c) Tendons are non-fibrous tissue and fragile.
(d) Cartilage is a form of connective tissue.
4. Which of the following features is NOT applicable to plant tissues?
(a) Growth is confined to specific regions.
(b) Most of the tissues are dead.
(c) Energy requirement is less.
(d) Growth is uniform.
5. Match the Column A with the Column B. Choose the correct option.
Column A
M. Fluid connective tissue
N. Filling of space inside the organs
O. Striated muscle
P. Adipose tissue
Q. Surface of joints
R. Stratified squamous epithelium
(a) M—iv; N—iii; O—v; P—i; Q—ii; R—vi
(c) M—iii; N—iv; O—v; P—i; Q—ii; R—vi
Column B
i. Subcutaneous layer
ii. Cartilage
iii. Skeletal muscle
iv. Areolar tissue
v. Blood
vi. Skin
(b) M—i; N—iv; O—iii; P—v; Q—ii; R—vi
(d) M—v; N—iv; O—iii; P—i; Q—ii; R—vi
6. Match the Column A with the Column B. Choose the correct option.
(a) The nature of matrix differs according to the function of the tissue.
(b) Fats are stored below the skin and in between the internal organs.
(c) Epithelial tissues have intercellular spaces between them.
(d) Cells of striated muscles are multinucleate and unbranched.
9. Which of the following is not applicable to meristematic tissues?
(a) Cells of meristematic tissue are very active.
(b) Cells have dense cytoplasm and thin cellulose walls.
(c) Cells have prominent nuclei.
(d) Cells have vacuoles inside.
10. Ranu is studying two plants: Plant A shows rapid vertical growth at its tips, while Plant B shows thickening of the stem and regrowth from the base after being grazed by animals.
Which combination of meristematic tissues is actively involved in these processes?
(a) Apical meristem in Plant A, lateral and intercalary meristem in Plant B.
(b) Apical meristem in Plant A, only lateral meristem in Plant B.
(c) Intercalary meristem in Plant A, apical and lateral meristem in Plant B.
(d) Lateral meristem in Plant A, apical and intercalary meristem in Plant B.
11. Which of the following statement is false regarding sclerenchymatous tissues?
(a) All are dead cells.
(b) They have long and narrow cells and the walls are thickened due to lignin.
(c) They have no intercellular spaces.
(d) They have large air cavities present that help them float.
12. Look at the figure. Identify A and B.
(a) A–Parenchyma, B–Epidermis
(b) A–Epidermis, B–Stomata
(c) A–Cork cells, B–Broken epidermis
(d) A–Meristem, B–Stomata
13. Observe the diagram below. Identify A, B, C and D.
14. Following are some features about the kind of animal tissues. On the basis of the features, identify the kind of tissue.
Description: The skin, lining of the mouth, lining of blood vessels, lung alveoli and kidney tubules are all made of this tissue. They are tightly packed. They have no intercellular spaces. They are separated by a fibrous basement membrane.
(a) Fibrous connective tissue
(c) Epithelial tissue
(b) Fluid connective tissue
(d) Skeletal muscular tissue
15. What would happen if the intercalated discs in cardiac muscle were absent?
(a) Heart muscles would not contract simultaneously. (b) The muscle would lose its striations.
(c) Blood vessels would not form.
(d) The heart would stop producing energy.
16. Which of the following statements about cartilage is false?
(a) It is a connective tissue.
(c) It is rigid and brittle.
17. Look at the Venn diagram. Choose the correct option about S.
(a) It is bone.
(b) It is areolar connective tissue.
(c) It is cartilage.
(d) It is ligament.
18. Choose the correctly matched pair.
(a) Inner lining of salivary ducts – Ciliated epithelium
(b) Moist surface of buccal cavity – Glandular epithelium
(c) Tubular parts of nephron – Cuboidal epithelium
(d) Inner surface of bronchioles – Squamous epithelium
(b) It lacks blood vessels.
(d) It provides cushioning in joints.
This tissue provides support, protection and enables movement. This tissue is flexible S
19. A person with a torn ligament in the knee is advised to undergo physiotherapy. Why is physiotherapy important for ligament recovery?
(a) It reduces the need for surgery.
(b) It strengthens the surrounding muscles and restores joint stability.
(c) It prevents the ligament from re-tearing.
(d) It eliminates the need for any other treatment.
Fig. 6.22
Fig. 6.23
Fig. 6.24
20. The image shows the structure of two types of muscles that are
present in two different locations in the body.
On the basis of their location, what can be concluded about their functions?
(a) Both show voluntary movements.
(b) Both help in movement of the body.
(c) Both protect internal body organs.
(d) Both show involuntary movements.
21. The image shows the location of a connective tissue B. Which of the following is the function of B?
(a) It allows the bones to move.
(b) It prevents the bone from bending.
(c) It provides strength to the bones.
(d) It connects the bone to other bones.
22. The image shows a kind of plant tissue.
Transverse Section
Section
Which characteristics of these tissues provides strength to the plants parts?
(a) Presence of large spaces between the cells
(b) Presence of thin dead cells
(c) Presence of cells with regular shape
(d) Presence of thick walls without intercellular spaces
23. The image shows the structure of a specialised epithelium. What will be the function of this epithelium?
(a) Secretion of substances
(b) Transport of substances across permeable surface
(c) Protection from wear and tear
(d) Restrict movement of the tissue
Fig. 6.25
Fig. 6.27
Simple pit pair
Longitudinal
Narrow lumen
Lignified thick wall
Fig. 6.28
Mucus
Goblet cell
Ciliated cell
Basal cell
Fig. 6.26
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): The cells of areolar tissues are filled with fat globules.
Reason (R): Areolar tissues help in the insulation of the body.
2. Assertion (A): Xylem vessels are dead cells, yet they are vital for the survival of plants.
Reason (R): Xylem vessels transport water and minerals from the roots to other parts of the plant through passive transport.
3. Assertion (A): Phloem translocation in plants is a bidirectional process.
Reason (R): Phloem is responsible for the transport of water in plants.
4. Assertion (A): Cardiac muscles are involuntary in action.
Reason (R): Cardiac muscles lack striations and have intercalated discs.
5. Assertion (A): Meristematic tissues are responsible for the primary and secondary growth of plants.
Reason (R): Permanent tissues lose the ability to divide and perform specialised functions.
6. Assertion (A): Collenchyma cells provide mechanical support and flexibility to plant parts.
Reason (R): Collenchyma cells have thickened secondary cell walls made of lignin.
Case-Based/Source-Based/Passage-Based Questions
1. Study the following table and answer the following questions.
The following table provides data on the basis of structure, locations and functions of different types of plant tissues: Structure
Thin cell wall, large vacuole Cortex of roots and stems
Thick lignified walls Around vascular bundles, seeds Provides rigidity and strength
Dead cells, tubular structure Roots, stems, leaves
Conducts water and minerals
Living cells with sieve plates Roots, stems, leaves Transports food from the leaves to other parts
(a) Which tissue provides both flexibility and mechanical support to the plant? Explain how.
(b) Identify the plant tissue responsible for food storage, based on its structure and location.
(c) Why is sclerenchyma considered unsuitable for food storage?
(d) Which tissue in the table is composed of dead cells and plays a critical role in upward water conduction?
2. Look at the diagram and answer the following questions.
Fig. 6.29 Structure of a neuron
Nucleus
Cell body
Dendrite
Axon
Node of ranvier Axon terminals
Myelin sheath Synapse
(a) What is the significance of the myelin sheath in the structure of a neuron, and how does it affect nerve impulse transmission?
(b) Which part of the neuron is responsible for receiving signals from other neurons? Describe its structural adaptation for this function.
(c) What would happen if the nodes of Ranvier were absent in the axon?
(d) Which part of the neuron plays a role in transmitting signals to the next cell, and what process occurs at this junction?
3. Read the paragraph and answer the questions.
Plants consist of various tissues specialised for different functions. Meristematic tissues are found in regions of growth such as the tips of roots and stems. These cells divide continuously and are undifferentiated. Over time, some meristematic cells differentiate to form permanent tissues.
Permanent tissues include simple tissues like parenchyma, collenchyma and sclerenchyma, and complex tissues like xylem and phloem. Parenchyma cells are living and store food, while collenchyma cells provide mechanical support and flexibility. Sclerenchyma cells, on the other hand, are dead at maturity and provide rigidity. In vascular plants, xylem and phloem work together as a transport system. Xylem transports water and minerals through dead tubular cells, while phloem transports food through living cells. The adaptability of plant tissues allows plants to grow in diverse environments, but their survival depends on the efficient functioning of these specialised tissues.
(a) Why is differentiation necessary for plant growth?
(b) Draw the diagram of two simple tissues that provide mechanical strength to the plants.
(c) What is the key difference between xylem and phloem, based on the paragraph?
(d) How does the adaptability of plant tissues contribute to their survival in diverse environments?
4. Read the paragraph and answer the questions.
Mona was analysing a blood sample under a microscope and observed different types of white blood cells (WBCs). She noted that some cells were small and round with a large nucleus occupying most of the cell. Others were irregularly shaped with lobed nuclei and granular cytoplasm. These cells showed different behaviours: some engulfed bacteria and other harmful particles, while others released chemicals like histamines during allergic reactions. Mona concluded that these WBCs play a critical role in defending the body against infections and maintaining immunity.
(a) Which type of WBC, observed in the sample, is responsible for engulfing bacteria through phagocytosis?
(b) Identify the WBC with a large nucleus that occupies most of the cell and its primary function.
(i) Basophils; releases histamines during allergic reactions.
(ii) Lymphocytes; responsible for immune responses and producing antibodies.
(iii) Neutrophils; performs phagocytosis.
(iv) Monocytes; transforms into macrophages.
(c) What would happen if the number of basophils in the blood increases significantly?
(i) The body’s ability to engulf bacteria would increase.
(ii) Allergic reactions and inflammation would become more severe due to excessive histamine release.
(iii) Antibody production would increase.
(iv) The body would lose the ability to fight infections.
(d) What is the main difference between neutrophils and lymphocytes in terms of their role in immunity?
Answers
Multiple Choice Questions
Assertion-Reason Based Questions
Case-Based/Source-Based/Passage-Based Questions
1. (a) Collenchyma provides both flexibility and mechanical support to the plant.
(b) Parenchyma
(c) Sclerenchyma has lignified tissues with thick walls.
(d) It is xylem.
2. (a) It insulates the axon and speeds up impulse transmission through saltatory conduction.
(b) Dendrites
(c) The impulse would travel slowly because saltatory conduction would not occur.
(d) Axon terminals
3. (a) Cell differentiation is a vital stage in plant growth because it allows the cells to mature and perform advanced functions.
(b)
(c) Most of the xylem tissues are dead cells, involved in conduction of water and minerals. Most of the phloem tissues are loving cells, involved in food transport.
(d) The adaptability of plant tissues contribute to their survival in diverse environments by facilitating the differentiation of tissues for specialised functions.
4. (a) Neutrophils
(b) (ii) Lymphocytes; responsible for immune responses and producing antibodies.
(c) (ii) Allergic reactions and inflammation would become more severe due to excessive histamine release.
(d) Neutrophils are part of innate immunity, while lymphocytes are part of adaptive immunity.
Practice Questions
Multiple Choice Questions
1. Which of the following is NOT a characteristic of meristematic tissue?
(a) Cells have thin cellulose walls. (b) Cells are living and undifferentiated.
(c) They divide continuously to form new cells. (d) Cells have large vacuoles to store food.
Lumen
Fig. 6.30 Cross section of collenchyma and sclerenchyma
2. Which plant tissue allows aquatic plants to float on water, and what is the structural feature responsible for this property?
(c) Aerenchyma; large air cavities in the tissue (d) Sclerenchyma; lignified cell walls
3. Which connective tissue in animals lacks blood supply and heals slowly due to its structure?
(a) Ligament
(c) Bone
(b) Cartilage
(d) Tendon
4. Which statement correctly describes xylem and phloem?
(a) Xylem transports water, and its cells are living.
(b) Phloem transports food, and its cells are dead.
(c) Xylem consists of dead cells, while phloem consists of living cells.
(d) Both xylem and phloem are made up of dead cells.
5. Which type of epithelial tissue allows gaseous exchange in the lungs?
(a) Squamous epithelium
(c) Columnar epithelium
Assertion-Reason
Based Questions
(b) Cuboidal epithelium
(d) Ciliated epithelium
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as (a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false.
(d) A is false but R is true.
6. Assertion (A): Neurons transmit nerve impulses throughout the body.
Reason (R): Neurons have a myelin sheath that insulates the axon and increases the speed of impulse conduction.
7. Assertion (A): Ligaments connect bones to bones and provide flexibility.
Reason (R): Ligaments contain a high proportion of elastic fibres, making them strong and elastic.
Very
Short
Answer
Questions (30-50 words)
8. Explain the role of the nervous tissue in transmitting signals in the body.
9. Differentiate between collenchyma and sclerenchyma.
10. Sate two functions of epidermis.
11. State the reason for the following:
(a) Cartilage is considered as a connective tissue.
(b) Collenchyma tissue allows plants to bend.
12. Discuss the process of phagocytosis in neutrophils.
Short Answer Questions (50-80 words)
13. What are the structural adaptations of the epithelial tissue for protection?
14. Differentiate between xylem and phloem in terms of their function.
15. Draw a labelled diagram of unstriated muscle and mention its occurrence, structure and functions.
16. Which type of muscle is found in the iris of eye. Are they involuntary or voluntary? In what way, they are different from striated muscle with respect to the number of nuclei?
17. Explain the structure and function of phloem tissue in plants.
Long Answer Questions (100-120 words)
18. Differentiate between
(a) plant tissues and animal tissues
(b) epithelial tissue and connective tissues
19. Define nervous tissue and state its function. Discuss the structure of a typical neuron with the help of a well-labelled diagram.
20. Answer the following questions.
(a) What are the similarities and differences between cartilage and bone?
(b) What are the similarities and differences between adipose tissues and blood tissues?
Brain Charge
1. Complete the flow chart.
epithelium Columnar epithelium
2. Solve the crossword puzzle below.
connective tissue
3. Responsible for secondary growth
6. Conducting tissue for water and minerals
7. Tissue meant for growth
8. Thick walls with lignin
10. Found in major parts of higher plants
1. Associated with sieve tubes
2. Parenchyma with air spaces
4. Outermost layer of cells
5. Provides flexible support to plants
9. Conducts food produced by photosynthesis
3. Fill the gaps in the following table.
Plant Tissue Animal Tissue
Tissue organisation is targeted towards habit of plants.
Organisation is simple.
Tissue organisation is targeted towards of animals.
Organisation is complex.
Many of the tissues are dead. For example, Cork Most of the tissues are
Growth is to certain areas.
Growth is not limited to areas.
Less maintenance energy required. maintenance energy required.
Plants grow throughout life. After reaching maturity stage animals do not show further growth.
Challenge Yourself
1. How would plant growth be affected if meristematic tissues were distributed throughout the plant body?
2. In what way do phloem and xylem tissues demonstrate the principle of division of labour in plants? Can you draw parallels between this and a human organ system?
3. Fill the correct words in the following sentences. The matrix of bone is in the form of thin concentric rings, called ____________. Bone cells, called ____________, are present in fluid-filled spaces called ____________. These fluid-filled spaces of the bone communicate with each other by a network of fine canals, called ____________.
4. A screw is inserted in the trunk of a tree at a height of 50 cm from the ground level. After two years, where will the screw be present?
5. Why do you think some epithelial tissues have cilia, while others do not? Provide examples of where ciliated epithelium is found and its significance.
6. The root tips of a plant were cut and the plant was replanted. What will happen to the plant and why?
7. If you were to design an artificial tissue for medical or agricultural purposes, what properties would you prioritise? Why?
Answers
Practice Questions
1. (d) Meristematic tissue cells lack large vacuoles because they are actively dividing, and require dense cytoplasm for cell division.
2. (c) Aerenchyma tissue contains air spaces that provide buoyancy to aquatic plants.
3. (b) Cartilages lack direct blood supply, making it slow to heal.
4. (c) Xylem vessels and tracheids are non-living, while phloem sieve tube elements are living.
5. (a) Squamous epithelium is thin and flat, facilitating efficient diffusion of gases.
1. If meristematic tissues were distributed throughout the plant body, uncontrolled and disorganised growth would occur, leading to structural instability. The plant would lack proper differentiation, disrupting essential functions like transport, support and photosynthesis.
2. Phloem and xylem demonstrate division of labour by performing specialised functions—xylem transports water and minerals, while phloem transports food. This is similar to the circulatory system in humans, where blood vessels (arteries and veins) transport oxygen, nutrients and waste to maintain body functions.
3. Lamellae, Osteocytes, Lacunae, Canaliculi
4. The screw will remain in the same position even after two years. This is because a plant or tree grows from its tip (stem or root), not from the point at which it joins the ground. So, the tree will grow but the screw will remain in the same place on the tree trunk.
5. Some epithelial tissues have cilia to help in the movement of substances over their surface, while others do not need this function. Ciliated epithelium is essential for protection, filtration and movement of substances in the body.
6. The plant will die within a few days of replanting. This is because, as the root tips are cut, the roots will not grow due to the absence of meristematic tissue. As a result, proper absorption of water and minerals will not occur.
7. When designing an artificial tissue for medical or agricultural purposes, the key properties would include biocompatibility, flexibility and strength, self-repair and regeneration and biodegradability.
SELF-ASSESSMENT
Time: 1.5 Hour
Scan me for Solutions
Max. Marks: 50
Multiple Choice Questions (10 × 1 = 10 Marks)
1. Which of the following best explains why cartilage heals more slowly than bones?
(a) Cartilage has fewer blood vessels than bones.
(b) Cartilage lacks collagen fibres.
(c) Bone cells are more active than cartilage cells.
(d) Cartilage cells cannot divide.
2. Which of these options are correct?
(i) Both sclerenchyma and parenchyma have dead tissues.
(ii) Blood is a fibrous connective tissue.
(iii) Cartilage is found in joints, nose and ear lobes.
(iv) The cells of meristematic tissues are very active.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv) (d) (iv) and (i)
3. Why do xylem vessels not collapse even under extreme tension during water transport?
(a) Presence of lignin in their walls
(c) Thick cellulose layers
(b) Presence of pectin
(d) Suberin coating
4. What would happen if smooth muscles in the walls of blood vessels fail to contract?
(a) Blood flow would increase uncontrollably.
(b) Blood pressure would drop.
(c) Oxygen transport would stop completely.
(d) Blood pressure would decrease and blood flow would increase.
5. Which of the following tissues in plants can perform both storage and photosynthesis?
(a) Xylem parenchyma
(c) Chlorenchyma
(b) Collenchyma
(d) Phloem parenchyma
6. Which of the following correctly explains the role of myelin sheath in neurons?
(a) It accelerates the regeneration of damaged neurons.
(b) It provides energy to the neuron.
(c) It insulates the axon and speeds up nerve impulse conduction.
(d) It increases the size of the nerve cell.
7. Observe the diagram below. It shows a section of an old stem.
Cork cambium forms cork on the outer side and secondary cortex on the inner side. What is the role of cork in the image?
(a) Absorption of water
(b) Prevention of water loss and protection
(c) Transport of nutrients
(d) Photosynthesis
Fig. 6.31
Epidermis Cork cambium Cortex
8. Which of the following is NOT a characteristic of meristematic tissue?
(a) Cells have thin cellulose walls. (b) Cells are living and undifferentiated.
(c) They divide continuously to form new cells. (d) Cells have large vacuoles to store food.
9. Find out incorrect sentence.
(a) Parenchymatous tissues have intercellular spaces.
(b) Collenchymatous tissues are irregularly thickened at corners.
(c) Apical and intercalary meristems are permanent tissues.
(d) Meristematic tissues, in its early stage, lack vacuoles
10. What would happen if the intercalated discs were absent in cardiac muscle?
(a) Heart muscles would not contract simultaneously.
(b) The muscle would lose its striations.
(c) Blood vessels would not form.
(d) The heart would stop producing energy.
Assertion–Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): Tendrils and stems of climbers bend in windy conditions.
Reason (R): The cells of collenchyma allow the plants to bend without breaking.
12. Assertion (A): Phloem transports food in plants, both upwards and downwards.
Reason (R): Phloem contains sieve tubes and companion cells that help in the active transport of food.
13. Assertion (A): Guard cells regulate the opening and closing of stomata.
Reason (R): Changes in turgor pressure of guard cells, due to water movement, control the stomatal aperture.
14. Assertion (A): Ligaments connect bones to muscles.
Reason (R): Ligaments are composed of elastic connective tissue and provide strength and flexibility to joints.
15. Look at the diagram and answer the following questions.
Nucleus
(a) Q in the picture represents (i) Tracheids (ii) Vessels (iii) Companion cells (iv) Xylem fibre
(b) What is the role of P in plants?
(c) How is Q related to P?
(d) What is R? What is its speciality in the plant?
16. Read the following paragraph and answer the following questions.
Myra is an athlete who spends a lot of time jogging and running. During her training sessions, she often feels fatigue in her legs after intense workouts. This fatigue gradually disappears after rest. Her coach explains that her muscles are undergoing contraction and relaxation continuously, which is causing the fatigue.
(a) Which type of muscle is primarily involved in Myra’ leg movements?
(b) Why does Myra feel fatigue in her muscles after intense workouts?
(c) Why does the fatigue disappear after rest? Justify.
(d) How is skeletal muscle different from smooth muscle?
Very Short Answer Questions (30-50 words)
17. Differentiate between simple and complex permanent tissues in plants.
18. What is the role of collagen in connective tissues?
(3 × 2 = 6 Marks)
19. Name the tissue that lines the inner surface of blood vessels and organs. What is the role of that tissue?
Short Answer Questions (50-80 words)
20. Draw and label different elements of phloem.
21. Differentiate between xylem and phloem in the context of their functions.
22. What is neuron? Describe the structure and function of a neuron.
23. What is the name of bone cells? Describe the functions of bone.
Long Answer Questions (80-120 words)
(4 × 3 = 12 Marks)
(2 × 5 = 10 Marks)
24. Describe the differences between cardiac, skeletal and smooth muscles in animals with the help of a labelled diagram.
25. Answer the following questions.
(a) Discuss the types of meristematic tissue and their roles in plant growth.
(b) Discuss the function of squamous epithelial tissues, its importance in providing protection and exchange of substances.
7 Motion
Motion is all around us, from a car moving on the street to the Earth rotating around the Sun. This chapter introduces the basic concepts of motion in physics: distance, displacement, speed, velocity, and acceleration. By understanding these concepts, students can connect their everyday experiences to scientific principles. The chapter explains the differences between uniform and non-uniform motion, showcases graphical representation of motion using distance-time and velocity-time graphs, and introduces equations of motion to simplify motion calculations. Through this, students become equipped to analyze motion in real life and appreciate the science behind every moving object.
Uniform Motion
Object covers equal distance in equal intervals of time.
Motion is always in a straight line.
Non-uniform Motion
Object covers unequal distance in equal intervals of time.
Motion may oy may not be in a straight line.
Uniform Circular Motion
Object moves in a circular path with a uniform speed.
Speed, v = 2πr t
Distance
Total length of the path traveled by an object during motion.
It has magnitude but no direction.
SI unit: Metre (m)
Speed
Distance travelled by an object in unit time.
It is independent of direction of motion. It is velocity without direction.
Avg. Speed = Distance Time
SI unit: Metres second (m/s)
Rate of Change of Speed
Rate of change of speed of an object per unit time.
Avg. rate of change of speed
= Change in speed time
SI unit: Metres per second squared (m/s2)
Motion
The change in position of an object with respect to time and a reference point.
Parameters of Motion
Equations of Motion
Equation 1: v = u + at
Equation 2: s = ut + 1 2 at2
Equation 3: 2as = v2 - u2
Graphs of Motion
Displacement
Shortest distance between initial and final position of object.
It has both magnitude and direction. SI unit: Metre (m)
Velocity
Displacement of an object in unit time. It depends on direction of motion. It can be change by changing the direction of motion.
Avg. Velocity = Dispalcement Time
SI unit: Metres per second (m/s)
Acceleration
Rate of change of velocity of an object per unit time.
Avg. Acceleration = Change in Velocity time
SI unit: Metres per second squared (m/s2)
Distance or DisplacementTime Graph
Y axis: Distance/Displacement X axis: Time
Slope of graph at any point give speed/velocity at that point.
If graph is a straight line, motion is uniform.
Speed or Velocity-Time Graph
Y axis: Speed or Velocity X axis: Time
Slope of graph at any point gives the Acceleration at that point.
Area under graph gives the displacement of the object. If graph is a straight line, velocity is uniform.
Chapter at a Glance
• Motion is a change of position, describable in terms of distance or displacement.
• Different types of motion include uniform motion (constant speed) and non-uniform motion (speed changes).
• Distance of a body is the length of path travelled by the body and is measured in metres (m).
• Displacement of a body is the minimum distance between the initial and final position of the body and is measured in metres (m).
• Distance has only magnitude but displacement has both magnitude and direction.
• Speed quantifies how quickly an object moves, typically expressed as distance covered per unit time in metres per second (m/s).
• Speed of an object can be positive or zero but it cannot be negative.
• Velocity is a basically speed with direction, measured as displacement/time in metres per second (m/s).
• Velocity can be positive, negative or zero.
• Acceleration describes the rate of change of velocity over time, typically expressed as change in velocity per unit time in metres per second squared (m/s2).
• Uniform acceleration occurs when an object’s velocity changes at a steady rate, displayed as a straight line on a velocity-time graph.
• Non-Uniform acceleration occurs when an object’s velocity changes unequally in equal intervals of time.
• When velocity and acceleration are in same direction, the speed increases and acceleration is considered to be positive.
• When velocity and acceleration are in opposite direction, the speed decreases and acceleration is considered to be negative.
• Three Key equations of motion for uniformly accelerated motion include
○ 1st equation: v = u + at
○ 2nd equation: s = ut + 1 2 at2
○ 3rd equation: 2as = v2 - u2
• In uniform circular motion, the object moves along a circular path at a constant speed with constant acceleration.
• The formula v = 2πr t is used to calculate the speed of an object in circular motion.
• Graphs are visual representation between two quantities on two axes.
• The distance/displacement-time graph depicts distance/displacement changes over time, enabling distinctions between uniform and non-uniform motions. This graph can be used to calculate speed/velocity of the object.
• The speed/velocity-time graphs show how speed/velocity varies with time. This graph can be used to calculate acceleration and displacement of the object. Velocity-time graph for uniform motion of a car
Velocity-time graphs of an object in non-uniformly accelerated motion.
Important Formulae
• Speed = Distance time
• Velocity, v = Displacement (s) time (t) = (s2 - s1) (t2 - t1)
• Acceleration, a = Change in Velocity time = (v - u) t
• Average velocity, v av = initial velocity + final velocity 2 = v + u 2
• 1st equation of motion, v = u + at
• 2nd equation of motion, s = ut + 1 2 at2
• 3rd equation of motion, 2as = v2 - u2
• Speed of uniform circular motion, v = 2πr T
NCERT Zone
Intext Questions
1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Ans. Yes, an object can have zero displacement even if it has moved through a distance. This happens when the object returns to its initial position after moving. For example, if a person walks 5 meters forward and then 5 meters back to their starting point, the total distance traveled is 10 meters, but the displacement is zero because the starting and ending positions are the same.
2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position?
Ans. Given,
Sides of square field = 10 m time for one-round trip = 40 s
total time of travel = 2 minutes 20 seconds or 140 seconds
perimeter of the square field = 4 × side = 4 × 10 m = 40 m.
In 40 seconds, the farmer completes one round.
In 140 seconds, the farmer completes 140 40 = 3.5 rounds.
After completing 3 rounds, the farmer returns to the initial position (A) with zero displacement from the start. The displacement in next 0.5 round is equal to the diagonal of the square field AC. From Pythagoras Theorem, AB2 + BC2 = AC2
22221010200102 m. ACABBC
3. Which of the following is true for displacement? (a) It cannot be zero. (b) Its magnitude is greater than the distance travelled by the object.
Ans. Neither (a) nor (b) is true. Displacement can be zero when the initial and final positions coincide. Its magnitude is usually less than or equal to the distance traveled, never greater.
4. Distinguish between speed and velocity.
Ans.
Speed
Velocity
1. Rate of change of distance. 1. Rate of change of displacement.
2. It has magnitude but no direction. 2. It has both magnitude and direction.
3. Always positive or zero but never negative. 3. May be positive, zero or negative
5. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Ans. The magnitude of average velocity equals average speed when an object moves in a straight line without changing direction.
6. What does the odometer of an automobile measure?
Ans. An odometer measures the total distance travelled by an automobile.
7. What does the path of an object look like when it is in uniform motion?
Ans. In uniform motion, the object moves in a straight line at constant speed, hence its path will be along a straight line. If plotted on a distance-time graph; it can be represented as a straight line.
8. During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 108 ms-1 .
Ans. We calculate distance as speed × time or v × t
The speed of light, v = 3 × 108 ms-1
The time, t = 5 minutes = 5 × 60 s = 300 s distance = v × t = 3 × 108 ms-1 × 300 s = 900 × 108 m = 9 × 1010 m.
9. When will you say a body is in (a) uniform acceleration? (b) non-uniform acceleration?
Ans. (a) A body is in uniform acceleration when its velocity changes by an equal amount in equal intervals of time.
(b) A body is in non-uniform acceleration when its velocity changes by unequal amounts in equal intervals of time, reflecting a variable rate of acceleration.
10. A bus decreases its speed from 80 km h-1 to 60 km h-1 in 5 s. Find the acceleration of the bus.
Ans. Given,
Initial velocity, u = 80 km/h
Final velocity, v = 60 km/h
We know, 1 km/h = 1000 m 3600 s = 5 18 m/s
Converting to speeds in m/s:
u = 80 × 5 18 = 200 9 m/s
v = 60 × 5 18 = 50 3 m/s
Using the 1st equation of motion, v = u + at
a = v - u t = 50 3200 9 5 = -10 9 m/s2
11. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h-1 in 10 minutes. Find its acceleration.
Ans. Given,
The initial speed, u = 0 m/s time, t = 10 × 60 = 600 s final speed, v = 40 kmh-1 = 40 × 5 18 m/s = 100 9 m/s.
Using acceleration formula,
a = v - u t = 100 9 - 0 600 = 100 9 × 600 = 1 54 m/s2 or 1.85 cm/s2
12. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Ans. In uniform motion, the distance-time graph is a straight line, indicating constant speed.
For non-uniform motion, the graph is a curve, reflecting varying speed as the steepness of curve changes over time.
13. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Ans. When the distance-time graph is a straight line parallel to the time axis as shown in Fig. 7.3, it indicates that the object is at rest. There is no change in distance over time.
14. What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Ans. If a speed-time graph is a straight line parallel to the time axis, it indicates that the object is moving at a constant speed. There is no acceleration, so the speed remains unchanged over time.
15. What is the quantity which is measured by the area occupied below the velocity-time graph?
Ans. The area occupied by the velocity-time graph and the time axis gives the magnitude of the displacement.
16. A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Ans. (a) Given, initial speed, u = 0 (bus starts from rest) acceleration, a = 0.1 m/s2 and t = 2 min or 120 s,
Fig. 7.3
Fig. 7.4
Using 1st equation of motion, v = u + at
v = 0 + 0.1 × 120 = 12 m/s.
(b) Using 2nd equation of motion, s = ut + 1 2 at2
17. A train is travelling at a speed of 90 kmh-1. Brakes are applied so as to produce a uniform acceleration of -0.5 ms-2. Find how far the train will go before it is brought to rest.
Ans. Given,
acceleration, a =-0.5 m s-2
Final velocity, v = 0 (train comes to rest)
initial velocity, u = 90 km/h = 90 × 1000 m 60 × 60 s = 25 m/s.
Using 3rd equation of motion, 2as = v2 - u2
s = v2 - u2 2a = 0 - (25)2 2 × (- 0.5) = -625 -1.0 = 625 m
18. A trolley, while going down an inclined plane, has an acceleration of 2 cms-2. What will be its velocity 3 s after the start?
Ans. Given acceleration, a = 2 cm/s2 initial velocity, u = 0 m/s and time, t = 3 s.
Using 1st equation of motion, v = u + at.
v = 0 + 2 × 3 = 6 cm/s.
The trolley’s velocity after 3 seconds is 6 cm/s.
19. A racing car has a uniform acceleration of 4 ms-2. What distance will it cover in 10 s after start?
Ans. Given,
initial velocity, u = 0 m/s
acceleration, a = 4 m/s2 and time t = 10 s.
Using 2nd equation of motion, s = ut + 1 2 at2 .
s = 0 × 10 + 1 2 × 4 × (10)2 = 0 + 200 = 200 m Therefore, the car covers 200 meters in 10 seconds.
20. A stone is thrown in a vertically upward direction with a velocity of 5 ms-1. If the acceleration of the stone during its motion is 10 ms-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Ans. Given, initial velocity, u =+5 m/s (upwards) acceleration, a =-10 m/s2 (downwards) and final velocity, v = 0 (at maximum height, ball will momentarily stop and then fall back)
Using the formula v2 = u2 + 2as 02 = 52 + 2 × (-10) × s
0 = 25 - 20s ⇒ s = 25 20 = 5 4 = 1.25 m.
For time, using 1st equation of motion, v = u + at,
0 = 5 + (-10)t ⇒ 10t = 5 ⇒ t = 5 10 = 0.5 seconds. Maximum height attained: 1.25 m, time taken: 0.5 s.
NCERT
Exercises
1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Ans. Given,
the diameter of circular track = 200 m.
radius of circular track, r = 100 m.
Round-trip time = 40 s
Total time = 2 min 20 s or 140 s
Let the athlete start moving from A which is treated as reference point. In 40s, he completes one round.
Therefore in 140 seconds, he will complete 140 40 = 3.5 rounds
Therefore, the final position of athletes at the end of 2 minutes and 20 seconds or just after three and a half rounds is B.
The displacement at the end of 2 minutes and 20 seconds is AB; the shortest distance between initial and final position.
AB = Diameter of circular track = 200 m.
Distance covered = circumference × number of rounds
= 2π × 100 × 3.5 ≈ 2200 m or 2.2 km
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Ans.
(a) From A to B,
Distance = 300 m
Displacement = 300 m time = 150 s (2 min 30 sec),
Average speed = Distance Time = 300 150 = 2 m/s
Average velocities = Displacement Time = 300 150 = 2 m/s
(b) From A to C,
Total distance = 300 + 100 = 400 m
Displacement = 200 m time = 210 s
Average speed = 400 210 ≈ 1.90 m/s
Average velocities = 200 210 ≈ 0.95 m/s
3. Abdul, while driving to school, computes the average speed for his trip to be 20 kmh-1. On his return trip along the same route, there is less traffic and the average speed is 30 kmh-1. What is the average speed for Abdul’s trip?
Ans. Let the distance between the school and Abdul's home be d.
The time taken to reach school, t1 = d v1 hours, and time back, t2 = d v2 hours.
distance for the round trip is 2d, and
=
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 ms-2 for 8.0 s. How far does the boat travel during this time?
Ans. Given,
initial speed, u = 0
acceleration, a = 3.0 ms-2 time, t = 8 s
Using the 2nd equation of motion, s = ut + 1 2 at2
5. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Ans. Given,
Car 1:
initial speed, u1 = 52 km/h = 52 × 5 18 = 14.44 m/s
final speed = 0
stopping time, t1 = 5 s
Car 2:
initial speed, u2 = 3 km/h = 3 × 5 18 = 0.83 m/s
final speed = 0
stopping time, t2 = 10 s
Since the speed-time graph forms a triangle due to uniform deceleration, the displacement will be equal to the area under the triangle.
Distance = (12) × base (time) × height (initial speed)
Distance = (12) × u2 × t2 = (12) × 10 × 0.83 = 4.15 m
Therefore, car A travels further after the application of brakes.
7.7
6. Figure shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
Ans. (a) B travels the fastest as it covers the greatest distance in shortest time.
(b) No. They are never at the same point because all the objects are never in same place at the same time.
(c) B passes A at approximately 1.1 hours, at that time C was at approximately 8 km mark. C started the journey from 2 km mark.
Therefore, C has travels (8 - 2) km or 6 km, when B passes A.
(d) B passes C at approximately 0.75 hours, at that time B was at approximately 5.5 km mark. B started the journey from 0 km mark.
Therefore, B has travels 5.5 km, when it passes C.
7.8
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms-2, with what velocity will it strike the ground? After what time will it strike the ground?
Ans. Given,
initial velocity, u = 0
acceleration, a = 10 ms-2 displacement, S = 20 m
Using 3rd equation of motion, v2 = u2 + 2aS
v2 = 0 + 2 × 10 × 20 = 400
==40020 m/s v
Using 1st equation of motion, v = u + at
20 = 0 + 10t
t = 20 10 = 2 seconds
The ball strikes the ground at 20 m/s after 2 seconds.
8. The speed-time graph for a car is shown in Fig. 7.9.
(a) Find how far does the car travel in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Ans. (a)
Fig. 7.10
(b) After 6 seconds the graph becomes a horizontal line in the speed-time graph, indicating that the speed remains constant.
9. State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity
(b) an object moving with an acceleration but with uniform speed.
(c) an object moving in a certain direction with an acceleration in the perpendicular direction.
Fig.
Fig. 7.9
Ans. (a) This situation is possible. For instance, when an object is thrown vertically upwards, at its highest point, the velocity becomes zero, but acceleration due to gravity remains constant at 9.8 m/s2 in downward direction.
(b) This is possible in circular motion, where an object moves at uniform speed while changing direction, resulting in a centripetal acceleration towards the center of the circle.
(c) This is possible in uniform circular motion. For example, a car moving at constant speed around a circular track has a centripetal acceleration that is perpendicular to the direction of the car’s motion (directed towards the center of the circular path).
10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Ans. Given,
Radius of circular orbit, r = 42250 km
Time period of revolution, T = 24 hours or 24 × 3600 s = 86400 s
Speed of satellite, v =
Multiple Choice Questions
1. Motion is best described as:
(a) Change in shape
(c) Change in temperature
(b) Change in colour
(d) Change in position with time
2. A reference point in describing motion is best defined as:
(a) The furthest point reached
(b) The moving object itself
(c) An arbitrary point from which positions are measured
(d) The point where speed is maximum
3. A car accelerates uniformly. Which of the following remains constant during its motion?
(a) Velocity
(c) Acceleration
(b) Displacement
(d) Distance covered
4. For an object in uniform motion, its displacement is given by: (a) s = 1 2 at2
s = ut + 1 2 at2 (c) s = ut
5. In a distance-time graph, a curved line indicates:
(a) Constant acceleration
(c) Uniform speed
s = u + at
(b) Non-uniform speed
(d) Rest
6. A horizontal line on a velocity-time graph indicates:
(a) Non-uniform motion
(c) Uniform velocity
7. Retardation of a body is expressed in:
(b) Uniform acceleration
(d) Zero displacement
(a) m (b) ms-1 (c) -ms-2 (d) ms-2
8. For an object moving with non-uniform speed along a straight path, which condition allows its velocity to be momentarily zero even though its speed is non-zero?
(a) When the displacement exactly equals zero.
(b) Only when the object momentarily changes its direction.
(c) When the velocity-time curve is a straight line.
(d) When the net force acting on the object is zero.
9. The sound of thunder is heard approximately 6 s after the flash of lightning is seen. If speed of sound in air is 346 m/s, then distance of point of lightning is (a) 1200 m (b) 2076 m (c) 3460 m (d) 965 m
10. A student jogs on a track of 150 m long. The student starts jogging on the track and reaches the end of the track in 1 minute 30 seconds and comes back to the middle of the track in 1 minute. What is the average velocity of the student?
(a) 1.5 m/s (b) 0.5 m/s (c) 0.25 m/s (d) 0 m/s
11. The numerical ratio of displacement to distance for moving object is (NCERT Exemplar) (a) always less than 1 (b) always equal to 1
(c) always more than 1 (d) equal to or less than 1
12. A car travels 5 km towards north then turns right and travels 3 km further, the car again turns right and travel 1 km and comes to rest. What is the distance travelled and displacement of the car?
(a) Distance : 7 km and displacement : 9 km
(b) Distance : 9 km and displacement : 5 km
(c) Distance : 5 km and displacement : 9 km
(d) Distance : 9 km and displacement : 7 km
13. Odometer of automobiles records:
(a) average speed
(b) instantaneous speed
(c) distance travelled (d) acceleration
14. The table shows the distance covered by three cars A, B and C at different time of a day.
Which option classifies the cars in uniform motion and non-uniform motion?
(a) Uniform Motion Non-uniform Motion
Car B Car A
Car C
(c) Uniform Motion Non-uniform Motion
Car B Car A Car C
(b) Uniform Motion Non-uniform Motion
Car A Car C
Car B
(d) Uniform Motion Non-uniform Motion Car A
15. Which of the following figures represents uniform motion of moving object correctly? (NCERT Exemplar) (a)
Fig. 7.11
16. A body moving along a straight line at 10 m/s undergoes an acceleration of 2 m/s2. After 2 s, its speed will be
(a) 6 m/s (b) 8 m/s (c) 12 m/s (d) 14 m/s
17. The velocity time graph of a moving particle is shown in Fig. 7.12. The acceleration of the particle is
(a) 2.5 m/s2
(b) 5 m/s2
(c) 10 m/s2
(d) 20 m/s2
18. From the given v - t graph in Fig. 7.13, it can be inferred that the object is (NCERT Exemplar)
(a) in uniform motion.
(b) at rest.
(c) in non-uniform motion.
(d) moving with uniform acceleration.
19. Area under a v - t graph represents a physical quantity which has the unit (NCERT Exemplar) (a) m2 (b) m (c) m3
20. The graph in Fig. 7.14 shows the change in velocity of a car with time.
Fig. 7.14
Which portion of the graph shows the negative acceleration for the car? (a) AB (b) CD (c) OA (d) DE
-1
21. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is (NCERT Exemplar)
(a) u g (b) u2 2g (c) u2 g (d) u 2g
22. The velocity-time (v – t) graph of a body is shown in Fig. 7.15. The distance travelled is
Fig. 7.15 (a) 112 m (b) 160 m (c) 56 m (d) 80 m
23. The graph in Fig. 7.16 shows the rate of change of the speed of the car. What will be the velocity of the car at 40 s?
(s)
Fig. 7.16
24. Four cars A, B, Cand D are moving on a levelled road. Their distance versus time graphs is shown in Fig. 7.17. Choose the correct statement.
(a) Car A is faster than car D
(b) Car B is the slowest
(c) Car D is faster than car C
(d) Car C is the slowest
Fig. 7.17 Time (s) Distance (m)
25. A particle is moving in a circular path of radius ‘r ‘. The displacement after half a circle would be (NCERT Exemplar)
Answers
Constructed Response Questions
Very Short Answer Questions (30-50 words)
1. Explain the differences between uniform and non-uniform acceleration with suitable examples.
Ans. Uniform acceleration occurs when an object’s velocity changes consistently over equal time intervals, while non-uniform acceleration features irregular changes. For instance, a freely falling object exhibits uniform acceleration under gravity, whereas a car in city traffic demonstrates non-uniform acceleration due to varying speeds and stops.
2. A motorcyclist drives from A to B with a uniform speed of 30 kmh-1 and returns back with a speed of 20 kmh-1. Find its average speed. (NCERT Exemplar)
Ans. Given,
Speed during onward journey = 30 kmh-1
Speed during onward journey = 20 kmh-1
Let the one-way distance be d km.
Time taken from A to B = d 30 hours,
Time taken from B to A = d 20 hours.
Total distance = 2d km,
Total time = d 30 + d 20 = 5d 60 = d 12
Thus,
3. Differentiate between speed and velocity.
Ans.
Speed is the rate of change of distance with respect to time. Velocity is the rate of change of displacement with respect to time.
It has only magnitude. It has both magnitude and direction. Speed does not depend on direction of motion. Velocity depends on the direction of motion. It is always positive or zero. It can be positive, negative, or zero depending on direction.
4. Define uniform circular motion and mention two real-life examples.
Ans. Uniform circular motion refers to movement along the circumference of a circle at a constant speed. Common examples of uniform circular motion are
(a) a car moving along a circular track at constant speed, and
(b) the moon revolving around the earth.
5. Provide one example for each of the following situations:
(a) An object moves such that the distance travelled and the displacement have the same magnitude.
(b) An object moves with constant speed but experiences non-zero acceleration.
Ans. (a) A car moving in a straight line from point A to point B without changing direction.
Since the path is straight and there is no change in direction, distance and displacement are equal.
(b) Moon moving in an orbit around the Earth at a constant speed. Although the speed of Moon remains constant, its direction of motion changes continuously, leading to a non-zero acceleration (centripetal acceleration).
6. Define the following:
(a) Uniform motion
(b) Uniform acceleration
Ans. (a) An object is said to be in uniform motion if it travels in a straight line and covers equal distances in equal intervals of time, regardless of how small those time intervals are.
(b) Acceleration of an object is said to be uniform if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time.
7. A ball is thrown vertically upward from a point P on the ground. It reaches the highest point Q and then falls back to the same point P.
(a) What are the net displacement and total distance travelled by the ball?
(b) At which point during the motion is the speed of the ball zero?
Ans. (a) The ball travels from point P to point Q and then back to point P. Since the ball returns to its initial position, its displacement will be zero. Distance travelled will twice the distance between position P and Q.
(b) The speed of the ball will be zero at maximum height or point Q.
8. A honeybee flies from point A to point B, and then takes a right turn to move towards point C (as shown in Fig. 7.18).
Draw the resultant displacement in the diagram.
If the distance AB = 3 m and BC = 4 m, calculate the magnitude of the resultant displacement.
Ans. Since, displacement is the shortest distance between two points, so AC, as shown in the diagram, will be the resultant displacement.
Using Pythagoras theorem
AC2 = AB2 + BC2
2222345 m ACABBC
A
7.18
7.19 C
9. Study the given velocity-time graph of two bodies A and B and answer the following questions.
(a) Which body starts from rest and which body has some initial velocity?
(b) Which body is having lesser acceleration?
Ans. (a) Body A starts from rest, as t = 0, v = 0 but body B has some initial velocity as t = 0.
(b) Acceleration is lesser for body A. Because v - t graph for B is steeper than A.
Fig. 7.20
Fig.
10. Two particles are moving with constant speed v, Such that they are always at constant distance d apart and their velocities are in opposite direction. After how much time, will they return to their initial positions?
Ans. The two particles moving with constant speed are always at a constant distance d, they must be at the two ends of diameter of a circular path as shown in Fig. 7.21. Each particle will return to its initial position after completing the circular path i.e. distance travelled by each particle = 2πr = πd.
Time = Distance Speed = πd v
Short Answer Questions (50-80 words)
1. The velocity-time graph (see Fig. 7.22) shows the motion of a cyclist. Find:
(a) its acceleration, (b) its velocity and
(c) the distance covered by the cyclist in 15 s.
7.21 v v d
(NCERT Exemplar)
Fig. 7.22
Ans. (a) From the given graph, it is clear that velocity is not changing with time, i.e. acceleration is zero. (b) As there is no change in the velocity with time, hence the velocity remain 20 m/s.
(c) Distance covered = Velocity × Time = 20 × 15 = 300 m
2. Differentiate between distance and displacement with example.
Ans.
Distance
Displacement
Total path length covered by a moving object. The shortest straight-line distance from the initial to final position.
It has only magnitude. It has both magnitude and direction. It is always positive. It can be positive, negative, or zero.
It depends on the path taken by the object. It is independent of path taken by the object but depends only on initial and final position.
Example: A ball is thrown vertically upward, rises to a height H, and then returns back to the ground.
Distance travelled = 2H
Displacement = 0 (same initial and final position)
Fig.
3. Identify the following graphs and answer the questions.
(a) What do you infer from the graph where velocity – time graph is parallel to the time axis?
(b) Which of the graphs indicate negative acceleration? Why?
(c) Which of the graphs represent a body moving with initial velocity not equal to zero but with constant acceleration? Justify your answer.
Ans. (a) Graph II: Since velocity is constant (uniform velocity), hence acceleration of the body is zero.
(b) Graph I: This is because velocity decreases with time.
(c) Graph III: This is because the graph does not pass through o, indicating a non-zero initial speed.
4. An object slides down an inclined plane with its velocity increasing uniformly from 10 cm/s to 15 cm/s within 2 seconds. Determine the acceleration of the object. Further, calculate the total distance it covers if it continues to slide down for 1 more second.
The object slides down with an acceleration of 2.5 cm/s2 and travels a distance of 41.25 cm down the incline in 3 seconds.
5. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in Fig. 7.25. Plot a velocity–time graph for the same. (NCERT Exemplar)
Fig. 7.23
Fig. 7.25
Fig. 7.24
Velocity
Velocity-time graph for this motion is shown in Fig.7.26.
6. An object is moving with a uniform speed in a circle of radius r. Calculate the distance and displacement,
(a) when it completes half the circle,
(b) when it completes full circle,
(c) what type of motion does the object possess?
Ans. (a) When an object completes half the circle, then the distance travelled by an object = 1 2 × circumference of a circle = 1 2 × 2πr = πr
Since the current position of the object is diametrically opposite to the initial position, therefore the displacement of object = 2r
(b) When an object completes full circle, then distance travelled by an object = 2πr
Displacement travelled by an object = 0
[since initial and final positions are same]
(c) Direction of motion of an object changes continuously in the circle, hence its velocity changes and its motion is accelerated motion.
7. How will the equations of motion for an object moving with a uniform velocity change?
(NCERT Exemplar)
Ans. We know that, the equations of uniformly accelerated motion are
(a) v = u + at
(b) s = ut + 1 2 at2
(c) v2 = u2 + 2as where, u = initial velocity, v = final velocity, a = acceleration, t = time and s = distance.
For an object moving with uniform velocity (velocity which is not changing with time), acceleration a = 0. So, equations of motion will become (putting a = 0 in above equations)
(a) v = u
(b) s = ut
(c) v2 = u2
8. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
Ans. Let u = initial velocity, v = final velocity, a = acceleration, t = time and s = distance. using 2nd equation of motion, s = ut + 1 2 at2
(NCERT Exemplar)
Distance travelled in 5 seconds, s5 = u × 5 + 1 2 a × 52 or s = 5u + 25 2 a
Similarly, distance travelled in 4 seconds, s4 = 4u + 16 2 a
Distance travelled in the interval between 4th and 5th seconds
= (s5 - s4) = (5u + 25 2 a) - (4u + 16 2 a)
= (u + 9 2 a) m
9. Consider the situation shown in Fig. 7.27.
(a) What is the position of a particle when it is at P1, and when it is at P2?
(b) Are the two positions the same?
(c) Are the two distances of the particle from the origin the same? Fig. 7.27
-1
Ans. (a) The position of the particle is x = 2 m when it is at P1 and x =-2 m when it is at P2.
(b) Two positions are not the same.
(c) The distance of the particles from the origin in the two positions are the same, and equal to 2 m.
10. (a) A body is moving with a velocity of 10 m/s. If the motion is uniform, what will be the velocity after 10 s?
(b) Can a body have constant speed but variable velocity? If yes, explain by giving an example.
Ans. (a) As the motion is uniform, the velocity remains 10 m/s after 10 s.
(b) Yes, a body in uniform circular motion has constant speed but due to the change in the direction of motion, its velocity changes at every point.
11. (a) When is the acceleration taken as negative?
(b) What is uniform acceleration?
(c) Give an example of non-uniform acceleration.
Ans. (a) Acceleration is taken as negative if direction of motion is opposite to the direction of velocity.
(b) Acceleration of an object is said to be uniform if it travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time. For example, motion of a freely falling body.
(c) A car travelling along a straight road increases its speed by unequal amounts in equal intervals of time.
12. The brakes applied to a car produce an acceleration of 6 ms-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.
Ans. Given,
a =-6 ms-2; t = 2 s and v = 0 ms-1
We know that
v = u + at
0 = u + (-6 ms-2) × 2 s
u = 12 ms-1
s = ut + 1 2 at2
= (12 ms-1) × (2 s) + 1 2 × (-6 ms-2) × (2 s)2 = 24 m - 12 m = 12 m
Thus, the car will move 12 m before it stops after the application of brakes.
Long Answer Question (80–120 words)
1. Deduce the following equations of motion.
(a) s = ut + 1 2 at2
(b) v2 = u2 + 2as
Ans. (a) Consider a body which starts with initial velocity u and due to uniform acceleration a, its final velocity becomes v after time t. Then, its average velocity is given by
v avg = initial velocity + final velocity 2 = u + v 2
Therefore, the distance covered by the body in time t is given by distance,
s = average velocity × time or s = u + v 2 × t
⇒ s = u + (u + at) 2 × t [∴ v = u + at]
∴ s = 2 ut + at2 2
s = ut + 1 2 at2
(b) Again using average velocity,
s = u + v 2 × t (i)
From 1st equation of motion, v = u + at
t = v - u a (ii)
Substituting the value of t in Eq. (i), we get
s =
∴ v2 = u2 + 2as
2. Suppose a squirrel is moving at a steady speed from the base of a tree towards some nuts. It then stays in the same position for a while, eating the nuts, before returning to the tree at the same speed. A graph can be plotted with distance on the y-axis and the time on x-axis.
Observe the graph in Fig. 7.28 carefully and answer the following questions.
(a) Which part of the graph shows the squirrel moving away from the tree?
(b) Name the point on the graph which is 6 m away from the base of the tree.
(c) Which part of the graph shows that the squirrel is not moving?
Fig. 7.28
(d) Which part of the graph shows that the squirrel is returning to the tree?
(e) Calculate the speed of the squirrel from the graph during its journey.
Ans. (a) Part AB
(b) Point B and C
(c) Part BC
(d) Part CD
(e) Total distance travelled = 6 m + 6 m = 12 m
Time = 11 s.
Speed = Distance Time = 12 11 = 1.09 m/s.
3. Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of : uu22 12 .
(Assume upward acceleration is -g and downward acceleration to be +g). (NCERT Exemplar)
Ans. Given,
initial speeds = u1 and u2 acceleration =-g
Using 3rd equation of motion, v2 - u2 = 2as
∴
But at highest point v = 0.
Therefore, h =
For first ball,
= and for second ball,
Thus,
4. The driver of train A travelling at a speed of 54 kmh-1 applies brakes and retards the train uniformly. The train stops in 5 seconds. Another train B is travelling on the parallel track with a speed of 36 kmh-1. Its driver applies the brakes and the train retards uniformly; train B stops in 10 seconds. Plot speed-time graphs for both the trains on the same axis. Which of the trains travelled farther after the brakes were applied?
Distance travelled by train A = Area under line RS = Area of DORS = 1 2 × OR × OS = 1 2 × 15 ms-1 × 5 s = 37.5 m
Distance travelled by train B = Area under line PQ
= Area of DOPQ
= 1 2 × OP × OQ = 1 2 × 10 ms-1 × 10 s = 50 m
Thus, train B travelled farther after the brakes were applied.
5. Give one example of each of the following situations:
(a) Uniformly accelerated motion
(b) Motion with uniform retardation
(c) Accelerated motion with uniform magnitude of velocity
(d) Motion in a direction with acceleration in a perpendicular direction
(e) Motion in which the velocity-time (v – t) graph is a horizontal line parallel to the X-axis Ans.
(a) A freely falling object (e.g., a stone dropped from a tower) accelerates uniformly due to gravity.
(b) An object thrown vertically upwards slows down uniformly due to the gravitational pull acting against its motion.
(c) A planet revolving around the sun in a circular orbit moves with constant speed but continuously changes direction, resulting in uniform acceleration.
(d) A bullet fired horizontally from a rifle experiences acceleration vertically downwards due to gravity, which is perpendicular to its horizontal motion.
(e) A car moving at constant speed along a straight road represents motion with constant velocity, hence its v - t graph is a horizontal line.
Competency Based Questions
Multiple Choice Questions (Choose the most appropriate option/options)
1. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal? (NCERT Exemplar)
(a) If the car moving on straight road
(c) The pendulum is moving to and from
(b) If the car is moving in circular path
(d) The earth is revolving around the sun
2. Which condition best distinguishes uniform acceleration from non-uniform acceleration?
(a) In uniform acceleration, the rate of change of velocity remains constant.
(b) Uniform acceleration implies zero net force on the object.
(c) Non-uniform acceleration shows equal changes in velocity over equal intervals.
(d) In non-uniform acceleration, the magnitude of velocity remains constant.
3. The velocity-time (v - t) graph of a moving particle is shown in Fig. 7.30. The acceleration is maximum for segment
(a) AB (b) BC
(c) CD (d) equal for all parts
4. Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms-1. It implies that the boy is (NCERT Exemplar)
(a) at rest.
(c) in accelerated motion.
7.30
(b) moving with no acceleration.
(d) moving with uniform velocity.
5. A students ties a stone to a thread of length 1 m and starts swinging it in a circular motion. The stone completes 200 rotations in 1 minute 40 seconds. With what speed the stone is moving?
(a) π m/s (b) 9π m/s (c) 4π m/s (d) 2π m/s
6. A car travels 13 km towards north then turns right and travels 5 km further, the car again turns right and travel 1 km and comes to rest. What is the distance travelled and displacement of the car?
(a) Distance : 19 km and displacement : 12 km
(b) Distance : 19 km and displacement : 13 km
(c) Distance : 12 km and displacement : 19 km
(d) Distance : 13 km and displacement : 19 km
7. Four cars A, B, C and D are moving on a levelled road. Their speed versus time graphs are shown in Fig. 7.31. Choose the correct statement.
(a) Car A always moves faster than car D.
(b) Car B has minimum acceleration.
(c) Car D travels a greater distance than car C.
(d) Car C has the travelled the maximum distance.
8. A person rides a motor bike at the speed of 30 m/s. The person applies the brake and the velocity of motor bike comes down to 10 m/s in 3 s. What is the magnitude of acceleration of motor bike?
9. A car moves towards South with a speed of 20 ms-1. It changes its direction to East and moves with speed of 20 ms-1. What is its velocity now?
(a) 20 ms-1 South-East
(b) 1 202 ms South-East
(c) 10 ms-1 Southh-East (d) 40 ms-1 South-East
10. Fig. 7.32 shows the velocity-time graph of a moving body, the total distance covered by the body from 2 s to 8 s is
(a) 28 m
(b) 56 m
(c) 14 m
(d) 35 m
11. The speed–time graphs of cars P and Q are shown in Fig. 7.33. Determine the time at which both cars will have covered the same distance.
(a) 3 sec
(b) 5 sec
(c) 6 sec
(d) 8 sec
7.31 Fig. 7.32
12. A car travels half the distance at a speed of v m/s and the rest of the journey at a speed of 2v m/s. what will be average speed of the car throughout the journey?
(a) 1.33 v m/s (b) 1.5 v m/s (c) 1.66 v m/s (d) 1.75 v m/s
13. Two balls A and B are thrown vertically upwards with speed of 20 m/s and 30 m/s respectively. What will be the ratio of maximum height achieved by ball A and B.
(a) 1:2 (b) 2:3 (c) 4:9 (d) 8:27
14. Consider the motion of two cars A and B.
Car A: u = 10 m/s and a = 0
Car B: u = 0 m/s and a = 5 m/s2
Assuming that they both start from same position, when will they meet again?
(a) 1 sec (b) 2 sec (c) 4 sec (d) 5 sec
15. Two balls X and Y, place 100 m apart, are moving towards each other with speeds of 10 m/s each. Balls X and Y experience acceleration of -2 m/s2 and 2 m/s2, respectively. If the balls collide at time t = t0, then speed of X at t = t0 s will be (a) 0 m/s (b) 10 m/s (c) 20 m/s (d) 30 m/s
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): Displacement can be zero even when an object covers a significant distance.
Reason (R): Because displacement depends on the net change in position rather than the total path length traveled.
2. Assertion (A): An object in uniform circular motion is always accelerating.
Reason (R): Even though its speed remains constant, its velocity changes continuously due to the change in direction.
3. Assertion (A): Average speed is always greater than or equal to average velocity.
Reason (R): The total distance is always equal to or greater than the magnitude of displacement.
4. Assertion (A): The acceleration of a body is equal to the slope of velocity – time graph.
Reason (R): Acceleration is the rate of change in displacement.
5. Assertion (A): Velocity incorporates both magnitude and direction, unlike speed.
Reason (R): Because velocity is defined as the rate of change of displacement, taking into account the object’s directional movement.
6. Assertion (A): Motion of moon around the earth is a circular motion.
Reason (R): In a circular motion, the distance of object from a fixed point remains the same.
7. Assertion (A): An odometer measures the total distance traveled by a vehicle.
Reason (R): It accumulates all movement irrespective of direction, thereby ignoring the concept of displacement.
1. Imagine a car with a leaky engine that drips oil at a regular rate of 50 drops per second. As the car travels through city, it would leave a trace of oil on the street that would reveal information about the motion of the car. The trace of the oil in four different situations are shown in Fig. 7.34. (The positions are given in centimeters.)
(a) Identify the type of motion by the car in situations A, B, C and D.
(b) What is the ratio of speed of the car in situations A, B and C.
(i) 1:2:4 (ii) 2:1:4
(iii) 4:2:1 (iv) 2:4:1
(c) What is the acceleration of the car in situation D?
2. If you consider the velocity of a car each second, you can see what this means clearly. You can see that the distance travelled by car during each second increases. It travels further (in the same time interval) because its velocity is greater.
A car maker claims their new car accelerates more quickly than any other new car. A rival car maker is not pleased by this claim and issue a challenge. Each car in turn is tested on a straight track with a speed recorder fitted. The results are shown in the table below. They show that the cars reach different speed after 6 seconds which means that one accelerates more than the other.
The results are plotted on the speed – time graph in Fig. 7.35. When an object changes, its speed is said to accelerate. If the object slows down this is often described as deceleration.
(a) Which car accelerates more in the first 6 seconds?
(b) What will be the acceleration if the velocity of car Y is uniformly increased from 18 m/s to 28 m/s in 10 – 14 seconds time interval.
(c) On the basis of the speed time graph plotted above, find the acceleration of the car X and Y in first 6 seconds.
7.35
3. A train travels from one station to the next. The driver of train A starts from rest at time t = 0 and accelerates uniformly for the first 20 s. At time t = 20 s, train reaches its top speed of 25 m/s, then travels at this speed for further 30 s before decelerating uniformly to rest. Total time for the journey of train A is 60 s. Another train B is travelling on the parallel of train A with zero initial speed at t = 0 and then accelerates uniformly for first 10 s. At time t = 10 s it reaches its top speed of 30 m/s, then travels at this speed for further 20 s before decelerating uniformly to rest. Total time for the journey of train B is 80 s.
(a) Draw a speed-time graph for the motion of train A.
(b) Calculate the deceleration of train A, as it comes to rest.
(c) In which time interval, speed of train B is constant?
(d) What is the initial speed of trains A and B?
Fig.
Fig.
Answers
Multiple Choice Questions
Assertion–Reason Based Questions
Case-Based/Source-Based/Passage-Based
Questions
1. (a) A, B and C show uniform motion. D shows non-uniform/uniform accelerated motion.
(b) ii) 2:1:4
(c) Time between each drop = 1 50 = 0.02 s
u = 0.02 0.02 = 1 m/s
v = 0.04 0.02 = 2 m/s
t = 7 × 0.02 = 0.14 s a = v - u t = 7.14 m/s2
2. (a) We can see that in the first 6 seconds, the speed of Y increases more than the speed of X So, Y accelerates more than X in the first 6 seconds.
(b) Acceleration, a = Change in velocity Time interval 281810 44 vu a t ===
= 2.5 m/s2
(c) For car X, a = 21 21 ( ) 15015 6 ) 60 ( vv tt === 2.5 m/s2
Therefore, a = 2.5 m/s2 For car Y, a = 21 21 ( ) 18018 6 () 06 vv tt === 3 m/s2 Therefore, a = 3 m/s2
3. (a) Speed-time graph for the motion of train A is shown in Fig. 7.36
(b) Given, change in velocity = 25 m/s Time interval (for deceleration) = 10 s
Deceleration = Change in velocity Time interval = 25 10 = 2.5 m/s2
Thus, deceleration of train A is 2.5 m/s2.
(c) Speed of train B is constant during time interval 10 s to 30 s, i.e. 20 s.
(d) Initial speed of trains A and B is zero as both trains start from rest.
(s)
Fig. 7.36
Numerical Questions
1. An engine increases the speed of a motorcycle from 5 m/s to 15 m/s in 5 seconds. Calculate its acceleration.
Ans. Given,
Initial velocity, u = 5 m/s
Final velocity, v = 15 m/s
Time, t = 5 s
Acceleration, a = (v - u) t
a = (15 - 5) 5 = 10 5 = 2 m/s2
2. A bus starting from rest accelerates uniformly to reach 20 m/s in 10 seconds. Compute its acceleration and the distance covered in this time.
Ans. Given,
Initial velocity, u = 0
Final velocity, v = 20 m/s
Time, t = 10 s
Acceleration, a = (v - u) t
a = (20 - 0) 10 = 2 m/s2
Distance, s = ut + 1 2 at2
s = 0 + 1 2 × 2 × (10)2 = 100 m
3. A cyclist covers 30 m in the first 5 seconds and then 70 m in the next 5 seconds. Determine average acceleration for the first interval and the second interval.
Ans. First 5 seconds: s = 30 m, t = 5 s, u = 0
s = 1 2 at2
30 = 0.5 × a × 25
a = (30 × 2) 25 = 2.4 m/s2
Final speed after 5 s: v = u + at = 0 + 2.4 × 5 = 12 m/s
Next 5 seconds: s = 70 m, t = 5 s, u = 12 m/s
s = ut + 1 2 at2
70 = 12 × 5 + 0.5 × a’ × 25
a’ = (70 - 60) 12.5 = 0.8 m/s2
4. A car decelerates uniformly from 25 m/s to 0 m/s in 5 seconds. Calculate stopping distance.
Ans. Given,
Initial velocity, u = 25 m/s
Final velocity, v = 0 m/s
Time, t = 5 s
We know that, a = (v - u) t = (0 - 25) 5 =-5 m/s2
Using 3rd equation of motion, v2 = u2 + 2as
0 = (25)2 + 2(-5)s
s = 625 10 = 62.5 m
5. An object covers 20 m in first 5 s, 40 m in next 5 s, and 60 m in final 5 s. Find average speed.
Ans. Total distance: 20 + 40 + 60 = 120 m
Total time: 5 + 5 + 5 = 15 s
Average speed = Total distance Total time = 120 15 = 8 m/s
6. A car starts from rest with acceleration 5 m/s2 for 8 s, then moves with constant speed. Find total distance travelled in 12 s.
Ans. Given: u = 0, a = 5 m/s2, t1 = 8 s, t2 = 4 s
Final speed after 8 s
v = u + at = 0 + 5 × 8 = 40 m/s
Distance in first 8 s, s1 = 1 2 × 5 × (82) = 160 m
Distance in next 4 s, s2 = 40 × 4 = 160 m
Total distance = s1 + s2 = 160 + 160 = 320 m
7. An object is sliding down an inclined plane. The velocity changes at constant rate from 20 cm/s to 40 cm/s in 2 s. What is its acceleration?
Ans. Given, v = 40 cm/s, u = 20 cm/s, t = 2 s
a = v - u t
a = 40 - 20 2 cm/s2
a = 10 cm/s2
8. An electron moving with a velocity of 5 × 104 m/s enters into a uniform electric field and acquired uniform acceleration 104m/s2 in the direction of the initial motion. (NCERT Exemplar)
(a) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(b) How much distance an electron would cover in this time?
Ans. Given, initial velocity of the electron, u = 5 × 104 m/s
Acceleration, a = 104 m/s2
(a) Final velocity, v = 2u = 2 × 5 × 104 m/s2
= 105 m/s
using v = u + at t = v - u a
= 10 × 104 - 5 × 104 104 = 5 × 104 104 = 5 s
(b) Using s = ut + 1 2 at2
= (5 × 104) × 5 + 1 2 × 104 × (5)2
= 25 × 104 + 25 2 × 104
= 37.5 × 104 m
9. A boy is moving 3 m North, then 4 m East and finally 6 m South. Calculate the distance travelled and the displacement.
Ans. As shown in the Fig. 7.37, the boy starts from O and moves 3 m North (= OA), 4 m East (= AB) and 6 m South (= BC)
Distance travelled = OA + AB + BC
= 3 m + 4 m + 6 m = 13 m
Displacement = OC, where, OC = hypotenuse of DODC
OC 22
22 22 () 4(63)5 m ODDC ABBCBD =+
10. A car moves with a speed of 30 km/h for half an hour, 25 km/h for one hour and 40 km/h for two hours. Calculate average speed of the car.
Ans. Time taken to travel, t1 = 0.5 h, t2 = 1 h, t3 = 2 h
Total time, t = t1 + t2 + t3 = 0.5 + 1 + 2 = 3.5 h
Speeds, v1 = 30 km/h, v2 = 25 km/h, v3 = 40 km/h
Distances,
s1 = v1t1 = 30 × 0.5 = 15 km
s2 = v2t2 = 25 × 1 = 25 km
s3 = v3t3 = 40 × 2 = 80 km
Total distance, s = s1 + s2 + s3 = 15 + 25 + 80 = 120 km
Average speed = Total distance Total time = s t = 120 3.5 = 34.3 km/h
11. A ball is gently dropped from a height of 45 m. If its velocity increases uniformly at the rate of 10 ms-2, with what velocity will it strike the ground? After what time, will it strike the ground?
Ans. Initial velocity of ball, u = 0
Height, h = 45 m, Acceleration, a = 10 ms-2 ,
Final velocity, v = ?, Time, t = ?
We know that,
∴ final velocity, v = 30 ms-1
v2 = u2 + 2as s = h
v2 = 0 + 2 × 10 × 45 = 900 m/s
v2 = 900 m/s
v = u + at
30 = 0 + 10 × t t = 30 10 = 3 s
∴ time of motion = 3 s
12. The velocity-time graph shows the motion of a cyclist. Find
(a) its acceleration (b) its velocity and (c) the distance covered by the cyclist in 15 seconds.
(NCERT Exemplar)
Ans. (a) Since velocity is not changing acceleration is equal to zero. 0 v a
(b) Reading the graph, velocity = 20 ms-1 (constant)
(c) Distance covered in 15 seconds = Area of ABNO
= v × t
= 20 × 15 = 300 m Fig. 7.38
13. A cheetah is the fastest land animal and can achieve a peak velocity of 100 km/h up to distances less than 500 m. If a cheetah spots its prey at a distance of 100 m, what is the minimum time it will take to catch its prey, if the average velocity attained by it is 90 km/h?
Ans. Average velocity = 90 km/h = 90 km 1 h = 90 × 1000 m 60 × 60 s = 25 ms-1
Also, Average velocity = Displacement Time Taken
Therefore, Cheetah’s displacement is equal to 100 m.
Time taken = 100 25 = 4s
14. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time? (NCERT Exemplar)
Ans. Initial difference in height = (150 - 100) m = 50 m
Distance travelled by first body in 2 s, h1 = 0 + 1 2 g(2)2 = 2g
Distance travelled by another body in 2 s, h2 = 0 + 1 2 g(2)2 = 2g
After 2 s, height at which the first body will be h′ 1 = 150 - 2g
After 2 s, height at which the second body will be = h′ 2 = 100 - 2g
Thus, after 2 s, difference in height = (150 - 2g) - (100 - 2g) = 50 m = initial difference in height
Thus, difference in height does not vary with time.
15. A particle moves in a circle with O as centre and AO = OB = 5 cm, as shown in the Fig. 7.39. It starts from A. Calculate:
(a) the distance covered, and (b) the displacement, when it reaches B.
Ans. (a) Distance covered = π × OA = π × 5 = 5π cm
(b) Displacement = AB = 2 × OB = 2 × 5 = 10 cm
The displacement is 10 cm along AB.
Practice Questions
Multiple Choice Questions
7.39
1. A bridge is 400 m long. A 150 m long train crosses the bridge at a speed of 50 m/s. Time taken by the train to cross it.
(a) 5 s (c) 6 s (b) 8 s (d) 11 s
2. When two bodies moves uniformly towards each other, then they cross each other at the speed of 10 m/s. If both the bodies move in the same direction, then they cross each other at the speed of 6 m/s. The speed of both bodies are
(a) 8 m/s, 2 m/s
(b) 6 m/s, 2 m/s (c) 8 m/s, 4 m/s
(d) 6 m/s, 4 m/s
3. If a body is moving on a circular path of radius 21 cm with velocity of 2 m/s, then time taken by the body to complete half revolution is
(a) 11 s (b) 44 s
(c) 22 s (d) 33 s
Fig.
4. A body moving along a circular path has (a) a constant speed (b) a constant velocity (c) zero acceleration (d) none of these
5. The graph in Fig. 7.40 shows the rate of change of speed of a train. How much distance the train will cover at 60 seconds if it continues to accelerate with same acceleration?
(a) 8500 m
(b) 4500 m
(c) 6000 m
(d) 2000 m
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
6. Assertion (A): Displacement of body may be zero, when distance travelled by it is not zero.
Reason (R): The displacement is the longer distance between initial and final positions.
7. Assertion (A): A body can have acceleration even its speed is constant.
Reason (R): In uniform circular motion, speed of body is constant but its velocity continuously changes.
Very Short Answer Questions (30-50 words)
8. Study the given velocity-time graph (Fig. 7.41) of two bodies A and B and answer the following questions.
(a) Which body starts from rest and which body has some initial velocity?
(b) Which body is having lesser acceleration?
9. A honey bee travels from A to B and then moves from B to C (as shown in Fig. 7.42). If AB = 4 m and BC = 3 m, then calculate the displacement of bee. Also, show the resultant displacement in the diagram. B A C
Fig. 7.42
(s) Fig. 7.40 Fig. 7.41
10. What are the characteristics of distance-time graph for an object moving with a non-uniform speed.
11. Suppose a ball is thrown vertically upwards from a position P above the ground. It rises to the highest point Q and returns to the same point P. If PQ = 25 m, then what is the net displacement and distance travelled by the ball?
12. A body starts to slide over a horizontal surface with an initial velocity of 0.5 m/s. Due to friction, its velocity decreases at the rate 0.5 m/s2. How much time will it take for the body to stop?
Short
13. The velocity-time graph of a body is shown in Fig. 7.43
(a) State the kind of motion represented by OP and PQ.
(b) Find the velocity of the body after 5 s.
(c) What is the negative acceleration of the body?
(d) Find the distance travelled between 10th and 30th s.
14. (a) State a condition under which a body moves in such a way that the magnitude of its average velocity is equal to its average speed.
(b) A train starting from rest moves with a uniform acceleration of 0.2 m/s2 for 5 minutes. Calculate the final velocity acquired and the distance travelled in this time.
15. A car acquires a velocity of 72 km per hour in 10 seconds starting from rest. Find (a) the acceleration, (b) the distance travelled in this time, and (c) the average velocity.
16. (a) What is uniform acceleration? Write the equations of motion for uniformly accelerated motion.
(b) If a car accelerates uniformly from rest and covers 100 m in 10 seconds, calculate its acceleration.
17. A ball is thrown vertically upward with a velocity of 15 m/s. Calculate:
(a) the time taken to reach the highest point, (b) the maximum height attained by the ball, and (c) the total time taken to return to the thrower’s hand. (Take g = 10 m/s2)
Long Answer Questions (80-120 words)
18. Study the speed-time graph of a body given in Fig. 7.44 and answer the following questions
(a) What type of motion is represented by PQ?
(b) What type of motion is represented OP?
(c) What type of motion is represented QR?
(d) Calculate the retardation of the body?
(e) Find the distance travelled by the body from P to Q.
19. Account for the following:
7.44
(a) Name the quantity which is measured by the area occupied below the velocity-time graph.
(b) An object is moving in a certain direction with an acceleration in the perpendicular directions.
(c) Under what condition is the magnitude of average velocity of an object equal to its average speed?
(d) An example of uniformly accelerated motion.
(e) A body is moving along a circular path of radius (R). What will be the distance and displacement of the body when it completes half revolution?
20. Figure 7.45 illustrates a velocity-time (v - t) graph of a car moving along a straight path. Examine the graph and determine the following:
(a) What is the acceleration of the car during the time interval AB?
(b) What is the velocity of the car at point C?
(c) How much distance does the car travel between 6 and 8 seconds?
(d) What is the average velocity for the entire journey?
Additional Numericals
1. The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip took 8 hours, calculate the average speed of the car in kmh-1 and ms-1 .
2. A train moving with velocity of 54 kmh-1 is accelerated so that its velocity becomes 72 kmh-1 in 15 seconds. Find the acceleration of the train.
3. A car travelling at a velocity of 10 ms-1 due north speeds up uniformly to a velocity of 25 ms-1 in 5 s. Calculate the acceleration during this period.
4. The brakes applied to a car produce a negative acceleration of 6 ms-2. If the car takes 2 seconds to stop after applying the brakes, calculate the distance it travels during this time.
5. A powerful motorcycle can accelerate from rest to 28 m/s in only 4 s.
(a) What is its average acceleration?
(b) How far does it travel in that time?
(CBSE 2016)
6. Name a device that measures distance travelled by automobiles. A body travels a distance of 15 m from A to B and then moves a distance of 20 m at right angle to AB. Calculate the total distance travelled and the displacement.
7. A boy runs for 10 min at a uniform speed of 9 km/h. At what speed should he run for the next 20 min so that the average speed comes to 12 km/h?
8. A cyclist completes one round of a circular track of radius 50 m in 40 seconds. Calculate:
(a) the cyclist’s speed, (b) the displacement after half a round.
9. A ball is thrown vertically upwards with a velocity of 30 m/s. Calculate:
(a) the maximum height reached, and (b) the time taken to reach the maximum height. (Use g = 10 ms-2)
10. On a day when there is no wind, an athlete runs a 100 m race in 14.2 s. A sketch graph showing her speed during the race is given in Fig. 7.46.
(a) Calculate the acceleration of the athlete during the first 3.0 s of the race.
(b) Calculate the speed with which she crosses the finishing line.
(c) Suggest two differences that might be seen in the graph if there had been a strong wind opposing the runners in the race.
(CBSE 2014)
Brain Charge
1. Crossword Puzzle Across
4. The total length of the path traveled by an object during motion
5. The rate of change of displacement with time
7. The quantity representing how fast an object moves regardless of direction
8. The rate of change of velocity of an object
9. A quantity having only magnitude, like speed Down
1. A physical quantity having both magnitude and direction
2. The property of an object to resist change in its state of motion
3. Acceleration that remains constant in magnitude and direction
4. The shortest straight-line distance between initial and final positions
6. Motion of an object along a circular path at constant speed
2. Word Puzzle
Use the first alphabet of each answer to the puzzle and combine them to spell a word.
1. A measure of the amount of matter in an object.
2. The curved path followed by a moving object, like a planet.
3. The measure of how long an event lasts.
4. The property of an object to resist any change in its state of motion.
5. Repetitive to-and-fro motion around a mean position.
6. The scientist who formulated laws describing motion.
Challenge Yourself
1. A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?
2. Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h-1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s2. If after 50s, the guard of B just brushes past the driver of A, what was the original distance between them?
3. The speed-time graphs of two cars are represented by P and Q as shown in Fig. 7.47:
(a) Find the difference in the distance travelled by the two cars after 4 s?
(b) At what time, P moves with the thrice the speed of Q?
(c) What type of motion cars P and Q are undergoing?
Answers
Practice Questions
1.
(s)
Fig. 7.47
Scan me for detailed Solutions
13. (a) OP represents uniform acceleration, PQ represents constant velocity.
(b) Velocity after 5 s = 10 ms-1 . Velocity after 40 s = zero.
(c) Negative acceleration = Slope of QC =-2 ms-2
(d) 400 m
15. (a) 2 m/s2 (b) 100 m (c) 10 m/s 16. (b) 2 m/s2 17. (a) 1.5 s (b) 11.25 m (c) 3 s
20. (a) 4 m/s2 (b) 16 m/s (c) 40 m (d) 144 m
Additional Numericals
1. 50 km h-1, 13.9 ms-1 2. 3 ms-1, 0 ms-1 3. 0.33 ms-2 4. 3 ms-2 5. 12 m
6. (a) 7 ms-2 (c) 56 m 7. 25 m 8. 13.5 km/h 9. (a) 7.85 m/s (b) 1.23 m/s-2 (c) 100 m
10. (a) 2.67 m/s2 (b) 8 m/s
Brain Charge
1. Crossword Puzzle
2. Word Puzzle Motion
1. Mass 2. Orbit
3. Time
4. Inertia
5. Oscillation
6. Newton
Challenge Yourself
1. The jeep will catch up with the bike 1.0 km away from the turning.
2. 1250 m
3. (a) 0 (b) t = 6 s.
(c) Car P and Q undergoes uniform acceleration and uniform motion respectively.
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 1.5 Hour Max. Marks: 50
Multiple Choice Questions
(10 × 1 = 10 Marks)
1. An object moving along a straight line exhibits an acceleration that increases linearly with time. Which of the following best describes its displacement after time t?
(a) Displacement remains constant as acceleration only affects velocity.
(b) Displacement follows a quadratic dependence on time due to the integration of the time-dependent acceleration.
(c) Displacement is directly proportional to time.
(d) Displacement is independent of acceleration if initial velocity is zero.
2. The graph in Fig. 7.48 shows the distance travelled by a car and the time taken by the car. Between which points the car travels the fastest? (CBSE QB)
(a) B to C (b) C to D (c) A to B (d) E to F
Fig. 7.48
3. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal? (NCERT Exemplar)
(a) If the car is moving on a straight road.
(b) If the car is moving in a circular path.
(c) The pendulum is moving to and from.
(d) The Earth is revolving around the Sun.
4. A ball is thrown vertically upwards with speed u. The ball goes up and then falls back. What will the speed – time graph of motion looks like?
(a) A curved line
(b) A straight line with a constant slope.
(c) Two straight lines forming a -shape.
(d) Two straight lines forming a V-shape.
5. Which statement best distinguishes speed from velocity?
(a) Speed can be negative, whereas velocity is always positive.
(b) There is no difference between speed and velocity.
(c) Speed is a scalar quantity, while velocity is a vector that includes direction.
(d) Velocity is always greater than speed.
6. For an object in uniform circular motion with constant speed, which of the following is true about its acceleration?
(a) It is directed tangentially to the circle.
(b) It is directed towards the centre of the circle, known as centripetal acceleration.
(c) It alternates between zero and non-zero values.
(d) It is zero because speed is constant.
7. In which of the following scenarios the velocity of a uniformly accelerating object can never be zero?
(a) u = 10 m/s and a =-2 m/s2.
(b) u =-5 m/s and a = 2 m/s2.
(c) a = 2 m/s2 and s = 0 at t = 4 sec.
(d) a = 2 m/s2 and s = 6 at t = 4 sec.
8. Identify the incorrect statement about an object in uniform circular motion with radius r:
(a) The speed remains constant.
(b) The object experiences a non-zero acceleration.
(c) The direction of acceleration does not change.
(d) The maximum displacement has a magnitude of 2r
9. The sound of thunder is heard approximately 6 s after the flash of lightning is seen. If speed of sound in air is 346 m/s, then distance of point of lightning is
(a) 1200 m
(b) 2076 m
(c) 3460 m
(d) 965 m
10. Fig. 7.49 shows v – t graph of two cars P and Q. The ratio of distance travelled by P and Q in 4 s is
7.49
(a) 2:1
(b) 3:2
(c) 4:3
(d) 3:4
Fig.
Assertion-Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false. (d) A is false, but R is true.
11. Assertion (A): An object moving with uniform velocity has zero acceleration.
Reason (R): Acceleration is the rate of change of velocity, and if velocity is constant, there is no change in velocity.
12. Assertion (A): Average speed and average velocity are always equal in motion along a straight line. Reason (R): Since both are based on distance and displacement respectively, they yield the same value when there is no change in direction.
13. Assertion (A): Moon is accelerating towards the Earth.
Reason (R): Motion of moon around the earth is a circular motion.
14. Assertion (A): An object can change velocity without changing speed.
Reason (R): Velocity can only be changed by changing the speed.
15. Distance travelled by a train and time taken by it is shown in the following table Time 10:00 am 10:30 am 10:40 am 11:00 am 11:15 am 11:30 am
(a) Plot distance-time graph.
(b) When is the train travelling at the highest speed?
(c) What is the average speed of the train?
(d) At what distance does the train slow down?
16. Two drivers, Driver X and Driver Y, are participating in a race. The velocity-time graph of both drivers is shown in Fig. 7.50.
Driver X accelerates uniformly for 10 seconds, then maintains a constant velocity for 20 seconds, and finally decelerates uniformly to rest over the next 10 seconds.
Driver Y accelerates uniformly to a higher maximum velocity in 20 seconds, then decelerates uniformly back to rest over the next 20 seconds.
Using the velocity-time graph
(a) What is the accelerations of Driver X and Driver Y in first 20 secs?
(b) Calculate the Distance travelled by Driver X in 30 s.
(c) Calculate the average speed of Driver Y over the entire 40 seconds. Fig. 7.50
Very
Short Answer Questions (30-50 words)
17. A food packet is dropped from a plane going at an altitude of 500 m. how long will it take to reach the ground ?
18. What is acceleration? State its SI unit.
19. Fig. 7.51 shows distance – time graphs of two objects A and B. Which object is moving with a greater speed when both are moving? Justify your answer.
Short
Answer Questions (50-80 words)
20. Velocity-time graph for the motion of the car is shown in Fig. 7.52. What does the nature of the graph show? Also find the acceleration of the car.
21. Explain the difference between speed and velocity with examples.
22. A stone dropped in a well hits the water surface after 2 s. What is the depth of the well and with what speed will the stone hit the water surface? Assume the uniform acceleration due to gravity to be 9.8 ms-2
23. What is displacement? How is it different from distance? Illustrate with an example.
Long Answer Questions (80-120 words)
(3 × 2 = 6 Marks)
(4 × 3 = 12 Marks)
(2 × 5 = 10 Marks)
24. Rohit is riding his bicycle in a straight path. He starts from rest and accelerates uniformly for 6 seconds, reaching a speed of 12 m/s. Then, he continues at this constant speed for 10 seconds before applying brakes and coming to rest uniformly in 4 seconds.
(a) What is the acceleration of Rohit’s bicycle during the first 6 seconds?
(b) How far does Rohit travel while accelerating?
(c) Calculate the deceleration when Rohit applies the brakes. Find the total distance travelled by Rohit during the entire trip
25. A roller coaster starts from rest at the top of a hill 40 m high and moves down a slope, accelerating uniformly. It takes 5 seconds to reach the bottom of the hill. Assuming no friction or air resistance, calculate:
(a) the acceleration of the roller coaster,
(b) its speed at the bottom, and
(c) how long it would take to climb back up the same hill if it decelerates uniformly at half the magnitude of the acceleration during descent.
Fig. 7.51
Fig. 7.52
8 Force and Laws of Motion
Force and motion are important ideas in science that help us understand how things move. A force is a push or a pull, and it can make things move, stop or change direction. There are three important rules about motion, together called Newton�s Laws. These rules explain many things we see every day, such as why a ball rolls when kicked or why a car stops when we apply brakes.
Balanced Forces
Net effect produced by a number of forces is zero, resulting in no change in the state of rest or motion.
Unbalanced Forces
Net effect produced by a number of forces is non-zero, resulting in a change in speed or direction of motion.
Force
Any push, pull or hit which changes or tends to change the position/ direction/state of rest or uniform motion/shape or size of the body.
Motion
Continuous change in the position of the body with time relative to a reference point.
Newton�s Laws of Motion
Newton presented three fundamental laws that govern the motion of objects.
Newton�s 1st Law
A body continues in its state of rest or of uniform motion in a straight line unless an external force acts on it.
Inertia
Property of a body by virtue of which it opposes any change in its state of rest or uniform motion in a straight line.
Inertia of Rest
Due to this inertia, a body at rest tends to remain at rest.
Inertia of Motion
Due to this inertia, a body in uniform motion tends to continue its motion.
Newton�s 2nd Law
The rate of change of momentum of a body is directly proportional to the force applied and the change in momentum is along the direction of the applied force (or F = ma).
Newton�s 3rd Law
To every action, there is an equal and opposite reaction; but the action and reaction forces act on different bodies, simultaneously.
Mass
It is the measure of an object�s inertia.
Momentum
It is the product of mass and velocity (p = m.v).
Chapter at a Glance
• A force can change
º the state of rest or uniform motion of an object.
º the direction and speed of a moving object.
º the shape and size of an object
• Balanced forces cannot change the state of rest or uniform motion.
• By observing the motion of objects on an inclined plane, Galileo deduced that objects move with a constant speed when no force acts on them
• Newton studied Galileo’s ideas on force and motion, and presented three fundamental laws that govern the motion of objects. These three laws are known as Newton’s laws of motion.
• First law of motion: An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.
• The natural tendency of objects to resist a change in their state of rest or of uniform motion is called inertia.
• The mass of an object is a measure of its inertia. Its SI unit is kilogram (kg).
• Larger the mass, greater is the inertia.
• Force of friction always opposes the motion of objects.
• The momentum of an object is the product of its mass and velocity, and has the same direction as that of the velocity. Its SI unit is kg ms–1
• Second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
F ∝ m(v u) t
• The second law of motion gives us a relation between force, mass and acceleration.
F = m × a
• The SI unit of force is kg ms–2. This is also known as newton and represented by the symbol N.
• A force of one newton produces an acceleration of 1 ms–2 on an object of mass 1 kg.
1 N = 1 Kg 1 ms–2 = 1 kg.ms–2
• Third law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies.
• To calculate Net Force, add all the forces that are in the same direction and then subtract the forces that are in the opposite direction.
• No force is required to move a body with constant velocity on a smooth surface.
Formulae
1. Momentum (p) = m × v
2. Force (F) = t pfinal − pinitial = m(v u) t = m × a
3. Equations of motion:
(a) v = u + at
(b) s = ut + 1 2 at2
(c) 2as = v2 − u2
NCERT Zone
Intext Questions
1. Which of the following has more inertia:
(a) a rubber ball and a stone of the same size?
(b) a bicycle and a train?
(c) a five-rupees coin and a one-rupee coin?
Ans. Objects with larger mass have a greater inertia.
(a) Out of a rubber ball and a stone of the same size, the stone has greater mass; hence, it has more inertia than the rubber ball.
(b) Out of a bicycle and a train, the train has greater mass; hence, it has more inertia than the bicycle.
(c) Out of a five-rupee coin and a one-rupee coin, the five-rupee coin has greater mass; hence, it has more inertia than the one-rupee coin.
2. In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case.
Ans. Whenever a player interacts with the football, they apply a force to it. In the given example, three agents/players interact with the football four times:
(a) When a player kicks the ball to their teammate, the velocity changes for the first time.
(b) When another player from the same team kicks the football towards the goal, the velocity changes for the second time.
(c) When the goalkeeper catches the ball, the velocity changes for the third time.
(d) The velocity changes for the fourth time when the goalkeeper kicks the football towards a player of his own team.
3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Ans. When the branch is shaken, it moves suddenly, but the leaves momentarily resist this motion due to their inertia of rest. As a result, the force applied causes the leaves to detach from the branch.
4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Ans. When we are inside a moving bus, our speed is equal to the speed of the bus. As the bus slows down and comes to a stop, the lower part of our body, in contact with the bus, comes to rest. However, the upper part of our body, due to inertia of motion, continues to move forward. As a result, we tend to fall in the forward direction. Conversely, when the bus accelerates from rest, our feet start moving, but the upper part of our body remains stationary due to inertia of rest. Therefore, we fall backwards.
NCERT Exercises
1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on.
Ans. Yes, it is possible for an object to move with non-zero velocity while experiencing a net zero external force. If the external force experienced by a moving body is equal in magnitude and opposite in direction to the frictional force exerted by the ground, then the object will move with a constant velocity without any change in the direction of motion.
2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Ans. When a carpet is beaten with a stick, the carpet experiences a sudden force and starts moving.
However, the dust particles on the carpet momentarily resist this motion due to their inertia of rest. As a result, the dust particles separate from the moving carpet and fall off.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Ans. When a bus starts suddenly, the lower part of the luggage on the roof, being in contact with the bus, moves forward with the bus�s speed. However, the upper part, due to inertia of rest, tends to stay stationary. This causes the upper part to lag behind, making the luggage fall backwards. Therefore, it is recommended to secure any luggage kept on the roof of a bus with a rope.
4. A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(a) the batsman did not hit the ball hard enough.
(b) velocity is proportional to the force exerted on the ball.
(c) there is a force on the ball opposing the motion.
(d) there is no unbalanced force on the ball, so the ball would want to come to rest.
Ans. (c) there is a force on the ball opposing the motion.
When a ball rolls on the ground, it experiences a frictional force exerted by the ground that acts in the direction opposite to its motion. This opposing force gradually reduces the ball�s speed, eventually bringing it to a stop.
5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Ans. Given,
Mass (m) = 7 tonnes or 7 × 1000 kg = 7000 kg
Distance (s) = 400 m
Time (t) = 20 sec
Since the truck starts from rest, therefore initial speed (u) = 0;
Let’s calculate the acceleration of the truck, using the distance-time formula,
s = ut + 1 2 at2
400 = 0 × 20 + 1 2 × a × 202
400 = 1 2 × a × 202
a = 2 × 400 202
Therefore, a = 2 ms 2
Now,
To calculate the force acting on the truck, we can use the formula F = m a
Putting the values of mass and acceleration, we get
F = (7000 × 2) kg.ms−2 = 14,000 kg.ms−2 or 14,000 N
6. A stone of 1 kg is thrown with a velocity of 20 ms-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Ans. Given,
Mass (m) = 1 kg
Initial velocity (u) = 20 ms 1
Distance travelled (S) = 50 m
Since the stone finally comes to rest, therefore, final velocity (v) = 0 ms 1
To calculate acceleration (a), we can use the formula
2.a S = v2 – u2
Putting the given values in this formula, we get
2.a.50 = 02 – 202
100.a = –400
a = –4 ms–2
Now, to calculate the force of friction we can use the formula
F = m.a
F = 1 kg (–4) ms–2 = –4 kg.ms–2 or –4 N
(Important: Negative sign indicates that the direction of frictional force is opposite to the direction of motion.)
7. A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:
(a) the net accelerating force, and (b) the acceleration of the train.
Ans. Given,
Force exerted by engine (Fengine) = 40,000 N
Force of friction (Ffriction) = 5,000 N
Total Mass (m) = mass of engine + mass of each wagon × number of wagons m = 8000 kg + 2000kg × 5 = 18,000 kg
Net accelerating force or Net force (Fnet) = Fengine – Ffriction = 40,000 N – 5,000 N = 35,000 N
Now,
acceleration of the train = Fnet m = 35000 N 18000 kg = 1.94 m/s2
8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms–2 ?
Ans. Given,
Mass of vehicle (m) = 1500 kg
Acceleration (a) = –1.7 ms–2
Force required to stop the vehicle = m.a = 1500 kg × (–1.7 ms–2) = –2550 kg ms–2 or –2550 N
(Note: Negative sign indicates that the direction of force is opposite to the direction of motion.)
9. What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)2 (b) mv2 (c) 1 2 mv2 (d) mv
Ans. (d) mv
Momentum of an object is the product of its mass and velocity.
10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Ans. In order for the wooden cabinet to move across the floor at a constant velocity, net force (Fnet) acting on the wooden cabinet should be zero.
Fnet = 0 = Fexternal – Ffriction
Therefore,
Ffriction = Ffriction = 200 N
11. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Ans. The student�s logic is incorrect because action and reaction forces act on different bodies, not the same one.
The actual reason the truck does not move is because the truck has a large mass, giving it significant inertia. Additionally, since it is parked on the ground, a frictional force exists between the truck and the surface. The force exerted by a person on the truck is insufficient to overcome this frictional force. As a result, the forces acting on the truck—the applied force and the opposing frictional force—balance each other. This balance of forces results in a net force of zero, which prevents the truck from moving.
12. A hockey ball of mass 200 g travelling at 10 ms–1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms–1. Calculate the magnitude of change of momentum that occurred in the motion of the hockey ball by the force applied by the hockey stick.
Ans. Taking the right direction as positive, we get the velocities of the ball before and after the strike.
Vbefore = +10 ms–1 Vafter = –5 ms–1
Since the given mass of the ball is 200g or 0.2 kg, Therefore,
Momentum of ball before strike (pbefore) = m.Vbefore = 0.2 kg × (+10 ms–1) = 2 kg.ms–1
Momentum of ball after strike (pafter) = m Vafter = 0.2 kg × (–5 ms–1) = –1 kg.ms–1
13. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Ans. Given,
Mass of bullet (m) = 10g or 0.01kg
Initial velocity (u) = 150 ms-1
Time to stop (t) = 0.03 s
Since the bullet finally stops, therefore, final velocity (v) = 0 ms-1
To calculate distance travelled by the bullet, we first need to calculate acceleration (a) using u, v and t.
Using the first equation of motion, v = u + a t
Putting the values in this equation, we get
0 = 150 + a × 0.03
a = –150 0.03 ms-2 = –5000 ms-2
Now, to calculate the distance travelled by the bullet, we can use the third equation of motion, i.e.
2.a S = v2 – u2
Putting the values, we get 2 × (–5000 ms-2) × S = 02 – (150 ms-1)2
Fig. 8.1
On simplifying,
S = 22500 10000 m = 2.25 m
14. An object of mass 1 kg travelling in a straight line with a velocity of 10 m s–1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object. Ans. Given,
Mass of object (mo) = 1 kg
Mass of wooden block (mwb) = 5 kg
Initial velocity of object (uo) = 10 ms-1
Initial velocity of wooden block (uwb) = 0 ms-1
Since, after the collision, both objects move together in the same line, their final velocities will be same. Let the final velocities of the block be v ms-1.
Now, consider the object and block together. They experience no force other than collision between them. During the collision, both the object and block experience the same magnitude of force for the same duration of time but in opposite direction (from Newton’s 3rd law of motion).
So, we can confidently say that the change in momentum in both the object and block is the same in magnitude but only opposite in direction, effectively cancelling out each other. So, for the object and the block, when considered together, collision does not bring about any change in the momentum
Considering this, we can say that the sum of momentum of the object and block remains unchanged by collison. Therefore,
Total momentum before collision = Total momentum after collision
To calculate final velocity, we can use the same equation,
pbefore = pafter
10 kg.ms-1 = 1 × v + 5 × v
On solving, we get
v = 10 6 or 1.67 ms-1
Note: During collision, there is no change in the sum of momentums of the objects. This is an example of conservation of momentum.
15. An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms–1 to 8 ms–1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Ans. Given,
Mass of object (m) = 100 kg
Initial velocity (u) = 5 ms-1
Final velocity (v) = 8 ms-1
Time of acceleration = 6 s
Initial momentum of the object (pinitial) = m.u = 100 kg × 5 ms-1 = 500 kg.ms-1
Final momentum of the object (pfinal) = m.v = 100 kg 8 ms-1 = 800 kg.ms-1
To calculate force exerted on the object, we can use Newton’s 2nd law of motion,
F = change in momentum time = t pfinal − pinitial
Therefore,
F = (800 kg.ms-1 − 500 kg.ms-1) 6 s = 50 N
16. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result, the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Ans. Rahul’s explanation is correct, while Akhtar’s and Kiran’s explanation are incorrect. The force applied by the insect on the windshield and the force applied by the windshield on the insect are an action-reaction pair and, according to Newton’s 3rd law of motion, both must be same in magnitude but opposite in direction.
Now, using Newton�s 2nd law of motion, we can say that, since both the insect and the windscreen experience the same magnitude of force for the same duration of time, change in the magnitude of momentum will be same for both the insect and the windscreen but it will be opposite in direction.
17. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms–2.
Ans. Given,
Mass of dumbbell (m) = 10 kg
Downward acceleration (a) = 10 ms-2
Distance to be travelled by dumbbell (S) = 80 cm or 0.8 m
Since the dumbbell falls from a height, its initial velocity (u) = 0 ms-1
Let final velocity of the dumb-bell be v ms-1
To calculate final momentum, we need to find final velocity (v)
Using the 3rd equation of motion, 2.a S = v2 − u2
We get,
v = (2.a.S) 1 2 = (2 × 10 × (0.8)) 1 2 = 4 ms-1
Now, final momentum of the dumbbell = m.v = 10 kg × 4 ms-1 = 40 kgms-1
Additional Exercises
1. The following is the distance-time table of an object in motion:
Ans.
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b) What do you infer about the forces acting on the object?
(a) Since average velocity is changing non-uniformly, therefore acceleration is also non-uniform. Since change in average speed in a unit interval is increasing, therefore acceleration is also increasing.
(b) Since mass of the object is constant and acceleration is non-uniform and increasing, therefore force acting on the object must also be non-uniform and increasing.
2. Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 ms-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
Ans. Given,
Mass of motorcar (m) = 1200 kg
Acceleration of car when three persons are pushing it (a) = 0.2 ms-2
Let force applied by each person be F Newton.
Now, since the car moves with constant velocity when two persons are pushing, this means that net force acting on the car is zero. This is possible only when the force of friction experienced by the motorcar is equal to the force exerted by two persons.
i.e. Fnet = 0 = 2F – Ffriction
Therefore, Ffriction = 2F
Now, when three persons push, the motorcar experiences two forces: (i) total push force (Fpush = 3F) and (ii) force of friction (2F).
Therefore, net force experienced by the motorcar = Fpush − Ffriction = 3F − 2F = F
From Newton’s 2nd law of motion,
F = m.a = 1200 kg × 0.2 ms-2 = 240 kg.ms-2 or 240 N
3. A hammer of mass 500 g, moving at 50 ms-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?
Ans. Given,
Mass of the hammer (m) = 500 g or 0.5 kg
Initial velocity of hammer (u) = 50 ms-1
Time to stop (t) = 0.01 s
Since the hammer stops finally, therefore final velocity of the hammer (v) = 0 ms-1
Using Newton’s 2nd law of motion,
F = m(v u) t = 0.5 kg (50 ms-1 – 0 ms-1) 0.01 s
On solving, we get F = 2,500 kg.ms-2 or 2,500 N
A4. A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.
Ans. Given,
Mass of motorcar (m) = 1200 kg
Initial velocity of motorcar (u) = 90 km/hr or 90 × 1000 m 3600 s = 25 ms-1
Final velocity of motorcar (v) = 18 km/hr or 18 × 1000 m 3600 s = 5 ms-1
Time (t) = 4 s
Let acceleration be a ms-2
Using 1st equation of motion, a = (v u) t = (25 ms-1 – 5 ms-1) 4 s = 5 ms-2
Now,
Change in momentum = Final momentum – Initial momentum = m.v – m.u = m × (v − u) = 1200 kg × (5 ms-1 – 25 ms-1) = 24000 kg.ms-1
For calculating the magnitude of force, we can use Newton’s 2nd law of motion, F = m a = 1200 kg × 5 ms-2 = 6000 kg.ms-2 or 6000 N
Multiple Choice Questions
1. The inertia of an object depends on its: (a) speed. (b) mass. (c) shape. (d) volume.
2. The unit of momentum is: (More than one answer may be correct) (a) kg.m/s (b) N (c) m/s² (d) N.s
3. Which of the following situations demonstrates the application of Newton's third law of motion? (More than one answer may be correct)
(a) A rolling ball stops after some time on a rough surface.
(b) A rocket is launched into space.
(c) A swimmer pushes water backwards while moving forward. (d) A car accelerates when the engine is turned on.
4. A force of 50 N is applied to an object of mass 10 kg. The acceleration produced is: (a) 0.5 m/s² (b) 5 m/s² (c) 10 m/s² (d) 50 m/s²
5. If the net force on an object is zero, the object will: (a) accelerate. (b) remain in its state of rest or uniform motion. (c) change direction. (d) move faster.
6. In the Fig. 8.2, if horizontal forces F1 and F2 cancel each other out, what is the net force experienced by the body?
(a) 0 N
(b) 10 N upwards
(c) 10 N downwards
(d) 5 N horizontally
7. Momentum of an object will remain the same, if the (a) mass is doubled and the velocity remains the same.
(b) mass remains the same and the velocity is doubled.
(c) mass is doubled and the velocity is halved.
(d) mass and velocity are both doubled.
8.2
8. A force of 10 N is applied to push a block along a smooth surface. What will be the acceleration of a 2 kg block?
(a) 5 m/s² (b) 10 m/s² (c) 20 m/s² (d) None of these
9. A truck with a mass of 5000 kg changes its velocity from 20 m/s to 10 m/s in 5 seconds. Find the force applied.
(a) 5000 N (b) −5000 N (c) −10,000 N (d) −50,000 N
10. Action and reaction forces act on:
(a) the same object.
(b) different objects.
(c) both same and different objects.
(d) None of the above.
11. A ball is kicked with a force of 15 N and travels along the path shown below in Fig. 8.3. Identify the force that acts on the ball after it leaves the foot.
(a) Frictional force
(b) Gravitational force
(c) Applied force
(d) Reaction force
12. A cannon recoils backwards after a cannonball is fired due to:
(a) Newton’s 1st law of motion.
(b) Newton’s 2nd law of motion.
(c) Newton’s 3rd law of motion.
(d) balanced force.
13. Explain why a passenger jerks forward when a moving bus suddenly stops.
(a) Due to unbalanced force
(b) Due to inertia of motion
(c) Due to inertia of rest
(d) Due to momentum conservation
8.3
14. If the engine of a car applies a forward force of 500 N and friction opposes it with 200 N, what is the acceleration of a 1000 kg car?
(a) 0.3 m/s² (b) 0.5 m/s² (c) 0.7 m/s² (d) 1 m/s²
Fig.
Fig.
15. Why does a balloon move forward when air is released from it?
(a) Conservation of energy
(b) Balanced force
(c) Newton�s third law
(d) Inertia
16. Fig. 8.4 shows the velocity-time relation for a moving object of mass 5 kg. Calculate the force experienced by the object during the first 5 seconds.
(a) −15 N
(b) 15 N
(c) −20 N
(d) 20 N
Fig. 8.4
17. When a moving car suddenly stops, the luggage on the roof may slide forward. This happens due to:
(a) balanced force.
(c) inertia of rest.
(b) inertia of motion.
(d) action-reaction forces.
18. Two objects, one of mass 3 kg and the other of mass 6 kg, are moving with the same velocity of 4 m/s. What is the ratio of their momentums?
(a) 1 : 2 (b) 1 : 3 (c) 2 : 3 (d) 3 : 4
19. Rocket works on the principle of conservation of (NCERT Exemplar) (a) mass.
(b) energy.
(c) momentum.
(d) velocity.
20. Fig. 8.5 is a velocity-time graph for a moving object. What will be the ratio of accelerations during 0–5 seconds and 5–10 seconds?
(a) 0
(b) 2.5
(c) 0.4
(d) None of these
Fig. 8.5
21. Why does a cricket player move their hands backwards while catching a fast-moving ball?
(a) To reduce momentum.
(b) To reduce the force of impact by increasing the time of impact.
(c) To increase velocity.
(d) To conserve energy.
22. A truck applies a braking force of 10,000 N to stop. If the truck’s mass is 5000 kg, what is the retardation produced?
(a) −2 m/s² (b) 2 m/s² (c) 5 m/s² (d) −5 m/s²
23. A boat will ___________ when a person jumps out of the boat in the forward direction.
24. A bird is sitting on a branch. When it flies away by pushing the branch down, the branch moves slightly. Which of the following best explains why the branch moves?
(a) The branch moves because of the bird�s weight.
(b) The branch moves due to the force the bird exerts on it.
(c) The branch moves due to wind resistance.
(d) The branch moves because of gravitational pull.
Answers
Constructed Response Questions
Very Short Answer Questions (30-50 words)
1. State Newton�s First Law of Motion and explain its significance with an example.
Ans. Newton’s First Law states that an object remains at rest or in motion unless an external force acts on it. This is due to inertia. For example, passengers move forward when a car suddenly stops because their bodies want to continue moving. It shows the tendency of objects to resist changes in motion.
2. Explain the concept of balanced forces and describe two effects of applying an unbalanced force on a body.
Ans. Balanced forces cancel each other out and do not cause any change in motion. When unbalanced forces act, they can move a stationary object or change the speed or direction of a moving object. For example, pushing a box makes it move, or speeding up a bicycle on a slope.
3. Define inertia and explain why athletes run before taking a jump.
Ans. Inertia is the tendency of an object to resist changes in its state of motion. Athletes run before jumping to gain speed and momentum. This helps their body stay in motion longer, allowing them to jump farther. It is inertia of motion that supports a longer leap.
4. Name the physical quantity that corresponds to the rate of change of momentum. Also, explain how a sand or cushioned surface reduces the impact during a fall in a high jump.
Ans. Force is the rate of change of momentum. A sand or cushioned surface increases the time over which the body comes to rest during a fall. This longer time reduces the force of impact, making the landing safer and less painful for the athlete.
5. State and explain Newton�s Third Law of Motion. How does it apply to a person walking on the road?
Ans. Newton’s Third Law states that for every action, there is an equal and opposite reaction. While walking, we push the ground backward with our feet. In response, the ground pushes us forward with equal force. This reaction force moves our body and helps us walk.
6. How does inertia explain why passengers feel a jerk when a moving car suddenly stops?
Ans. When a car stops suddenly, the passengers� bodies continue to move forward due to inertia. Their bodies resist the sudden change in motion, which causes them to feel a jerk. This happens because objects in motion want to keep moving unless something stops them.
Fig. 8.6
7. Why does a person jumping out of a boat cause the boat to move in the opposite direction?
Ans. When a person jumps forward out of a boat, they push the boat backward. According to Newton’s Third Law, the boat pushes back with equal and opposite force. As a result, the boat moves in the opposite direction of the person jumping out.
8. Why does a car driver need to keep pressing the accelerator to maintain a constant speed?
Ans. A car slows down due to friction from the road and air. To balance this force and keep the car moving at the same speed, the driver must press the accelerator. This provides enough force from the engine to overcome friction and maintain motion.
9. Does a standing woman carrying a matka full of water on her head need to apply any force? Justify your answer with a reason.
Ans. Yes, the woman must apply an upward force with her neck and body to balance the downward weight of the matka. Without this force, the matka would fall due to gravity. Her body provides an equal and opposite force to keep it steady on her head.
10. If two equal and opposite forces act on a body, will it always remain at rest? Justify your answer.
Ans. If equal and opposite forces act on a body, they cancel out. This means there is no net force, so the body remains at rest or continues moving at constant speed. These are called balanced forces. Only unbalanced forces can change the state of motion.
11. A horse continues to apply a force in order to move a cart with a constant speed. Explain why.
(NCERT Exemplar)
Ans. When a cart moves on the ground, its motion is opposed by the frictional force exerted by the ground. To maintain a constant speed, this frictional force must be counteracted by an equal force, which is supplied by the horse.
12. A vehicle of mass 1000 kg is moving with a velocity of 15 m/s. What is the momentum of the vehicle?
Ans. Given,
Mass of the vehicle (m) = 1000 kg
Velocity of the vehicle (v) = 15 m/s
By definition,
Momentum = mass.velocity = m.v = 1000 kg × 15 m/s = 15,000 kg.m/s
13. A force of 10 N is applied to a body of mass 2 kg. What will be the acceleration produced in the body?
Ans. Given,
Force exerted (F) = 10 N
Mass of the body (m) = 2 kg
From Newton’s 2nd law of motion,
Acceleration (a) = F m = (10 N) (2 kg) = 5 ms-2
14. A car accelerates from 0 to 20 m/s in 10 seconds. If the mass of the car is 800 kg, calculate the force required to produce this acceleration.
Ans. Given,
Mass of the car (m) = 800 kg
Initial velocity of the car (u) = 0 m/s
Final velocity of the car (v) = 20 m/s
Time (t) = 10 sec
Fig. 8.7
Using Newtons 2nd law of motion,
F = m(v u)
t We get,
F = [800 kg × (20 m/s – 0)] 10 s = 1600 kg.m/s2 or 1600 N
Short Answer Questions (50-80 words)
1. What is force? Explain the effects of force on an object.
Ans. Force is a push or pull acting on an object due to its interaction with another object. It can change the state of motion of an object, such as making a stationary object move, stopping a moving object or altering its speed or direction. Force can also deform or change the shape of an object, depending on its magnitude and direction. For example, pressing a spring compresses it, illustrating the effect of force on an object.
2. Differentiate between balanced and unbalanced forces with examples. Ans.
Balanced Forces
Forces that are equal in magnitude and opposite in direction.
Do not cause any change in the state of motion of the object.
Net force is zero.
Ex. A book resting on a table (weight balanced by normal reaction force).
Unbalanced Forces
Forces that are unequal in magnitude or not directly opposite in direction.
Cause a change in the state of motion of the object (start, stop, or accelerate).
Net force is not zero.
Ex. A person pushing a stationary box which then starts moving.
3. State Newton’s Second Law of Motion. How does it explain the relationship between force, mass and acceleration? Give an example.
Ans. Newton�s Second Law of Motion states that the force acting on an object is directly proportional to the rate of change in momentum. From this, we get the result, the net force on a body is equal to the product of its mass and acceleration (F = ma). This law explains that a greater force results in greater acceleration for a given mass and, for the same force, a heavier object will have less acceleration. For example, pushing an empty shopping cart is easier (greater acceleration) than pushing a fully loaded one because the loaded cart has more mass.
4. Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed v again before striking the ground. It implies that the magnitude of initial and final momentums of the ball are same. Yet, it is not an example of conservation of momentum. Explain why. (NCERT Exemplar)
Ans. The momentum in this case is not conserved because only the magnitude of the initial and final momentums is same, not the direction. The initial momentum was in an upward direction, while the final momentum is in a downward direction. Additionally, since an external force (gravitational force) is acting on the ball during the flight, its momentum cannot remain unchanged. (Force = change in momentum / time)
5. A truck of mass M is moved under a force F. If the truck is then loaded with an object equal to the mass of the truck and the driving force is halved, then how does the acceleration change? (NCERT Exemplar)
Ans. Assumption: Friction does not play any role in this scenario. Since the force is reduced by half and mass is doubled, the acceleration will be reduced to ¼ of the original acceleration.
From Newton’s 2nd law, F = m.a. ⇒ a = F m
In initial scenario,
a1 = F m
In new scenario, a2 = ( F 2 2m ) = 1 4 ( F m ) = 1 4 a1
6. Why does a cricket player playing with a leather ball use a helmet but one playing with a tennis ball does not?
Ans. For the same speed of both balls, the leather ball has greater momentum because it is heavier and harder than a tennis ball, making its impact more forceful. The helmet protects the player from the significant force of the leather ball, which could cause serious injuries. In contrast, a tennis ball is softer and lighter, causing much less force on impact, so a helmet is not typically necessary.
7. Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown in Fig. 8.8.
Calculate the acceleration and frictional force of the floor on the ball. (NCERT Exemplar)
Ans. Given,
Mass of the ball (m) = 50 g or 0.05 kg
Since v − t curve is a straight line, therefore acceleration will be uniform. We can calculate acceleration using the v − t curve.
Since the ball is rolling on the floor, we can assume that only frictional force is acting on the ball. Hence, the acceleration produced is due to friction only.
Therefore, Ffriction = m.a = 0.05 kg . (−10 ms-2) = −0.5 kg.ms-2 or –0.5 N Negative sign indicates that the direction of friction is opposite to the direction of motion.
8.8
8. Describe the principle of Newton�s Third Law of Motion and illustrate it with a practical example.
Ans. Newton�s Third Law of Motion asserts that for every action, there is an equal and opposite reaction. Essentially, this means that forces always occur in pairs. For instance, when a person sits on a chair, their body exerts a downward force on the chair and, in response, the chair exerts an upward force of equal magnitude against the person, effectively supporting them. This interaction illustrates the law�s application in everyday life.
9. Two identical bullets are fired one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?
Ans. When a bullet is fired, it exits the gun with significant momentum, imparting an equal and opposite momentum to the gun, causing it to recoil. If the bullets fired from two guns have the same momentum, the lighter gun will recoil at a higher speed because of its lower inertia. As a result, the lighter gun will exert a greater impact on the shooter’s shoulder.
10. By observing Fig. 8.9, answer the following questions in brief.
(a) When the thread tied to the balloon�s neck is removed, in which direction does the balloon move?
(b) Which of Newton�s law of motion is applicable here?
(c) What conclusion can be drawn from this activity?
Ans. (a) When the tied thread is removed, the air inside the balloon escapes towards the left and the balloon moves from left to right.
(b) Newton�s third law of motion is applicable here.
8.9
Fig.
Fig.
(c) The action of air exiting ballon causes a reaction in the form of the balloon’s motion in the opposite direction. The action and reaction forces are equal and opposite in direction.
11. Two friends on roller-skates are standing 5 m apart facing each other. One of them throws a ball of 2 kg towards the other, who catches it. How will this activity affect the position of the two? Explain your answer.
Ans. When the first person throws the ball, he exerts a forward force on the ball and, in response, he will experience an equal and opposite force (according to Newton�s third law), causing him to move backwards. When the second person catches the ball, he exerts a force on the ball to stop it, and the ball exerts an equal and opposite force on him, causing him to move backwards as well.
Since both individuals are moving backwards and away from each-other, the distance between them will increase.
12. An object experiences a force of 12 N and accelerates at 4 m/s². When the same object is subjected to a force of 30 N, what will be its acceleration?
Ans. Given,
Initial force (Fi) = 12 N
Initial acceleration (ai) = 4 m/s²
Final force (Ff) = 30 N
Let the mass of the object be m kg and final acceleration be a m/s2.
Now, using the result of Newton’s 2nd law of Motion, m = F a ,
Since mass of an object does not change, therefore, m = Fi ai = Ff a
Rearranging the terms we get,
a = ai × Ff Fi = 4 m/s² × 30 12 = 10 m/s²
13. A ball of mass 0.5 kg is thrown horizontally with an initial velocity of 10 m/s. An invisible force acting on the ball stops it in 5 seconds. Calculate the magnitude of this invisible force.
Ans. Given,
Mass of the ball (m) = 0.5 kg
Initial velocity of the ball (u) = 10 m/s
Final velocity of the ball (v) = 0 m/s
Time to stop (t) = 5 s
Let acceleration be a m/s2 and Force be F N
From 1st equation of motion,
a = (v u) t = (0 – 10 m/s) 5 s = −2 m/s2
Now, from Newton’s 2nd law of motion,
F = m a = 0.5 kg . (−2 m/s2) = −1 N
Note: Negative sign indicates that the direction of force is opposite to the motion.
14. Rajesh and Punit are on a frozen lake where Rajesh pushes Punit, who weighs 40 kg, with a force of 50 N. What acceleration does Punit experience? Additionally, does Rajesh, weighing 50 kg, experience movement, and if so, what will be his acceleration?
Ans. When Rajesh applies a force of 50 N to push Punit, Punit will experience an acceleration due to this force and his acceleration will be:
Acceleration (a) = force (F) mass (m)
For Punit, the acceleration is:
a p = F m = (50 N) (40 kg) = 1.25 m/s2
Due to Newton’s third law of motion, Rajesh will also experience the same magnitude of force but in opposite direction. Rajesh�s acceleration will be:
aR = F m = (50 N) (50 kg) = 1 m/s2
Thus, both friends will experience acceleration in opposite directions, with Punit accelerating at 1.25 m/s² and Rajesh at 1.0 m/s².
Long Answer Questions (80–120 words)
1. Describe Galileo’s experiment to demonstrate the motion of objects on an inclined plane.
Ans. Galileo deduced that objects move at a constant speed when no force acts on them by observing the motion of objects on an inclined plane. He rolled a marble down an inclined plane, noting that it accelerated as it descended. Galileo asserted that when a marble is released from one inclined plane, it rolls down and ascends the opposite plane to the same height from which it was released.
As the angle of inclination of the second plane was gradually reduced, the marble travelled greater distances before reaching the same height. Galileo hypothesized that if the second plane were made horizontal, the marble would continue moving indefinitely along the same path, attempting to reach the same height. In this scenario, the unbalanced force acting on the marble would be zero, implying that an external unbalanced force is required to change its motion. Here, the effect of friction was assumed to be negligible. Galileo’s hypothesis laid the foundation for the first law of motion.
2. Derive the mathematical relation of Newton’s second law of motion.
Ans. Consider an object of mass m moving along a straight line with an initial velocity u. It is uniformly accelerated to velocity v in time t by the application of a constant force F in time t.
Then, initial momentum of the object (Pi) = m.u
Final momentum of the object (Pf) = m v
Change in momentum = Pf – Pi = mv − mu = m × (v − u)
The rate of change in momentum = change in momentum time = m × (v u) t
According to Newton�s second law of motion,
F ∝ Rate of change in momentum
Therefore,
F = k × m × (v u) t where k is constant for proportionality
Now, from the 1st equation of motion, we know that acceleration (a) = (v u) t
Therefore,
F = k × m × a (1)
Putting m = 1 kg, a = 1 ms-2
F becomes 1 N. (by definition)
So,
1 N = k × 1 kg × 1 ms-2
k = 1
From equation (1), we have
F = m × a
This represents the second law of motion.
Thus, the second law of motion allows us to measure the force acting on an object as a product of its mass and acceleration.
(a) (b)
(c)
Fig. 8.10
3. (a) Derive the unit of force using the second law of motion.
(b) A force of 5 N produces an acceleration of 8 ms–2 on a mass m1 and an acceleration of 24 ms–2 on a mass m2. What acceleration would the same force provide if both the masses are tied together?
Ans. (a) From Newton’s 2nd law of motion, F = m.a
For 1 kg mass moving with an acceleration of 1 ms-2, F = (1 kg) × (1 ms-2) = 1 kg.ms-2
Therefore, the unit of force is kgms-2 or Newton (N).
(b) Given,
Force exerted = 5 N
Initial acceleration of mass m1 (a1) = 8 ms–2
Initial acceleration of mass m2 (a2) = 24 ms–2
Using the given data and Newton’s 2nd law of motion, we can calculate mass of m1 and m2, F = m.a
Therefore,
5 N = m1 × 8 ms–2 m1 = 5 8 kg
5 N = m2 × 24 ms–2 m2 = 5 24 kg
Now, the combined mass of m1 and m2 = 5 24 5 8 + = 15 + 5 24 = 5 6 kg
Since both masses are tied together, when one block is pulled, it will pull the other one with the same velocity and acceleration. The blocks will act as one block with the combined mass of both.
Therefore, the acceleration of the combined mass of m1 and m2 will be F = m.a
5 N = 5 6 kg × a
Therefore, Acceleration (a) = 6 ms-2
4. A bullet of 10 g strikes a sandbag at a speed of 103 m/s and gets embedded after travelling 5 cm.
Calculate (NCERT Exemplar)
(a) the resistive force exerted by the sand on the bullet.
(b) the time taken by the bullet to come to rest.
Ans. Given,
Mass of the bullet (m) = 10 g or 0.01 kg
Initial speed of the bullet (u) = 103 ms-1
Distance travelled by the bullet (S) = 5 cm or 0.05 m
Since the bullet finally stops, its final velocity (v) will be 0 ms-1
Let the acceleration be ‘a’ ms-2 and time be ‘t’ sec.
According to the 3rd equation of motion,
2.a.S = v2 – u2
Therefore,
a = (02 – (103 ms-2)2) (20.05 m) = 107 ms-2
(a) The resistive force exerted by the sandbag will be
F = m a = (0.01 kg) × (107 ms-2) = 105 kg.ms-2 or 105 N
(b) To calculate the time taken to stop the bullet, we can use the 1st equation of motion, a = (v u) t
Putting the known values in this equation, we get
107 N = (0 – 103 ms-1) t t = 10-4 sec or 0.0001 sec
5. A ball of mass 500 grams is dropped from a height of 5 metres. Calculate its momentum velocity just before hitting the ground. The ball experiences a downward force of 5 N during the fall. Also, calculate how much time the ball will take to hit the ground.
Ans. Given,
Mass of the ball (m) = 500 g or 0.5 kg
Distance from drop point to the ground (S) = 5 m
Since the ball is dropped, its initial velocity (u) = 0 ms-1
Acceleration of the ball (a) = F m = 5 0.5 = 10 ms-2
To calculate the final momentum of the ball, we need to know the final velocity of the ball. Using the 3rd equation of motion,
2.a.S = v2 – u2
We get, v = (u2 + 2.a.S) 1 2 = (02 + 2 × 10 ms-2 × 5 m) 1 2 = 10 ms-1
Now, final momentum = m v = 0.5 kg × 10 ms-1 = 5 kg.ms-1
To calculate the time the ball takes to hit the ground, we can use the 1st equation of motion, a = (v − u) t or t = (v − u) a
Therefore,
t = (10 ms-1 0) 10 ms-2 = 1 s
6. A 500 kg car is moving with a speed of 20 m/s. What is the momentum of the car at this moment? Now, if the speed is halved in the next 10 seconds, calculate the force experienced by the car and the distance travelled during this interval.
Ans. Given,
Mass of the car (m) = 500 kg
Initial speed of the car (u) = 20 m/s
Time (t) = 10 s
Final velocity (v) = 20 2 = 10 m/s
Now, using 1st equation of motion
Acceleration (a) = (v − u) t = (10 − 20) 10 = −1 ms−2
Therefore, Force (F) = m a = 500 × (−1) = −500 N
For distance travelled, we can use 2nd equation of motion
Displacement (S) = ut + 1 2 at2
S = (20 × 10) + ( 1 2 × −1 × 102) = 200 − 50 = 150 m
7. The velocity-time graph of a car of mass 800 kg is shown in Fig. 8.11. Observe the graph and answer the following questions.
(a) What is the net force experienced by the car in 0–5 s?
(b) What is the net force experienced by the car in 5–8 s?
(c) What is the net force experienced by the car in 8–10 sec?
(d) Calculate the total distance travelled by the car in 0–10 sec.
Ans. From the graph,
Initial velocity (v0) = 0
Velocity at 5 s (v5) = 40 m/s
Velocity at 8 s (v8) = 40 m/s
Velocity at 10 s (v10) = 0 m/s
Using the 1st equation of motion, a = (v − u) t
(a) acceleration in 0–5 s (a0−5) = (v5 − v0) 5 = (40 − 0) 5 = 8 m/s2
F = m.a = 800 × 8 = 6,400 N
(b) acceleration in 5–8 s (a5−8) = (v8 − v5) 3 = (40 − 40) 3 = 0 m/s2
F = m.a = 800 × 0 = 0 N
(c) acceleration in 8–10 s (a8−10) = (v10 − v8) 2 = (0 − 40) 2 = −20 m/s2
Multiple Choice Questions (Choose the most appropriate option/options)
1. When a car takes a sudden sharp turn at high speed, passengers tend to slide sideways. This is due to:
(a) Friction
(c) Centripetal force
(b) Inertia
(d) Gravity
2. A person standing on a stationary boat throws a heavy bag forward into the water. As a result, the boat moves backward. Which principle of physics does this situation illustrate?
(a) Newton�s First Law
(c) Newton�s Third Law
(b) Newton�s Second Law
(d) Law of Conservation of Energy
3. A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to (NCERT Exemplar)
(a) exert larger force on the ball
(b) reduce the force exerted by the ball on hands
(c) increase the rate of change of momentum
(d) decrease the rate of change of momentum
4. Why are athletes advised to come to a stop slowly after a race?
(a) To save energy
(c) To reduce injury
(b) To look professional
(d) To reduce impact force
5. When a carpet is beaten with a stick, dust comes out. Why?
(a) Stick breaks the carpet
(c) Inertia of dust
(b) Dust repels stick
(d) Air blows the dust
6. When two boys are standing on a wooden plank and one of them jumps off towards the left, what is likely to happen to the other boy?
(a) He also moves to the left.
(c) He remains stationary.
(b) He moves to the right.
(d) The information given is not enough to decide
7. Using soft mats in high jumps reduces injuries by:
(a) Decreasing impact speed
(c) Absorbing heat
8. Which object has more momentum?
(a) A 100 gm tennis ball moving at 200 km/hr.
(b) A 100 kg motor bike moving at 3 km/hr.
(b) Increasing time of impact
(d) Providing bounce
(c) A 2 kg space rock moving with at a speed of 80 m/s.
(d) A 1 μg dust particle moving at half the speed of light.
9. A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is (NCERT Exemplar)
(a) accelerated
(c) retarded.
(b) uniform
(d) along circular tracks
10. An object is falling near earth’s surface. Which of the following forces it does not experience?
(a) Gravitational forces
(c) Buoyant force
(b) Frictional force
(d) Atmospheric pressure force
11. A heavier object does not fall faster than a lighter object in a vacuum because:
(a) A vacuum does not have any air to resist motion.
(b) In a vacuum, both experience the same stopping force.
(c) In a vacuum, both experience the same gravitational force.
(d) In a vacuum, all objects fall with the same acceleration regardless of mass.
12. When a goal-keeper stops a penalty kick, in which direction does he apply a force?
(a) Forwards
(b) Backwards
(c) Side-ways
(d) Downwards
13. During which phase of journey shown in the given speed-time graph (Fig. 8.12), does the body experiences maximum force?
(a) I
(b) II
(c) III
(d) IV
14. A spaceship is moving in space with its engine fired. What will happen if its engines suddenly fails?
(a) It will stop.
(b) It will continue moving in the same direction with a constant velocity.
(c) It will move with same speed but its direction of motion changes.
(d) It will move direction and its speed will increase.
Fig. 8.12
15. A water tanker filled up to 2 3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would (NCERT Exemplar)
(a) move backward
(b) move forward
(c) be unaffected
(d) rise upwards
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false. (d) A is false but R is true.
1. Assertion (A): A body in motion will eventually stop if no external force acts on it.
Reason (R): Frictional force can stop the motion of a body.
2. Assertion (A): A rocket moves forward because gases are expelled backwards.
Reason (R): The action force exerted by the rocket on the gases creates an equal and opposite reaction force.
3. Assertion (A): If an object experiences multiple non-zero forces, it will always accelerate. Reason (R): When an object experiences a non-zero force, its acceleration will be non-zero.
4. Assertion (A): A person leans forward when the brakes of a moving vehicle are suddenly applied. Reason (R): The person’s lower body comes to rest, while the upper body continues moving due to inertia.
5. Assertion (A): A ball rolling on a perfectly smooth surface will never stop on its own.
Reason (R): The absence of friction does not affect the motion of the ball.
6. Assertion (A): A balanced force changes the state of motion of a body.
Reason (R): A balanced force cancels out the net force acting on a body.
Case-Based/Source-Based/Passage-Based
Questions
1. Read the given passage and answer the questions that follow:
A boy named Aryan is riding a bicycle on a straight road. He observes that when he pedals harder, the bicycle speeds up, but when he stops pedalling, the bicycle gradually slows down and comes to rest. Aryan wonders why this happens. One day, his friend explains to him about forces and Newton’s laws of motion, including friction and inertia.
Based on Aryan�s experience, answer the following questions:
(a) What is the role of friction in Aryan�s observation when he stops pedalling?
(i) Friction accelerates the bicycle.
(ii) Friction opposes the motion of the bicycle, causing it to slow down.
(iii) Friction has no effect on the bicycle’s motion.
(iv) Friction increases the speed of the bicycle.
(b) Which of Newton�s laws explains why Aryan needs to pedal harder to increase the speed of the bicycle?
(i) Newton�s First Law
(iii) Newton�s Third Law
(ii) Newton�s Second Law
(iv) All of the above
(c) Due to which property of the bicycle does it slow down when Aryan stops pedalling?
(i) Momentum
(iii) Action-reaction pair
(ii) Inertia
(iv) Gravitational force
(d) If Aryan applies the brakes suddenly, the bicycle stops, but Aryan’s body moves forward. This happens due to:
(i) Inertia of motion
(ii) Friction between the wheels and the road
(iii) Gravitational force
(iv) Momentum conservation
2. Read the given passage and answer the questions that follow:
Imagine you are an astronaut travelling to the Proxima Centauri, our closest star after the Sun, in a futuristic spaceship. Proxima Centauri is 4.2 light years away from Earth. The spaceship, having travelled half the distance, experiences the following scenarios: For each scenario, explain what you will experience (in terms of forces experienced and the subsequent changes in the speed of the spaceship). Assume that the spaceship has left the solar system and is currently travelling in empty space.
(a) Your spaceship’s engine is firing at full capacity.
(b) Your spaceship’s engine is running but suddenly runs out of fuel.
(c) Your spaceship is caught in the gravity of a planet.
(d) Your spaceship enters the atmosphere of a planet, similar to Earth.
3. The velocity-time graph of an F1 car, of mass 500 kg, is shown in Fig. 8.13. Observe the graph carefully and answer the following questions.
(a) What is the net force experienced by the car during acceleration?
(b) What will be the velocity of the car at the end of acceleration period?
(c) When the brakes are applied, the car experiences a retarding force of 10,000N. What will be the time taken by the car to come to rest once the brakes are applied?
(d) What distance is covered by the F1 car during the entire journey?
8.13
4. A police officer, sitting in his parked car (700 kg), sees a thief running away on his motorbike (200 kg) at a constant speed of 20 m/s. When the thief crosses the police officer, the officer starts accelerating his car at the rate of a m/s2. At the same time, the thief, upon seeing the police officer, starts accelerating his bike at the rate of 2 m/s2. Answer the following question based on this scenario.
(a) If the police officer catches up with the thief in 10 s, what should be the value of a?
(b) What is the distance the police officer’s car travels before catching up with the thief?
(c) What is the maximum speed achieved by the car and the bike during the chase?
(d) What is the ratio of the force exerted by the car’s engine and the bike’s engine?
Explanation: Friction acts opposite to the motion of the bicycle, causing it to decelerate and eventually stop. (b) (ii)
Explanation: Newton�s Second Law states that the acceleration of an object is directly proportional to the net force applied to it, explaining why pedalling harder (applying more force) increases the bicycle’s speed.
(c) (ii)
Explanation: The bicycle slows down due to inertia, which is the tendency of an object in motion to resist changes to its state unless acted upon by an external force (friction in this case).
(d) (i)
Explanation: Aryan’s body continues to move forward due to the inertia of motion, as his body resists the change in its state of motion, even though the bicycle has stopped.
2. (a) The engine produces thrust, which applies an unbalanced force on the spaceship. According to Newton�s Second Law, this force causes the spaceship to accelerate, meaning, its speed will keep increasing as long as the engine fires.
Force → Acceleration → Increase in Speed
(b) In the empty space (vacuum), there are no resistive forces like friction or air drag. Once the engine stops, no force acts on the spaceship. By Newton�s First Law, the spaceship will continue moving at a constant speed (inertia of motion).
No Force → Constant Speed
(c) The gravitational force of the planet pulls the spaceship towards it, causing the spaceship to accelerate towards the planet. This results in an increase in speed as it moves closer to the planet.
Gravitational Force → Acceleration towards Planet
(d) The spaceship experiences air resistance (a frictional force) as it enters the planet’s atmosphere. This air resistance opposes its motion and causes the spaceship to slow down.
Air Resistance → Decrease in Speed
3. (a) Since acceleration does not change in 0–10 seconds, force will also remain same.
Therefore, Fnet = m a = m (v − u) t = 500 × (50 − 0) 5 = 500 × 10 = 5000 N
(b) vt = u + a.t; v10 = 0 + 10 × 10 = 100 m/s
(c) Since the car stops eventually, vf = 0; t = (vf – v20) ( ( F m = (0 − 100) 500 −10000 ( ( = 5 s
(d) Distance travelled = area under v t curve = ( 1 2 × 10 × 100) + (10 × 100) + ( 1 2 × 5 × 100) = 1750 m
4. (a) Acceleration of the police officer’s car:
The thief’s initial speed is 20 m/s and rate of acceleration is 2 m/s².
The police officer starts from rest and accelerates at a m/s².
Distance travelled by the police officer in 10 seconds:
Spoliceman= 1 2 .a.t2 = 1 2 .a.102 = 50.a
To catch the thief in 10 seconds, the distances must be equal:
50.a = 300 a = 6 m/s2
(b) Distance travelled by the police officer’s car:
Distance travelled by the police officer’s car = 1 2 a c t2 = 1 2 × 6 × 102 = 300 m
(c) Maximum speed achieved by the car and the bike:
Car’s maximum speed: v car = 0 + 6 × 10 = 60 m/s
Bike’s maximum speed: vbike = 20 + 2 × 10 = 40 m/s
(d) Ratio of the forces exerted by the engines:
Force exerted by the car’s engine:
F car = 700 × 6 = 4200 N
Force exerted by the bike’s engine:
Fbike= 200 × 2 = 400 N
Ratio of forces:
F car Fbike = 4200 400 = 21 2
Numerical Questions
1. An object experiences a net force of 15 N and accelerates at 5 m/s². Calculate its mass.
Ans. Given,
Force (F) = 15 N
Acceleration (a) = 5 m/s2
Using Newton’s 2nd law of motion, F = m.a
m = F a = 15 5 = 3 kg
2. Two objects, one of mass 2 kg and the other of mass 4 kg, are acted upon by forces of F N and 2F N, respectively. Which object has a greater acceleration?
Ans. From Newton’s 2nd law of motion, F = m.a
For the 2 kg object: a2 = F m = F 2 m/s2
For the 4 kg object: a4 = F m = 2F 4 = F 2 m/s2
Both objects have the same acceleration.
3. A body of mass 2 kg is acted upon by two forces, 10 N and 20 N, in opposite directions. What is the net force on the body, and what will be its acceleration?
Ans. Given,
Mass of the body (m) = 2 kg
Forces F1 = 10 N and F2 = 20 N
Net force (Fnet) = F2 – F1 = 10 N
Now, using the Newton’s 2nd law of motion, F = m.a
We get, a = Fnet m = 10 N 2 kg = 5 m/s²
4. A force of 100 N is applied to an object, causing it to accelerate at 20 m/s². If friction opposes the motion with a force of 40 N, calculate the mass of the object.
Ans. Given,
Acceleration of the body (m) = 20 m/s2
Force (F) =100 N
Force of friction (Ffriction) = 40 N
Net force, Fnet = F – Ffriction = 100 − 40 = 60 N
m = Fnet a = 60 20 = 3 kg
5. A rocket with a mass of 1000 kg is pushed upward with a thrust force of 15,000 N. If the gravitational force acting on the rocket is 10,000 N, calculate the net acceleration of the rocket.
Ans. Given,
Mass of rocket (m) = 1,000 kg
Thrust force (Ft) = 15,000 N
Gravitational force (Fg) = 10,000 N
Net Force (Fnet) = Ft – F g = 15,000 N – 10,000 N = 5000 N
Now, using the result of Newton’s 2nd law of motion, a = F m
a = 5000 N 1000 kg = 5 m/s2
6. An object, initially at rest, is subjected to a force of 12 N for 4 seconds. If its mass is 3 kg, find the final velocity.
Ans. Given,
Mass (m) = 3 kg
Force (F) = 12 N
Time (t) = 4 s
Acceleration, a = F m = 12 3 = 4 m/s2
Using the 1st equation of motion, final velocity, v = u + at
v = 0 + (4 × 4) = 16 m/s.
7. A force of 15 N is applied on a body of mass 3 kg, initially at rest. Calculate the distance travelled by the body in 4 seconds.
Ans. Given,
Mass (m) = 3 kg
Force (F) = 15 N
Time (t) = 4 s
Acceleration, a = F m = 15 5 = 5 m/s2.
Distance, S = ut + 1 2 .at2 = 0 + ( 1 2 × 5 × 42) = 40 m.
8. A car moving with a uniform velocity of 25 m/s is brought to rest by applying brakes in 10 seconds. Find the magnitude of the force applied if the car has a mass of 800 kg.
Ans. Given,
mass (m) = 800 kg
initial speed (u) = 25 m/s
final speed (v) = 0 m/s
time (t) = 10 s
Acceleration, a = (v − u) t = (0 − 25) 10 = −2.5 m/s2
Force, F = m a = 800 × (−2.5) = −2000 N
9. A ball of mass 0.5 kg is thrown vertically upwards with an initial velocity of 20 m/s. Calculate the time taken to reach its highest point, if the downward gravitational force experienced by the ball is 5 N.
Ans. Given,
mass (m) = 0.5 kg
gravitational force (F) = 5 N
initial speed (u) = 20 m/s
Acceleration of the ball, a = F m = 5 0.5 = 10 m/s2
At the highest point, the ball will momentarily stop, therefore, v = 0
Using v = u + at
0 = 20 − 10t
10t = 20
t = 2 sec
10. A force of 5 N produces an acceleration of 5 m/s2 in a mass m1 and 25 m/s2 in a mass m2. If the same force is applied to both masses tied together, calculate the resulting acceleration.
Ans. Given,
force (F) = 5 N
acceleration of m1= 5 m/s2
acceleration of m2= 25 m/s2
Using m = F a ;
Mass of m1 = 5 5 = 1 kg
Mass of m2 = 5 25 = 0.2 kg
Therefore, when tied together, the combined mass of m1 and m2 will be m c = 1 + 0.2 = 1.2 kg
Hence, the combined acceleration produced by 5 N will be a = F m c = 5 1.2 = 50 12 = 4.167 m/s2
Practice Questions
Multiple Choice Questions
1. State the relation between mass, acceleration and force, as given by Newton’s Second Law of Motion.
(a) F = m a
(b) F = m × a
(c) a = m × F
(d) m = a F
2. Why is it easier to stop a moving cycle than a moving truck when both are travelling at the same speed?
(a) The bicycle is lighter, so it has less inertia.
(b) The bicycle is heavier, so it has more friction.
(c) The truck’s engine makes it harder to stop.
(d) Both have the same momentum at the same speed.
3. A force of 10 N acts on a mass of 2 kg. What is the acceleration produced?
(a) 0.2 m/s2
(b) 5 m/s2
(c) 2 m/s2
(d) 20 m/s2
4. A car takes 5 seconds to stop after the brakes are applied. If its initial velocity is 20 m/s, find its acceleration.
(a) −2 m/s2 (b) −4 m/s2 (c) −5 m/s2 (d) −10 m/s2
5. A cricket ball of mass 1.5 kg is moving with a velocity of 12 m/s. What is its momentum?
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
6. Assertion (A): A body will accelerate when acted upon by balanced forces.
Reason (R): Balanced forces cancel each other out, resulting in no net force on the body.
7. Assertion (A): If an object’s momentum changes, there must be a net force acting on it.
Reason (R): The net force on an object is equal to the rate of change of its momentum.
Very Short Answer Questions (30-50 words)
8. Why does a person fall while walking if someone trips them?
9. Why is it important to wear a seat belt in a moving car? Relate this to Newton�s laws of motion.
10. Provide two examples from the natural world where organisms utilise Newton�s Third Law of Motion for movement.
11. A hockey ball of mass 200 g, travelling at 10 m/s, is struck by a hockey stick, so it moves in the opposite direction with a velocity of 5 m/s. Calculate the change in momentum of the ball.
12. A truck of mass 3000 kg, moving with a velocity of 20 m/s, applies brakes and comes to rest in 5 s. Calculate the force exerted by the brakes.
Short Answer Questions (50-80 words)
13. Explain with a diagram how action and reaction forces operate during swimming.
14. A ball of mass 2 kg is dropped from a height of 10 m. Calculate the velocity with which it hits the ground and its momentum upon impact. (acceleration produced by Earth’s gravity = 10 m/s2)
15. Design an experiment to demonstrate Newton’s Third Law of Motion.
16. A force of 50 N is applied to an object of mass 10 kg at rest for 4 s. Calculate the velocity acquired by the object and the distance travelled during this time.
17. Two objects, one of mass 4 kg and the other of mass 6 kg, are acted upon by the same force for 2 s. Compare their accelerations and velocities after this time.
Long Answer Questions (80-120 words)
18. A car of mass 1000 kg, travelling at 72 km/h, comes to a halt in 10 s. Find the force exerted by the brakes and the distance covered before stopping.
19. Explain how rockets are launched into space. Relate your answer to Newton’s third law. Also, explain why a rocket continues to move in space even after the fuel is exhausted.
20. A stretched, unbreakable, massless and inextensible string connects two blocks of 2 kg and 3 kg, placed on a frictionless surface. Now, if one of the masses is pulled away from the other by an external force of 15N, what will be the acceleration of each block? (Hint: Net force on a string is always zero. A string only transfers force from one end to another, without any loss.)
Brain Charge
1. Crossword Puzzle ACROSS
3. The physical quantity defined as mass times velocity.
6. The rate of change of velocity with time. DOWN
1. The mutual force of attraction between two masses.
2. The tendency of an object to resist a change in its state of motion.
5. The SI unit of force
2. Word Puzzle
4. For every action, there is an equal and opposite
Use the first alphabet of each answer to the puzzle and combine them to spell a word.
1. Most abundant gas in our atmosphere.
2. A battery provides us with electrica ________________.
3. The SI unit of electrical energy.
4. What does a clock measure?
5. Which device in your car/bike measures the distance travelled?
6. An unpleasurable sound.
3. Propose a way to speed up a motor vehicle without using any fuel or manual force. If you think it is not possible, then state the reason.
Challenge Yourself
1. A ball is thrown vertically upwards with a velocity of 30 m/s. Find:
(a) The maximum height reached by the ball.
(b) The total time taken by the ball to return to the ground.
(c) If the ball bounces back with the same speed after hitting the ground, what will be the change in momentum of the ball? (Take g =10 m/s2 and ignore air resistance.)
2. A car at rest accelerates uniformly at the rate of 2 m/s2 for 10 seconds. After that, it maintains its velocity for 20 s and finally decelerates to rest in 10 s. Calculate:
(a) The total distance travelled by the car.
(b) The average velocity of the car over the entire journey.
(c) Draw the graph of force experienced by the car with respect to time.
3. A gun of mass 3 kg fires a bullet of mass 0.03 kg with a velocity of 100 m/s. What is the recoil velocity of the gun?
4. Two skaters push off from each other and start moving in opposite directions. If one skater has a mass of 60 kg and moves backwards at 2 m/s, while the other moves forward at 3 m/s, what should be the mass of the second skater?
5. Match the items in each column with the appropriate items in the other two columns. (Sample Answer: 1 – b – iii)
Column 1 (Situation)
Column 2 (Driving Force)
Column 3 (Speed Variation)
1. Bicycle moving downhill a. Friction i. Speed up
2. Ball dropped from a tower b. Gravity ii. Speed down
3. Rocket moving in outer space c. No Force iii. Constant speed
6. Two blocks A and B of masses m1 and m2, respectively, are placed on a smooth, frictionless surface and connected by a light inextensible string (Fig. 8.14). A force F is applied horizontally to block A.
(a) Derive an expression for the acceleration of the system in terms of F, m1, m2.
(b) Find the force exerted by the string connecting the two blocks.
8.14
(c) What would happen to the force exerted by the string if m2 were much larger than m1?
(d) If the force F is applied to block B instead of A, how would the acceleration change?
Answers
Practice Questions
1. (b) 2. (a) 3. (b) 4. (b) 5. (c) 6. (d)
7. (a) 11. 3 kg.m/s 12. −12,000 N
14. Velocity: 14 m/s; Momentum: 28 kg.m/s
16. Velocity: 20 m/s; Distance: 40 m
17. a = F m ; a1 a2 = m2 m1 = 6 4 = 3 2 ; vt = u + at ; v1 v2 = a1 a2 = 3 2
3. Solution: When a car travels downhill with its engine turned off, it accelerates naturally due to the gravitational pull exerted by Earth. This phenomenon occurs because gravity acts on the car, pulling it downwards along the slope.
A similar experience can be observed when riding a bicycle or a motorbike down a bridge or a sloping road. The steeper the slope, faster is the acceleration. This demonstrates how gravity, in the absence of any opposing force, such as brakes or friction, can cause an object to gain speed as it moves downhill.
Challenge Yourself
1. (a) Maximum height = 45 m
(b) Total time = 6 s
2. (a) Total distance = 600 m
(b) Average velocity = 15 m/s
3. Recoil velocity of the gun = –1 m/s
4. Mass of the second skater = 40 kg
5. 1 – b – i
2 – b – i
3 – c – iii
6. (a) Acceleration of the system = a = F (m1 + m2)
(b) Force by the string = T = (m2.F) (m1 + m2)
(c) If m2 > m1, force exerted by the string approaches F.
(d) Acceleration remains a = F (m1 + m2) ; no change.
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 1.5 Hour
Max. Marks: 50
Multiple Choice Questions (10 × 1 = 10 Marks)
1. Which of the following statements best describes the concept of inertia in terms of the force required to change an object�s state of motion?
(a) The more force applied, greater the inertia of the object.
(b) The more mass an object has, the greater its inertia, and the more force is needed to change its motion.
(c) Inertia only applies to objects that are at rest.
(d) Inertia is unrelated to an object�s mass or force.
2. The SI unit of force is:
(a) Kilogram
(c) Newton
(b) Metre/second
(d) Joule
3. A ball is thrown vertically upwards. At its highest point
(a) its velocity is zero, but acceleration is non-zero.
(b) both velocity and acceleration are zero.
(c) velocity is non-zero, but acceleration is zero.
(d) both velocity and acceleration are non-zero.
4. Which of the following is NOT an application of Newton�s Third Law?
(a) Recoil of a gun
(c) Launching of a rocket
5. When a body moves with uniform velocity,
(a) a net force acts on it.
(c) it has constant acceleration.
(b) Swimming
(d) A car moving uphill
(b) no net force acts on it.
(d) its momentum keeps increasing.
6. Why does an astronaut start moving in the opposite direction when he throws an object while floating in space?
(a) Due to the force applied by the astronaut�s muscles.
(b) Because the object repels the astronaut in the opposite direction.
(c) Due to Newton�s Third Law of Motion, the action of throwing causes an equal and opposite reaction.
(d) Because of the lack of gravity.
7. A ball of mass 2 kg is initially moving with a velocity of 10 m/s. After a force is applied for 4 seconds, its velocity increases to 20 m/s. What is the change in momentum of the ball?
(a) 20 kg.m/s
(c) 40 kg.m/s
8. Action and reaction forces
(b) 30 kg.m/s
(d) 50 kg.m/s
(a) act on the same object. (b) act on different objects. (c) cancel each other out. (d) always produce motion.
9. If a force of 50 N is applied to a stationary object, it
(a) remains stationary.
(b) moves with constant velocity.
(c) accelerates in the direction of the force. (d) decelerates.
10. A 10 kg object is subjected to a constant force of 50 N. If the object initially has a velocity of 5 m/s, what will be its velocity after 4 seconds?
(a) 10 m/s (b) 15 m/s (c) 20 m/s (d) 25 m/s
Assertion–Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): A heavy object falls faster than a lighter object under free fall.
Reason (R): Gravitational acceleration is independent of the mass of the object.
12. Assertion (A): A cricket ball requires more force to stop than a tennis ball moving at the same speed.
Reason (R): The cricket ball has more mass than the tennis ball.
13. Assertion (A): A body continues to move even after the applied force is removed.
Reason (R): Inertia resists any change in the state of motion of a body.
14. Assertion (A): A soccer player feels a backward push while kicking the ball.
Reason (R): The more mass an object has, more is its inertia.
15. Case 1: A cyclist is travelling along a straight road at a steady speed. As he approaches a red traffic light, he notices the signal and decides to stop. To reduce his speed and eventually bring the cycle to a halt, he applies the brakes. Despite the braking force, the cycle does not stop instantly but slows down gradually and eventually comes to rest.
Answer the following:
(a) Which force/forces acting on the bicycle are trying to stop it?
(i) Gravitational force
(ii) Only frictional force between the tires and the road
(iii) Only frictional force between the brake pads and tyre rim
(iv) Frictional forces between the road and tyres and between the brake pads and tyre rim
(b) Why does the cycle not stop immediately after the brakes are applied?
(i) The brakes are not strong enough to stop the motion instantly.
(ii) Because the brakes take time to generate friction.
(iii) The cycle’s inertia resists the change in motion.
(iv) The force applied by the brakes is less than the cyclist�s pedalling force.
(c) How does mass affect the stopping distance of the cycle?
(i) A heavier cycle has more stopping distance due to greater inertia.
(ii) A lighter cycle has more stopping distance due to reduced friction.
(iii) Mass does not affect the stopping distance of the cycle.
(iv) Heavier and lighter cycles stop at the same distance if the same force is applied.
(d) Can the cycle slow down without the application of brakes?
(i) No, brakes are the only way to reduce speed.
(ii) Yes, due to the frictional force between the tyres and the road
(iii) No, unless an external force, like a collision, occurs
(iv) Yes, but only if the cyclist stops pedalling completely
16. Case 2: A soccer player kicks a stationary ball along the ground. The ball moves forward until it comes to rest after a few seconds.
Answer the following:
(a) Explain the role of force in setting the ball in motion.
(i) Force increases the ball�s mass, enabling it to move.
(ii) Force overcomes the ball�s inertia and sets it in motion.
(iii) Force decreases the ball�s weight, causing it to move.
(iv) Force is required only for the ball to stop.
(b) Why does the ball stop after a few seconds?
(i) Because no external force acts on it.
(ii) Because the ball�s mass is too high to continue motion.
(iii) Because the player’s kick was not strong enough.
(iv) Because of friction between the ball and the ground.
(c) Which law of motion applies when the ball is at rest initially?
(i) Newton’s First Law of Motion
(ii) Newton’s Second Law of Motion
(iii) Newton’s Third Law of Motion
(iv) The law of conservation of energy
(d) Which law of motion applies when the ball finally stops?
(i) Newton’s First Law of Motion.
(ii) Newton’s Second Law of Motion.
(iii) Newton’s Third Law of Motion.
(iv) The law of conservation of energy.
Very Short Answer Questions (30-50 words)
(3 × 2 = 6 Marks)
17. Why does a coin, kept on a cardboard placed on glass, falls into the glass, when the cardboard is flicked with a finger?
18. How can we express a force of magnitude 1 Newton (N) in terms of grams, centimetres and seconds?
19. Explain the effect of unbalanced forces on an object.
Short Answer Questions (50-80 words)
20. Why is it necessary to wear seat belts in a moving vehicle?
21. (a) State Newton’s Second Law of Motion.
(4 × 3 = 12 Marks)
(b) What happens to the acceleration of an object if the net force acting on it is doubled while its mass is halved?
22. A 10 kg block is placed on a horizontal frictionless surface. Five forces are acting on the block:
(a) A force of 30 N to the right
(b) A force of 20 N to the left
(c) A force of 10 N to the right
(d) Weight of the block acting downwards
(e) Reaction force of the ground acting upwards
Draw a diagram showing the forces acting on the block. Also, calculate the net force acting on the block and the acceleration of the block.
23. A truck of mass 2000 kg is initially at rest. A constant force of 4000 N is applied to the truck for 5 seconds.
(a) What is the acceleration of the truck?
(b) What will be the final velocity of the truck after 5 seconds?
(c) How much distance does the truck cover during this time?
24. Explain the process and principle behind a rocket launch. Include a diagram in your explanation.
25. A 5 kg box is sliding on a horizontal surface with an initial velocity of 10 m/s. A constant force of 25 N is applied in the opposite direction of motion, causing the box to decelerate.
(a) What is the acceleration of the box?
(b) How long does it take for the box to come to rest?
(c) How will the acceleration and stopping time change, if the applied force is reduced by 10 N?
9 Gravitation
This chapter delves into the fascinating dynamics of gravitation and the forces that govern the motion of celestial bodies and everyday objects. It begins by introducing the universal law of gravitation, which explains how all masses, from planets to moons, exert an attractive force on each other. Key concepts such as gravitational force, centripetal force, free fall, and the relationship between mass and weight are explored, providing insight into both celestial movements and everyday experiences. Additionally, the chapter discusses the principles of buoyancy, uncovering why objects float or sink in fluids, and presents Archimedes’ principle, illustrating its real-world applications. Through these interconnected principles, we deepen our understanding of the natural forces that shape the world around us.
Gravitational Force
• The force with which the earth pulls the bodies towards it.
• It dose not need any contact between two bodies.
Gravitational Constant
• The gravitational force between two objects of unit mass each separated by a unit distance from each other.
F = G
• It is a universal constant.
• Its unit is Nm2/kg2
• Its value is 6.67 × 10–11 Nm2/kg2
Centripetal Force
The force that keeps a body moving along the circular path acting towards the centre.
Free Fall
• When objects fall towards the earth under the gravitational force alone.
• In vacuum, all bodies fall towards earth with the same speed.
Acceleration due to Gravity (g)
Acceleration produced in a free falling body due to gravitational attraction of earth.
The value of g is 9.8 m/s2.
Equation of motion for free fall
(i) v = u + gt
(ii) h = ut + 1 2 gt2
(iii) v2 = u2 + 2gt
Universal Law ot Gravitation
• The attractive force between any two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of distance between them.
GRAVITATION
Mass
• It is the quantity of matter contained in a body.
• Its Sl unit is kg.
• It cannot be zero at any place.
Weight
• The force with which a body is attracted towards the centre of earth.
• Itʼs Sl unit is newton.
w = mg
• Weight of body changes from place to place.
• Weight of an object on moon is one-stxth of the weight on the earth.
Thrust, Pressure and Density
• Thrust Thrust is the force acting on an object perpendicular to its surface.
• Density The density of a substance is defined as mass unit volume.
• Pressure Pressure is the force acting perpendicularly on a unit area of an object.
• Pressure in Fluids All liquids and gases are together called fluids.
• Buoyancy The tendency of a liquid to exert an upward force on an object immersed in it is called buoyancy.
Factors Affecting Buoyant Force
(i) Density of the fluid.
(ii) Volume of object immersed in the liquid.
Floating or Sinking of Objects in Liquid
• When an object is immersed in a liquid, then following two forces act on it
Weight of the object which acts in downward direction.
Buoyant force which acts in upward direction.
(i) If the buoyant force or upthrust exerted by the liquid is less than the weight of the object, the object will sink
(ii) If the buoyant force is equal to the weight of the object, the object will float in the liquid.
(iii) If the buoyant force is more than the weight of the object, the object will rise in the liquid and then float.
Archimedes Principle
When an object is fully or partially immersed in a liquid, it experiences a buoyant force or upthrust which is equal to the weight of liquid displaced by the object.
Applications
• Designing ships and subrnarines
• Lactometer (a device used to determine purity ot milk)
• Hydrometer (a device used to determine the density of liquid)
Chapter at a Glance
• Gravitation is the force that attracts objects toward each other, and its strength depends on their masses and the distance between them.
• The law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Since this law holds true everywhere in the universe, it is called a universal law.
• Gravitational force is generally weak, except when very large masses are involved.
• The universal gravitational constant (G = 6.67 × 10–11 Nm2kg–2) is defined as the gravitational force between two bodies of unit masses separated by a unit distance from each other and placed anywhere in space.
• Centripetal force is essential for maintaining circular motion. It acts inward, toward the center of the circle, to maintain the object’s circular path.
• Weight is the gravitational force acting on an object due to its mass. The weight of an object changes depending on location. For instance, an object on the Moon has a lesser weight compared to Earth due to the Moon’s weaker gravitational force.
• The strength of gravity decreases with altitude and also varies across Earth’s surface, becoming weaker from the poles toward the equator.
• Free fall motion occurs when an object is subject to Earth’s gravitational attraction alone, accelerating at a constant rate of approximately 9.8 m/s², regardless of the object’s mass.
• Galileo’s experiments showed that in free fall, all objects experience the same rate of acceleration, regardless of their mass, as long as air resistance is negligible.
• Mass represents the amount of matter in an object, measured in kilograms, and it remains constant regardless of the object’s location.
• Weight of a body is the force exerted by gravity due the object’s mass. For instance, a 10 kg object weighs 98 N on Earth and less on the Moon.
• Since acceleration due to gravity slightly change from place to place on earth, the weight may also vary from place to place but the mass stays constant.
• Acceleration due to gravity (g) is approximately 9.8 m/s² on Earth and dictates the speed at which objects fall.
• Sign of acceleration due to gravity (g):
If an object falls vertically downwards, then acceleration due to gravity taken as positive.
If an object is thrown vertically upwards, then acceleration due to gravity taken as negative.
• Thrust is the force applied by an object perpendicular to a surface.
• Buoyancy is the upward force exerted by a fluid on an object submerged in it. It determines whether the object will float or sink based on its density relative to the fluid.
• Buoyancy principles are applied in the design of ships to ensure they displace enough water to stay afloat by balancing the upward buoyant force with their weight.
• Archimedes’ Principle states that the buoyant force acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. This principle is crucial for understanding the behavior of objects in fluids.
• Objects float when their density is less than the fluids, and they sink when their density is greater.
Formulae
1. Gravitational Force (F) 2 Mm G d
2. Weight of an Object (W) = m × g
3. Equations of Motion for free fall
• v = u + gt
• h = ut + 1 2 gt2
• v2 = u2 + 2gh
4. Acceleration due to gravity (g) = GM R2
5. Pressure = Thrust Area
6. Density = Mass Volume
NCERT Zone
Intext Questions
1. State the Universal Law of Gravitation.
Ans. The universal law of gravitation states that every object in the universe attracts every other object with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This law is universal as it applies to all objects, both celestial and terrestrial.
2. Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the Earth.
Ans. The formula to find the gravitational force between the Earth and an object on its surface is given by Newton’s universal law of gravitation:
where F = Gravitational force, G = Universal gravitational constant, M = Mass of the Earth, m = Mass of the object and R = Radius of the Earth.
3. What do you mean by free fall?
Ans. Free fall refers to the motion of an object where the only force acting upon it is gravity. In free fall, an object falls toward the Earth solely due to gravitational attraction, without any other forces like air resistance opposing its motion. During free fall, the object experiences uniform acceleration termed as the acceleration due to gravity (g).
4. What do you mean by acceleration due to gravity?
Ans. Acceleration due to gravity, denoted by g, is the acceleration experienced by an object due to the gravitational force of the Earth. This acceleration causes freely falling objects to increase their velocity by approximately 9.8 m/s² (near earth’s surface); however, it can vary slightly depending on location due to factors such as altitude and Earth’s rotation.
Acceleration due to gravity (g) = GM R2
5. What are the differences between the mass of an object and its weight?
Mass Weight
Quantity of matter contained in an object. Force with which an object is attracted towards the centre of the earth due to gravity.
The SI unit is kilograms (kg).
Mass is never zero.
The SI unit is newtons (N).
Weight is zero when gravitational force is zero. Example: at the centre of the earth or free space.
The value of mass remains constant everywhere. Weight changes depending on the strength of gravity at the location.
Mass has only magnitude.
Weight is a force and hence has both magnitude and direction.
6. Why is the weight of an object on the moon 1 6 of its weight on the earth?
Ans. The weight of an object on the Moon is 1 6 of its weight on Earth because the gravitational force of the Moon is weaker. This is due to the Moon’s smaller mass and radius compared to Earth. Thus, the gravitational acceleration (g) on the Moon is only about 1 6 of that on Earth, resulting in a correspondingly lower weight for the same mass.
7. Why is it difficult to hold a school bag having a strap made of a thin and strong string?
Ans. A school bag with a thin strap exerts more pressure on the shoulder because pressure is the force per unit area. A thin strap has a smaller area, leading to greater pressure for the same weight of the bag. This increased pressure can cause discomfort or pain, making it difficult to hold the bag.
8. What do you mean by buoyancy?
Ans. Buoyancy is the upward force exerted by a fluid on an object immersed in it. This force acts against gravity, making objects in the fluid feel lighter. The magnitude of the buoyant force is equal to the weight of the fluid displaced by the object. Buoyancy allows objects with appropriate densities to float or makes it easier to lift submerged objects.
9. Why does an object float or sink when placed on the surface of water?
Ans. An object floats or sinks in water depending on its weight and the buoyant force acting on it.
• If the buoyant force is equal to or greater than the weight of the object, it floats.
• If the buoyant force is less than the weight of the object, it sinks. This also depends on the density of the object.
• If the object is less dense than water, it floats.
• If the object is denser than water, it sinks.
This happens because of Archimedes’ Principle, which says that a body immersed in a fluid experiences an upward force equal to the weight of the fluid it displaces.
10. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
Ans. When we use a weighing machine, we are actually measuring the weight of the object, but we often refer to this measurement as the mass.
Since the weighing is done in air, there is an upward buoyant force exerted by the air on the object. Therefore, the weight measured by the machine is not the true weight.
Reading of weighing machine = True weight – Force of buoyancy.
So it is clear that the true weight must be more than the measured weight (42 kg. Hence, the mass of boy is more than 42 kg.
11. You have a bag of cotton and an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than the other. Can you say which one is heavier and why?
Ans. In reality, the bag of cotton is heavier.
This is because the upward thrust (buoyant force) due to air acting on an object is equal to the weight of the air displaced by the object. Since the volume of the bag of cotton is larger than that of the iron bar, it displaces more air.
As a result, the buoyant force on the cotton bag is greater, which reduces the apparent weight measured by the weighing machine more than for the iron bar. Therefore, the true weight of the cotton bag is actually more than that of the iron bar.
NCERT Exercises
1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Ans. Let the mass of objects be m1 and m2 and the original distance between the objects be r.
Force of gravitation, F = G M × m r2
∴ Original force of gravitation, Fi = G m1 × m2 r2
If distance is reduced to half, i.e. r’ = r 2ʹ
mm
then new force of gravitation, Ff = G 12 2
The force of gravitation becomes 4 times of the original value.
2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Ans. Although the gravitational force acting on an object increases with its mass, the acceleration produced by this force remains the same for all masses.
From Newton’s second law of motion, acceleration, a = F m 2 () GmM r ag m
= GM r2
Since G, M and r have fixed values, g will be constant. This proves that all objects fall at the same rate because acceleration due to gravity (g), approximately 9.8 m/s², is constant regardless of the mass of the object. Hence, a heavy object falls with same acceleration as light object.
3. What is the magnitude of the gravitational force between the Earth and a 1 kg object on its surface? (Mass of the Earth is 6 × 10²⁴ kg and radius of the Earth is 6.4 × 10⁶ m.)
Ans. Given:
G = 6.67 × 10⁻¹¹ N m²/kg² (gravitational constant)
M = 6 × 10²⁴ kg (mass of Earth)
m = 1 kg (mass of object)
R = 6.4 × 10⁶ m (radius of Earth)
The gravitational force, F = G M × m d2
F = (6.67 × 10⁻¹¹ × 6 × 1024 × 1) (6.4 × 106)² ≈ 9.8 N
Therefore, the gravitational force is approximately 9.8 N.
4. The Earth and the Moon are attracted to each other by gravitational force. Does the Earth attract the Moon with a force that is greater or smaller or the same as the force with which the Moon attracts the Earth? Why?
Ans. According to Newton’s third law of motion, every action has an equal and opposite reaction. The Earth attracts the Moon and the Moon attracts the Earth, these two forces are action-reaction pair. Hence, the gravitational force exerted by the Earth on the Moon is equal in magnitude and opposite in direction to the force exerted by the Moon on the Earth.
5. If the Moon attracts the Earth, why does the Earth not move towards the Moon?
Ans. Although the Moon attracts the Earth with an equal gravitational force as the Earth attracts the Moon, the Earth’s much larger mass means it experiences negligible acceleration. Thus, its movement towards the Moon is imperceptible, whereas the lighter Moon exhibits a noticeable orbit around the Earth due to its comparatively smaller mass.
6. What happens to the force between two objects if
(a) the mass of one object is doubled?
(b) the distance between the objects is doubled and tripled?
(c) the masses of both objects are doubled?
Ans. Let the mass of objects be m1 and m2 and the distance between objects be r.
Gravitational force, F = G m1 × m2 r2
(a) Given, new mass
If the mass of one object is doubled, the gravitational force doubles.
(b) Given, new distance r = 2r
then the new force,
If the distance is doubled, the force decreases to one-fourth. Similarly, if distance is tripled, the force reduces to one-ninth of the original.
(c) Given, new masses m1 = 2m1 and m2 = 2m2 12
If both masses are doubled, the gravitational force becomes four times stronger.
7. What is the importance of the universal law of gravitation?
Ans. The Universal Law of Gravitation successfully explains several important natural phenomena, including:
1. The force that keeps us firmly bound to the Earth.
2. The Moon’s orbit around the Earth.
3. The motion of planets revolving around the Sun.
4. The occurrence of tides caused by the gravitational pull of the Moon and the Sun.
5. The flow of water in rivers, driven by the Earth’s gravitational attraction on the water.
8. What is the acceleration of free fall?
Ans. The acceleration of free fall refers to the acceleration caused by Earth’s gravitational pull on freely falling objects. It is also known as acceleration due to gravity (g), which is approximately 9.8 m/s² on Earth’s surface. This acceleration is independent of mass of the falling object.
9. What do we call the gravitational force between the Earth and an object?
Ans. The gravitational force between the Earth and an object is called the force of gravity (or simply gravity). In everyday physics problems, we also refer to this force as the object’s weight.
10. Amit buys a few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of the gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator.]
Ans. Amit’s friend may find the weight of gold to be slightly less at the equator than at the poles. This is because weight of an object is product of its mass and the acceleration due to gravity (g). This acceleration, g, is greater at the poles due to the Earth’s shape, causing objects to weigh more there compared to the equator where g is lesser, influencing the weight measurement.
w = mg (i)
w = G M × m R2 (ii)
From Eqs. (i) and (ii), we get
mg = GMm R2 g ∝ 1 R2 Or g ∝ 1 (Distance of an object from centre of the earth)2
11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans. A sheet of paper falls slower because it has a larger surface area and faces more air resistance than when it is crumpled into a ball. Air resistance significantly slows the paper, counteracting gravity, whereas a crumpled ball has reduced air resistance due to its smaller surface area, allowing it to fall faster.
12. Gravitational force on the surface of the moon is only 1 6 as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the Moon and on the Earth?
Ans. Given, mass of the object, m = 10 kg
Weight on the earth, w = mg
= 10 × 9.8 = 98 N
Since gravity on moon is 1 6 of the gravity on Earth, weight of an object on the moon will be equal to 1 6 of its weight on the earth.
therefore, weight of 10 kg object on moon = 1 6 × weight on earth
= 1 6 × 98 = 49 3 = 16.33 N
13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate (a) the maximum height to which it rises, (b) the total time it takes to return to the surface of the Earth.
Ans. (a) Given,
v = 0 m/s (at max height)
u = 49 m/s
a =-9.8 m/s²
Using 3rd equation of motion,
v² = u² + 2as
0 = 49² - 2 × 9.8 × h
⇒ h = 49 × 49 2 × 9.8 = 122.5 m
∴ Maximum height attained (h) = 122.5 m.
(b) Using 2nd equation of motion,
s = ut + 1 2 at2
0 = 49 × t1 2 × 9.8 × t2
⇒ 0 = 49 × t - 4.9 × t2
⇒ 0 = 49 × t × (1 - 0.1 × t)
∴ t = 0 or 10 s
At t = 0 sec, ball started its journey, therefore, it returns to the ground at t = 10 sec.
14. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Ans. Given,
u = 0 (ball is released)
a =+ 9.8 m/s²
s =+ 19.6 m.
Using the 3rd equation of motion:
v² = u² + 2as
v2 = 0 + 2 × 9.8 × 19.6 = 2 × 9.8 × 2 × 9.8
⇒ v = 2 × 9.8 = 19.6 m/s
Therefore, the final velocity is approximately 19.6 m/s.
15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Ans. Given,
u = 40 m/s
a =-10 m/s2
at max. height, v = 0
Using the 3rd equation of motion,
v² = u² + 2as
0 = 40² - 2 × 10 × h
h = 40 × 40 2 × 10 = 80 m
Maximum height achieved by stone is 80 m.
Since the stone moves up to the maximum height and then falls back to the starting point, therefore net displacement will be zero.
The total distance covered will be length of path, which is twice the maximum height, i.e. 2 × 80 = 160 m.
16. Calculate the force of gravitation between the Earth and the Sun, given that the mass of the Earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.
Ans. Given,
G = 6.674 × 10⁻¹¹ N m²/kg²
mass of the Sun, M s = 2 × 10³⁰ kg
mass of Earth, Me = 6 × 10²⁴ kg
distance between the earth and the sun, r = 1.5 × 10¹¹ m
The gravitational force (F) between the Earth and the Sun,
F = (G × M s × M e) r2 = (6.674 × 10⁻¹¹ × 2 × 10³⁰ × 6 × 10²⁴) (1.5 × 10¹¹)² = 3.56 × 1022 N
17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Ans. Given,
height of tower, h = 100 m
For the stone falling down:
u = 0
a = 10 m/s2
For the stone projected upwards:
u = 25 m/s
a =-10 m/s2
Let the time for both stones to meet be t sec.
Using 2nd equation of motion,
For the stone falling downward, distance travelled: s1 = ut + 1 2
the stone projected upwards, using
When both stones meet, the sum of distances travelled by both stones will be equal to height of tower.
s₁ + s₂ = 100
5t² + 25t - 5t² = 100
25t = 100 ⇒ t = 4 seconds.
At t = 4 s, the first stone has fallen s₁ = 5 × (4)² = 80 m.
Therefore, the two stones meet 80 m below the top of the tower (or 20 m above the ground) at t = 4 sec.
18. A ball thrown up vertically returns to the thrower after 6 s. Find (a) the velocity with which it was thrown up, (b) the maximum height it reaches, and (c) its position after 4 s. Ans. Given,
Total time of travel, T = 6 sec
For upwards motion, a =-9.8 m/s2
For downwards motion, a = 9.8 m/s2
Let the maximum height of ball during motion be h.
Since the total time of travel is 6 s, so the time to reach maximum height will be T 2 s or 3 s.
(a) Using the 1st equation of motion, v = u + at
0 = u - 9.8 × 3
u = 29.4 m/s
So, the initial velocity with which the ball was thrown up is 29.4 m/s.
(b) Using the 3rd equation of motion, v² = u² + 2as
0 = (29.4)2 - 2 × 9.8 × h
h = (29.4)2 2 × 9.8 = 44.1 m
So, the maximum height reached by the ball is 44.1 m.
(c) In initial 3 s, the ball will rise, then in the next 3 s it falls toward the earth.
∴ The position after 4 s = Distance covered in 1 s in the downward motion
From the 2nd equation of motion,
s = ut + 1 2 at2
s = 0 × 1 + 1 2 × 9.8 × 12 = 4.9 m
Height of ball from ground = (44.1 - 4.9) m = 39.2 m
Therefore, after 4 seconds, the ball will be 39.2 m above the ground, or 4.9 m below the top of the tower.
19. In what direction does the buoyant force on an object immersed in a liquid act?
Ans. The buoyant force on an object immersed in a liquid act in the upward direction, i.e. opposite to the force of gravity. This upward force makes objects feel lighter in the liquid.
20. Why does a block of plastic released under water come up to the surface of water?
Ans. Since most plastics are less dense than water, the upward buoyant force they experience is greater than their gravitational weight; consequently, once immersed, the plastic block accelerates upward until it reaches the surface and settles partly submerged, floating stably.
21. The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g/cm³, will the substance float or sink?
Ans. Given,
Mass of the substance = 50 g
Volume of the substance = 20 cm³
density of water, ρw = 1 g/cm³
If a substance’s density (ρs) is greater than water’s density (ρw), the upward buoyant force is smaller than the substance’s weight, so it sinks. If ρs is less than ρw, the buoyant force exceeds its weight, allowing the substance to float on the water’s surface.
The density of given substance, ρs = Mass Volume = 50 g 20 cm3 g/cm³
Since the density of the substance (2.5 g/cm³) is greater than the density of water (1 g/cm³), the substance will sink in water.
22. The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g/cm³? What will be the mass of the water displaced by this packet?
Ans. Given,
Mass of the packet = 500 g
Volume of the packet = 350 cm³
Density of water = 1 g/cm³
The density of sealed packet, ρsp = Mass Volume = 500 g 350 cm3 = 1.43 g/cm³
Since the density of the packet (1.43 g/cm³) is greater than the density of water (1 g/cm³), the packet will sink in water.
The mass of water displaced by the packet = the volume of the displaced water × density of water = 350 cm³ × 1 g/cm³ = 350 g.
Multiple Choice Questions
1. What is the relationship between gravitational force and the mass of the objects?
(a) Gravitational force is inversely proportional to mass
(b) Gravitational force is proportional to the product of masses
(c) Gravitational force is independent of mass
(d) Gravitational force is proportional to the square of mass
2. What happens to the gravitational force between two objects if the distance between them is halved?
(a) It increases by a factor of 2
(c) It decreases by a factor of 4
(b) It increases by a factor of 4
(d) It decreases by a factor of 2
3. Why does a heavy object not fall faster than a light object in the absence of air resistance?
(a) Both experience the same acceleration (b) Heavier objects experience more resistance (c) Lighter objects have more air resistance (d) All objects fall at the same speed
4. The weight of an object is the product of its mass and:
(a) the acceleration due to gravity. (b) the speed of light.
(c) the force of friction.
(d) the mass of the Earth.
5. If an object’s weight is measured on the Moon, it will be:
(a) The same as its weight on Earth. (b) Twice its weight on Earth.
(c) Six times its weight on Earth.
(d) One-sixth of its weight on Earth.
6. Law of gravitation gives the gravitational force between (NCERT Exemplar) (a) the earth and a point mass only (b) the earth and the sun only (c) any two bodies having some mass (d) two charged bodies only
7. What causes objects to float in a liquid?
(a) Buoyant force (b) Gravitational pull (c) Air resistance (d) Friction
8. The weakest force out of the following is (a) magnetic force (b) nuclear force (c) gravitational force (d) electric force
9. The atmosphere is held to the earth by (NCERT Exemplar) (a) gravity (b) wind (c) clouds (d) earth’s magnetic field
10. Why does the Earth not move towards the apple when it falls?
(a) The Earth’s mass is too large for any noticeable movement.
(b) The apple exerts no force on Earth.
(c) The distance is too great to detect the movement.
(d) The force of gravity is not strong enough.
11. A planet has the same radius as Earth but only half its density. The weight of an object on this planet would be:
(a) the same as on Earth
(c) half its weight on Earth
(b) twice its weight on Earth
(d) one-quarter its weight on Earth
12. What is the force of gravity between the Earth and an object with mass 2 kg on its surface?
(a) 1.96 N (b) 2 N
(c) 19.6 N (d) 20 N
13. Why does a steel ship stay afloat while an equal mass of steel shaped into a solid cube sinks?
(a) The ship displaces less water, creating a larger upward buoyant force.
(b) Steel used in ships has a special property that stops it from sinking.
(c) Because of its design, the ship pushes water aside, keeping it from sinking.
(d) The solid steel cube has a greater density than the ship’s overall average density.
14. Why does a crumpled piece of paper fall faster than a flat one?
(a) The crumpled paper has less surface area, which reduces air resistance.
(b) The crumpled paper is denser, so it accelerates faster.
(c) The flat paper creates a vacuum around it, making it fall slower.
(d) The crumpled paper has a greater mass, increasing its gravitational pull.
15. The force of attraction between two unit point masses separated by a unit distance is called (NCERT Exemplar)
(a) gravitational potential.
(c) gravitational field.
(b) acceleration due to gravity.
(d) universal gravitational constant.
16. As illustrated in Fig. 9.1, Earth moves in a nearly circular orbit around the Sun. At the instant shown, the centripetal force acting on Earth is directed
(a) away from the Sun.
(b) toward the Sun.
(c) along the tangent to Earth’s orbital path.
(d) perpendicular to the plane of Earth’s orbit.
17. The value of universal gravitational constant (G) was found out by
(a) Issac Newton (b) Johannes Kepler (c) Galileo Galilei (d) Henry Cavendish
18. A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(a) maximum when length and breadth form the base.
(b) maximum when breadth and width form the base.
(c) maximum when width and length form the base.
(d) the same in all the above three cases.
19. Why does a military tank have wide tracks?
(a) To increase the friction between the tank and the ground for better control.
(b) To reduce the pressure exerted on the ground, preventing the tank from sinking.
(c) To make the tank heavier and more stable during movement.
(d) To increase the speed of the tank on rough terrains.
20. The principle used to design submarines is:
(a) Archimedes’ principle (b) Newton’s law of motion
(c) The law of conservation of energy (d) The law of universal gravitation
21. A solid wooden cube weighing 4 kg and measuring 20 cm on each side is carefully placed so that one whole face rests flat on a horizontal tabletop. Assuming the entire lower face is in full contact with the table, calculate the pressure the block exerts on the surface (use g = 10 ms⁻²).
(a) 100 Pa (b) 250 Pa (c) 1000 Pa (d) 10000 Pa
22. Calculate the mass of an object whose weight is 49 N on Earth. (use g = 980 cm/s²)
(a) 0.05 kg (b) 0.5 kg (c) 5 kg (d) 50 kg
23. The value of ‘g’ on the Moon is: (a) 1.63 m/s² (b) 9.80 m/s² (c) 0.98 m/s² (d) 0 m/s²
Fig. 9.1
24. The pressure exerted by a nail on the surface it is driven into is:
(a) High, because the nail’s tip has a very small contact area.
(b) Low, because the nail is made of a strong metal which spreads out the force.
(c) Low, because the nail’s length distributes the force over a larger area.
(d) High, because the nail is heavy and presses strongly on the surface.
25. Which of the following best describes thrust?
(a) The force applied by an object parallel to a surface.
(b) The force applied by an object perpendicular to a surface.
(c) The force that opposes motion between two surfaces.
(d) The force due to gravity acting on an object.
Answer
Constructed Response Questions
Very Short Answer Questions (30–50 words)
1. Who formulated the universal law of gravitation? Why is ‘G‘ called the universal gravitational constant?
Ans. Isaac Newton formulated the universal law of gravitation. The constant ‘G‘ is universal because it is independent of the nature and sizes of bodies, the space where they are kept and at the time at which the force is considered.
2. (a) At what place on the Earth’s surface is the weight of a body (i) maximum (ii) minimum
(b) The Earth is acted upon by gravitation of the Sun, even though it does not fall into the Sun. Why? (NCERT Exemplar)
Ans. (a) (i) At the poles (ii) At the equator
(b) The gravitational force provides the necessary centripetal force that allows the Earth to move around the Sun along a stable orbit. Therefore, the Earth does not fall into the Sun.
3. Why does the formation of tides take place in the sea or ocean?
Ans. Tides arise because the Moon’s gravity pulls ocean water toward it, creating a bulge, while an opposite bulge forms on Earth’s far side due to inertia. The Sun’s gravitational attraction reinforces or partially cancels these bulges. Earth’s rotation carries coastlines beneath them, producing the regular rise and fall called tides.
4. Find the gravitational force between two protons kept at a separation of 1 femtometer (1 femtometer = 10–15 m). The mass of proton is 1.67 × 10–27 kg.
Ans. The gravitational force between the two protons is
F = Gm1m2 r2 = 6.67 × 10–11 × (1.67 × 10–27)2 (10-15)2 = 50 27 × 10–34 N
F = 1.85 × 10–34 N
5. How does the force of attraction between the two bodies depend upon their masses and distance between them? A student thought that two bricks tied together would fall faster than a single one under the action of gravity. Do you agree with his hypothesis or not? Comment.
Ans. According to law of gravitation, F ∝ m1 m2 and F ∝ 1 r2
Hypothesis of the student is not correct. The two bricks, like a single body, fall with the same speed to reach the ground at the same time in case of free fall. This is because acceleration due to gravity is independent of the mass of the falling body.
6. Identical packets are dropped from two aeroplanes – one above the equator and other above the north pole, both at height h. Assuming all conditions to be identical, will those packets take the same time to reach the surface of Earth? Justify your answer. (NCERT Exemplar)
Ans. The value of ‘g’ at the equator of the Earth is less than that at the poles. Therefore, the packets fall slowly at equator in comparison to the poles. Thus, the packets will remain in air for a longer time interval, when it is dropped at the equator.
7. (a) If the distance between two objects is increased four times, then by how many times will the mass of one of the objects be changed to maintain the same gravitational force? (b) Can a body have mass but no weight?
Ans. (a) Gravitational force between two particles m1 and m2 at a distance of r is
F = Gm1m2 r2
When distance is increased by four times,
F = Gm1m2 (4r)2 = Gm1m2 16r2 = F 16
Thus, one of the masses has to be increased 16 times so as to maintain the same gravitational force. (b) A body with mass can indeed be weightless when the acceleration due to gravity (g) is zero—for example, at Earth’s centre.
W = mg
8. Two objects of masses m1 and m2 having the same size are dropped simultaneously from heights h1 and h2, respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if (i) one of the objects is hollow and the other one is solid; and (ii) both of them are hollow, size remaining the same in each case? Give reasons.
Ans. Given,
initial speed of both objects, u = 0 dropping height of m1 = h2 dropping height of m2 = h2
Using 2nd equations of motion, s = ut + 1 2 at2
h1 = 1 2 gt2 1
h2 = 1 2 gt2 2
11 22 th th
(NCERT Exemplar)
Since the acceleration during free fall does not depend upon mass and size, the ratio will not change in either case.
9. A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration? Take g = 9.8 m/s2.
Ans. Given, Force, F = 20 N weight, W = 9.8 N
acceleration due to gravity, g = 9.8 m/s2
let mass of body be m kg.
we know, W = mg
⇒ m = W g = 9.8 9.8 = 1 kg
So, acceleration, a = F m = 20 1 = 20 m/s2
10. Calculate the value of acceleration due to gravity ‘g‘ using the relation between g and G.
Ans. We know that G = 6.67 × 10–11 Nm2 kg–2
Mass of the Earth, M e = 6 × 1024 kg
And radius of the Earth, R e = 6.4 × 106 m
We know, g = G × M e R2 e
∴ g = 6.67 × 10–11 × 6 × 10–24 (6.4 × 106)2 m/s2
⇒ g = 6.67 × 6 × 10 6.4 × 6.4 m/s2 ≈ 9.8 m/s2
11. Give reason.
(a) The railway tracks are laid on large sized concrete sleepers.
(b) The tip of the sewing needle is sharp.
Ans. (a) Railway tracks rest on large concrete sleepers because the sleepers spread the train’s weight over a wider area. A larger area means lower pressure on the ground and rails, preventing them from sinking or bending.
(b) A sewing needle has a very sharp tip so that the force you apply is concentrated on a tiny area. This creates a high pressure, allowing the needle to pierce fabric easily.
12. (a) What do you mean by buoyant force? What is the SI unit of buoyant force?
(b) State Archimede’s Principle.
Ans. (a) It is defined as the upward force exerted by a fluid on a body immersed in it. The SI unit of buoyant force is newton or N.
(b) Archimedes’ Principle states that the buoyant force acting on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.
13. An elephant weighing 50,000 newton stands on one foot of area 1000 cm2 as shown in Fig. 9.2. What is the pressure exerted on the ground?
Ans. Given,
Force (weight of elephant), W = 50000 N
Area, A = 1000 cm² = 1000 × 10⁻⁴ m² = 0.1 m²
Now, applying the formula: P = Force Area = 50000 N 0.1 m2 = 50000 N/m²
The pressure exerted by the elephant on the ground is: 5 × 10⁵ N/m²
Short Answer Question (50–80 words)
1. (a) When does an object show weightlessness?
(b) If the small and big stones are dropped from the roof of a house simultaneously, they will reach the ground at the same time. Why?
Ans. (a) Weightlessness is a state when an object does not weigh anything. It occurs only when a body is in a state of free fall under the effect of gravity alone.
Fig. 9.2
(b) The acceleration due to gravity does not depend upon the mass of the stone or body. Both the bodies fall with the same acceleration towards the surface of the Earth. Thus, a big stone will fall with the same acceleration as a small stone. So, both the stones will reach the ground at the same time when dropped simultaneously.
2. (a) What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?
(b) Suppose gravity of Earth suddenly becomes zero, then which direction will the Moon begin to move if no other celestial body affects it? (NCERT Exemplar)
Ans. (a) Gravitational force between the Sun and planet. This force depends on the product of the masses of the planet and Sun and the distance between them.
(b) The Moon will begin to move in a straight line in the direction in which it was moving at that instant because the circular motion of Moon is due to centripetal force provided by the gravitational force of the Earth.
3. Why does a body reach the ground quicker at poles than at the equator when dropped from the same height?
Ans. When you drop an object, the time it takes depends on g at that place. At the poles g is slightly higher than at the equator because the poles are nearer Earth’s centre and feel no outward push from Earth’s spin. With the same height and zero speed, the stronger pull at the poles makes the object speed up faster, so it reaches the ground sooner.
4. If the Earth’s radius were to become twice its original size while its mass remains unchanged, what would be the new weight of a person who currently weighs 980 N?
Ans. We know that F = GMm r2
As weight of a body is the force with which a body is attracted towards the Earth,
∴ W = GMm r2
If the radius of the Earth becomes twice of its original radius, then new weight will be W̕ʹ = GMm (2r)2 = GMm 4r2 = W 4 = 980 N 4 = 245 N
Therefore, new weight will be 245 N or one-fourth of the original weight.
5. On the Earth, a stone is thrown from a height in a direction parallel to the Earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why? (NCERT Exemplar)
Ans. For both the stones,
Initial velocity in downward direction, u = 0
Acceleration in downward direction = g
displacement, s = h
Using 2nd equation of motion, s = ut + 1 2 at2
h = ut + 1 2 gt2
⇒ h = 0 + 1 2 gt2 = 1 2 gt2 2h t g
Both stones will take the same time to reach the ground because the two stones fall from the same height.
6. On the Moon’s surface, the acceleration due to gravity is 1.67 ms–2. If the radius of the Moon is 1.74 × 106 m, calculate the mass of the Moon. (use G = 6.67 × 10–11 Nm2 kg–2)
Ans. Given,
acceleration due to gravity, g = 1.67 ms–2
radius of the Moon, R = 1.74 × 106 m
G = 6.67 × 10–11 Nm2 kg–2
we know,
g = GM R2 ⇒ M = gR2 G
M = 1.67 × (1.74 × 106)2 6.67 × 10–11 = 7.6 × 1022 kg
7. How does the weight of an object vary with respect to mass and radius of the Earth? In a hypothetical case, if the diameter of the Earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the Earth be affected? (NCERT Exemplar)
Ans. Weight of an object is directly proportional to the mass of the Earth and inversely proportional to the square of the radius of the Earth, i.e., Original weight, W0 = mg = m × G × M R2
Therefore, Weight of a body ∝ M R2
When hypothetically M becomes 4M and R becomes R 2,
Then new weight,
= 16 × m × G × M R2 = 16 × W0
The weight will become 16 times.
8. A ball weighing 4 kg of density 4000 kgm–3 is completely immersed in water of density 103 kgm–3 Find the force of buoyancy on it. (Given g = 10 ms–2) (NCERT Exemplar)
Ans. Given:
Mass of the ball, m = 4 kg
Density of the ball, ρb = 4000 kg/m³
Density of water, ρw = 1000 kg/m³
acceleration due to gravity, g = 10 m/s²
the volume of the ball, V = mass density = 4 1000 = 0.001 m³
According to Archimedes’ principle, the buoyant force is equal to the weight of the water displaced by the object.
the volume of water displaced = the volume of the ball = V
Now, weight of the water displaced = mass of displaced water × g = volume × density × g (mass = volume × density)
= 0.001 × 1000 × 10 = 10 N
So, the buoyant force acting on the ball is 10 newtons.
9. Tabulate the difference between acceleration due to gravity (g) and universal gravitational constant (G).
Acceleration due to Gravity (g)
It is the acceleration acquired by a body due to Earth’s gravitational pull.
Universal Gravitational Constant (G)
It is the constant of proportionality in Newton’s law of gravitation, equal to the force between two 1 kg masses placed 1 m apart.
It has both magnitude and direction. It has only magnitude, but no direction.
Its value varies from place to place on Earth and differs for other celestial bodies.
It is a universal constant with the same value everywhere in the universe. (i.e. 6.67 × 10–11 Nm2 kg–2).
10. (a) Density of glass is 3.5 g/cm3. What does it mean?
(b) Name the instrument which is used to determine the density of liquid.
(c) Give a simple activity to prove how the objects of density less than that of liquid float on it.
Ans. (a) 3.5 g/cm3 density of glass means that the 1 cm3 volume of glass weighs 3.5 g.
(b) The instrument which is used to determine the density of liquid is hydrometer.
(c) First, take a beaker and fill it with water. Then, put an iron nail and a cork into the water. We will observe that the iron nail sinks, while the cork floats on the surface. This happens because the density of the iron nail is greater than that of water, making it sink, whereas the cork has a density less than water, so it floats.
11. A ball is thrown vertically upwards with some speed u o m/s. Show that under the free fall, it will fall on the ground with same speed.
Ans. Given,
initial velocity = u o m/s
When the ball is thrown upwards, then it will reach certain height h with acceleration -g. At maximum height h, the final velocity (v) will be zero.
Using 3rd equation of motion, v2 = u2 + 2as
02 = u o 2 - 2gh
u o 2 = 2gh …(i)
when the ball starts to fall, then the initial velocity (u) will be zero. It will accelerate due to gravity, i.e. a = g and reach ground with speed (say v0). again using the equation, v2 = u2 + 2as
v2 = u2 + 2gh
v o 2 = 02 + 2gh = u o 2 (from (i))
Thus, the ball reaches the ground with the same speed.
12. (a) A piece of ice is placed on the surface of water, kept in a glass, so that when the ice floats, the water comes up to the brim of the glass. What will happen to the level of water when the ice melts, will it overflow?
(b) What will happen, if we will replace the water with (i) a liquid denser than water (ii) a liquid lighter than water.
Ans. (a) According to Archimedes’ principle when the ice melts, the volume of water formed will be equal to the volume of the displaced water. So, the level of water in the glass will remain unaffected.
(b) (i) When the liquid denser than water is filled in the glass, then the volume of water formed by the melting of ice will be more than the volume of liquid displaced, so the level of water will rise. Therefore, the water will overflow from the glass.
(ii) When the liquid lighter than water is filled in the glass, the level of water in the glass will come down. So, the water will not overflow from the glass.
13. Two bodies of masses 3 kg and 12 kg are placed at a distance 12 m. A third body of mass 0.5 kg is to be placed at such a point that the force acting on this body is zero. Find the position of that point.
Ans. Given, mass of first body, m1 = 3 kg and mass of second body, m2 = 12 kg
Distance between masses m1 and m2 = 12 m
Let the third body of mass m3 = 0.5 kg be placed at a distance of x from m1 as shown in figure.
If net force on m3 is zero, then, force acting on m3 due to m1 is equal and opposite to the force acting on m3 due to m2.
∴ F31 = F32
Gm3m1 x2 = Gm3m2 (12 - x)2
⇒ 3 x2 = 12 (12 - x)2
⇒ 2 12 x x
⇒ 12 - x = 2x
⇒ x = 4 m
The position of the required point is at a distance 4 m from the body with mass 3 kg.
Long Answer Question (80–120 words)
1. Velocity-time graph for the ball’s motion is shown in Fig. 9.4.
Observe the graph and answer the following questions.
Assume that g = 10 m/s2 and that there is no air resistance.
(a) In which direction is the ball moving at point C?
(b) At which point is the ball stationary?
(c) At which point is the ball at its maximum height?
(d) What is the ball’s acceleration at point A, B and C?
(e) At which point does the ball have the same speed as when it was thrown?
Ans. (a) Downward; Because downward velocity is positive.
(b) At point B; Because velocity is zero.
(c) At point B; Because at maximum height velocity is zero.
9.4
(d) Since v - t graph is a straight line, the acceleration must be constant throughout the journey. therefore, Acceleration at Point A = Acceleration at Point B = Acceleration at Point C = g = 10 ms–2
(e) At point C; initial speed (at point A) = 30 m/s and final speed (at point C) = 30 m/s
2. (a) Explain, with the help of an example, the difference between thrust and pressure. Which of the two has the same SI unit as force?
(b) Consider a wooden block of mass 5 kg with dimensions 40 cm × 20 cm × 10 cm. It is placed on a table first with its 20 cm × 10 cm face, and then with its 40 cm × 20 cm face. In which case will the pressure exerted by the block on the table be greater? Justify your answer with calculations.
Ans. (a) Pressing the flat end of a ballpoint pen against your hand with some force and then pressing the pointed tip of the same pen with the same force produces different sensations. The pointed tip causes more pain because the force acts on a smaller area, increasing the effect.
Fig.
The total force acting perpendicular to a surface is called thrust, while the force acting per unit area is called pressure. Thrust has the same SI unit as force, which is the newton (N).
(b) Given,
mass of block, m = 5 kg
Pressure = Force Area
Gravitational force applied by block on table = mg = 5 × 9.8 N
(i) When 20 cm × 10 cm is kept on the table:
Pressure = 5 × 9.8 N
0.2 × 0.1 m2 = 2450 Nm–2
(ii) When the block is kept on the table with its face 40 cm × 20 cm,
Pressure = 5 × 9.8 N
0.4 × 0.2 m2 = 612.5 Nm–2
Thus, the pressure exerted is greater when the block rests on the smaller face (20 cm × 10 cm).
3. A body weighs 25 kg on the surface of the earth. If the mass of the earth is 6 × 1024 kg, the radius of the earth is 6.4 × 106 m and the gravitational constant is 6.67 × 10–11 Nm2/kg2. Calculate.
(a) the mutual force of attraction between the body and the earth.
(b) the acceleration produced in the body.
(c) the acceleration produced in the earth.
Ans. Given, mass of earth, M e = 6 × 1024 kg
Mass of body, m = 25 kg
Radius of earth, R e = 6.4 × 106 m and gravitational constant,
G = 6.67 × 10–11 N – m2/kg2
(a) Mutual force, F = G M e R2 e m = 6.67 × 10⁻¹¹ × 6 × 1024 × 25 (6.4 × 106)² = 244 N
(b) Acceleration produced in the body,
a = F m = 244 25 = 9.8 m/s2
(c) Acceleration produced in the earth,
a = F m e = 244 6 × 1024
= 4.06 × 10–23 m/s2
4. (a) A person weighs 110.84 N on the moon, whose acceleration due to gravity is 1 6 of that the earth. If the value of g on the earth is 9.8 m/s2, then calculate
(i) acceleration due to moon’s gravity (g’) on the moon
(ii) mass of person on the moon
(iii) weight of person on the earth
(b) How is the value of g on the earth related to the mass of the earth and its radius? Derive it.
Ans. (a) (i) g on the moon is given by g’ = g 6 = 9.8 6 = 1.63 m/s2
(ii) Given,
Weight of person on the moon = 110.84 N
Mass of the person on the moon,
m = 110.84 1.63 = 68 kg
(iii) Weight of person on the earth = mg
= 68 × 9.8 = 666.4 Nm2/kg2
(b) Value of g on earth is related to the mass of the earth and its radius.
From Newton’s law of gravitation, F = GMm R2 (i)
Here, M is mass of earth, object having mass m is falling towards it and R is the distance between centre of earth and the object.
From second law of motion, force exerted on an object, F = ma
Since, a = g (i.e. acceleration due to gravity) F = mg (ii)
Equating RHS of Eqs. (i) and (ii), we get
mg = GMr R2
g = GM R2
5. (a) State the laws of floatation.
(b) Fig. 9.5 shows three identical blocks of wood floating in three different liquids, of densities d1, d2 and d3 respectively kept in vessels A, B and C. Which of these has the highest density? Give reason to justify your answer.
(c) The three liquids used are oil, water, and honey. Based on their properties, identify each liquid.
Ans. (a) A body floating freely in a fluid, must obey the following laws known as laws of floatation:
(i) A body floats only if its weight is equal to the weight of fluid displaced by its immersed part.
(ii) For the body to float in an upright position, the centre of gravity of the floating body and the centre of buoyancy of the fluid displaced by the immersed part of the body must lie on the same straight line.
1st law stated above is necessary for the body to float whereas the 2nd law stated above is necessary for the body to float in an upright position.
(b) Liquid C has the highest density. This is because volume of liquid displaced is least in liquid C of density d3.
(c) We can see that in vessel A volume of water displaced is maximum, therefore d1 <, d2 < d3. Among the given liquids, oil is least dense and honey is most dense. Hence, vessel A has oil, vessel B has water and vessel C has honey.
Competency Based Questions
Multiple Choice Questions (Choose the most appropriate option/options)
1. Fig. 9.6 shows a block of wood with height 2 cm, width 2 cm, and length 5 cm kept in two different orientations on a table. How does the thrust and pressure applied by the block onto the table vary in these two orientations?
(a) Thrust of 50 N remains same, while pressure in orientation A is 5 N/cm2 and in orientation B is 12.5 N/cm2
(b) Thrust of 50 N remains same, while pressure in orientation A is 12.5 N/cm2 and in orientation B is 5 N/cm2
(c) Pressure of 50 N remains same, while thrust in orientation A is 5 N/m2 and in orientation B is 12.5 N/m2.
(d) Pressure of 50 N remains same, while thrust in orientation A is 12.5 N/m2 and in orientation B is 5 N/m2
2. The weight of an object at the centre of the earth of radius R is (NCERT Exemplar)
(a) zero
(b) infinite
(c) R times the weight at the surface of the earth
(d) 1 R2 times the weight at the surface of the earth
3. Why the dam of water reservoir is thick at the bottom?
(a) Quantity of water increases with depth (b) Density of water increases with depth
(c) Pressure of water increases with depth (d) Temperature of water increases with depth
4. An object is put one by one in three liquids having different densities. The object floats with 1 9 , and 2 11 and 3 7 parts of their volumes outside the liquid surface in liquids of densities d1, d2 and d3
respectively. Which of the following statement is correct? (NCERT Exemplar) (a) d1 > d2 > d3 (b) d1 > d2 < d3 (c) d1 < d2 > d
5. “An apple falls from its tree onto the ground”. What does it say about the nature of gravitational force?
(a) It is a repulsive force, which acts at greater heights.
(b) It is a repulsive force, which is independent of height.
(c) It is an attractive force, which acts at greater heights.
(d) It is an attractive force, which is independent of height.
6. Three planets X, Y, and Z have the following properties:
• Planet X: radius = R, density = ρ
• Planet Y: radius = 2R, density = ρ
• Planet Z: radius = R, density = 2ρ
Assuming all planets are perfect spheres and neglecting rotational effects, compare the acceleration due to gravity (g) on their surfaces. Which of the following statements is correct? (a)
7. If the radius and the mass of Earth were to reduce by 1%, the acceleration due to gravity on Earth’s surface would (a) decrease (b) increase (c) remain unchanged (d) can’t be calculated
8. What would happen if our moon suddenly stops?
(a) It will remain stationary. (b) It will be repelled away from the Earth.
(c) It will start moving towards the Earth. (d) It will start moving towards the Sun.
9. Why do objects of different masses fall at the same rate in vacuum?
(a) Forces cannot applied in a vacuum.
(b) In vacuum all objects experience same air resistance.
(c) Acceleration due to gravity is independent of mass.
(d) Objects of different masses fall at the different rate in vacuum.
10. An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be (NCERT Exemplar)
(a) 2 N (b) 8 N
N
N
11. A ball is dropped from a height and the distance covered by the ball each second is recorded. Fig. 9.7 shows the distance the ball covers each second. (use g = 10 m/s2)
(a) In 1 second the ball has travelled 10 m.
(b) In between 1 and 2 sec, ball has travelled 10 m.
(c) In between 2 and 3 sec, ball has travelled 25 m.
(d) In 3 seconds, ball has travelled 90 m.
12. How does the buoyant force on an object floating in water changes, if its mass is doubled without changing the volume?
(a) It increases by a factor of 2.
(c) It increases by a factor of 4.
(b) It decreases by a factor of 2.
(d) It remains unchanged.
13. The value of quantity G in the law of gravitation (NCERT Exemplar)
(a) depends on mass of the earth only
(b) depends on radius of earth only
(c) depends on both mass and radius of the earth
(d) is independent of mass and radius of the earth
14. Which law of motion explains the occurrence of tides due to the gravitational pull of the Moon and the Sun? (CBSE QB)
(a) Newton’s third law of motion, as their gravitational pull affects the motion of Earth.
(b) Newton’s first law of motion, as their gravitational pull changes the shape of Earth.
(c) Newton’s third law of motion, as their gravitational pull is balanced by the weight of ocean water.
(d) Newton’s second law of motion, as their gravitational pull is uniformly applied on the Earth’s oceans.
15. If rQ > rP, how would the acceleration due to gravity (g) and weight of an object (W) compare at these two points? (CBSE QB)
(a) gP > gQ and WP > WQ
(b) gP < gQ and WP < WQ
(c) gP > gQ and WP < WQ
(d) gP < gQ and WP > WQ
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): The moon revolves around the earth due to gravitational force between moon and earth.
Reason (R): Gravitational force between moon and earth is calculated by Newton’s law of gravitation.
2. Assertion (A): At the centre of earth, a body has no centre of gravity.
Reason (R): g = 0 at the centre of earth.
3. Assertion (A): When a body is dipped into water, then upthrust force acting on the body is equal to the weight of water displaced by the body.
Reason (R): A body will sink in water, if density of body is less than or equal to density of water.
4. Assertion (A): To float, a body must displace liquid whose weight is greater than actual weight of the body.
Reason (R): During floating, the body will experience no net downward force.
5. Assertion: The value of acceleration due to gravity (g) is slightly less at the equator than at the poles.
Reason: The Earth is not a perfect sphere; it is slightly flattened at the poles and bulging at the equator.
6. Assertion: A nail with a sharp tip exerts more pressure than a nail with a blunt tip.
Reason: Pressure is inversely proportional to the area on which force is applied.
7. Assertion: A stone dropped from a height will take the same amount of time to reach the ground, whether it’s a small stone or a large stone.
Reason: The time taken to fall freely depends only on the acceleration due to gravity and the height of the fall.
8. Assertion (A): An object immersed in a liquid weigh less.
Reason (R): Upward buoyant force acts on an object when immersed in a fluid.
Case-Based/Source-Based/Passage-Based Questions
1. We know that all the planets in our solar system revolve around the Sun. Similarly, the Moon revolves around the Earth. Naturally, a question arises: which force is responsible for the revolution of planets and the Moon? Sir Isaac Newton conclusively proved that a gravitational force acts on the planets due to the Sun, which causes the planets to revolve around the Sun in their respective orbits.
Later, Newton generalized the concept of gravitational force and demonstrated that every object in the universe attracts every other object towards itself. The magnitude of this force depends on the masses of the objects and the distance between them.
(a) With what force does the Earth attract the Moon towards itself?
(b) Does the Moon also attract the Earth towards itself? If yes, with what force?
(c) Draw a neat diagram showing the force between the Earth and the Moon.
(d) What is the value of the gravitational constant G on the surface of moon?
2. According to Newton’s law of gravitation, any two objects in the universe attract each other. The gravitational force exerted by the Earth on objects near its surface is commonly referred to as gravity. When an object is released from a height, it falls towards the Earth due to the force of gravity and experiences an acceleration ‘g’, known as the acceleration due to gravity. It’s commonly accepted value is 9.8 ms–2, but it varies from place to place.
The force of gravity experienced by an object placed on or near the Earth’s surface is termed the weight of the object. Thus, the weight of an object is given by: W = mg
Where m is the mass of the object, and g is the acceleration due to gravity.
(a) Are the terms mass and weight the same? What is the weight of an object of mass 1 kg on the surface of the Earth?
(b) An object is taken from Mumbai to Manali. What would happen to its mass and its weight?
(c) How does the weight of an object change when it is taken from the North Pole of Earth to a place on the equator?
(d) Can the weight of an object be zero? If yes, suggest where?
3. Riya was watching a documentary on Arctic wildlife. She noticed that icebergs float in the ocean even though they are made of solid ice. Her science teacher later explained that it’s due to buoyant force acting on the iceberg, which is equal to the weight of the water displaced by the submerged part of the iceberg.
Based on this, answer the following:
(a) What is the principle that explains why the iceberg floats?
(b) Why is it dangerous for ships to navigate near icebergs, even when it appears to be a small one?
(c) How does the difference in density between ice and seawater explain why icebergs float on the ocean surface?
(d) What is the relation between weight of iceberg and the buoyant force?
Answers
Multiple Choice Questions
Assertion-Reason Based Questions
Case-Based/Source-Based/Passage-Based Questions
1. (a) The gravitational force F = GMEME R2 , where ME = mass of earth, MM = mass of moon and R = distance from centre of earth to centre of moon.
(b) Yes, the Moon does attract the Earth towards itself with the same force that the Earth exerts on the Moon. According to Newton’s third law of motion, every action has an equal and opposite reaction. (c)
(d) Value of gravitational constant G is always 6.67 × 10–11 N m2kg–2. It is a universal constant and its value does not change under any condition or at any location.
2. (a) Mass and weight are not the same.
• Mass of an object is a measure of its inertia and is constant everywhere, regardless of location.
• Weight is the force experienced by an object due to gravity and varies depending on the location (altitude, latitude, etc.).
Fig. 9.9
The weight of an object with a mass of 1 kg on the surface of the Earth is: W = mg = 1 kg × 9.8 m s–2 = 9.8 kg m s–2 = 9.8 N
Thus, the weight of a 1 kg object on Earth is 9.8 Newtons.
(b) The mass of the object remains unchanged because mass is independent of location. The weight of the object decreases because the value of g (acceleration due to gravity) decreases with an increase in altitude. As you move to Manali (higher altitudes), the force of gravity becomes weaker, so the object weighs less.
(c) Weight of a given object gradually decreases when it is taken from the north pole to equator on earth. In other words, the weight of the object is maximum at pole and minimum at the equator.
(d) Yes, the weight of an object can be zero at places where the acceleration due to gravity (g) is zero. For example, at the center of the Earth, the gravitational forces from all directions cancel out, resulting in a net gravitational force of zero.
3. (a) The principle is Archimedes’ Principle, which states that an object immersed in a fluid experience an upward buoyant force equal to the weight of the fluid displaced.
(b) It is dangerous because though an iceberg appears small, a large portion of the iceberg remains hidden underwater. Ships may collide with the submerged part, which can cause serious damage, as most of the iceberg’s volume is below the surface.
(c) Because ice has a lower density than seawater, it experiences a buoyant force greater than its own weight when partially submerged, causing it to float. In general, an object floats in a fluid if its density is less than the density of the fluid, and it sinks if its density is greater. This is why a large part of an iceberg remains underwater while only a small portion is visible above the surface.
(d) Since the iceberg is floating on the surface of the water, it means there is no net force acting on it. This implies that the upward buoyant force exactly balances the downward weight of the iceberg. In other words, the weight of the iceberg is equal to the buoyant force acting on it.
Numerical Questions
1. Explain what happens to the force between two objects
(a) The mass of an object is doubled.
(b) The distance between the objects is doubled.
Ans. (a) If the mass of an object is doubled, then
F1 = G(2m1) m2 r2
F1 = 2 ‧ Gm1 m2 r2
F1 = 2F
Therefore, the gravitational force will be two times.
(b) If the distance between the object is doubled, then
F2 = G ‧ m1m2 (2r)2
F2 = 1 4 . Gm1m2 r2
F2 = F 4
Therefore, the gravitational force will be one-fourth.
2. A firecracker is fired and it rises to a height of 1000 m. Find the (a) velocity by which it was released.
(b) time taken by it to reach the highest point. (Take, g = 9.8 m/s2 )
Ans. Given,
(a) Height, h = 1000 m
Final velocity, v = 0
Acceleration due to gravity, g =-9.8 m/s2 using 3rd equation of motion, v2 = u2 + 2as
v2 = u2 + 2gh
⇒
0 = u2 - 2 × 9.8 × 1000
29.8100019600140m/s u
Velocity by which firecracker is released = 140 m/s
(b) From the 1st equation of motion, v = u + at
v = u + gt
⇒ 0 = 140 - 9.8t t = 140 9.8 = 14.28 s
Time taken, t = 14.28 s
3. If the distance between two masses be increased by a factor of 6, by what factor would the mass of one of them have to be altered to maintain the same gravitational force? Would this be an increase or decrease in the mass? (CBSE 2015)
Ans. Let gravitational force between two masses m1 and m2 separated by a distance r be F,
Now, distance between the masses is increased by a factor of 6 i.e., rʹ = 6r and let one of the masses m1 changes to mʹ 1 such that force remains unchanged. Then
From (i) and (ii), we get
m1 = mʹ 1 36 or mʹ 1 = 36 m1 It means that one of the masses should be increased by a factor of 36.
4. Which will exert more pressure 100 kg of mass on an area of 10 m2 or 50 kg of mass on an area of 4 m2? (Take, g = 10 m/s2)
Ans. Given,
m1= 100 kg
m2 = 50 kg
A1 = 10 m2
A2 = 4 m2
F1 = m1g = 1000 N
F2 = m2g = 500 N
p1 = F1 A1 = 1000 N 10 m2 = 100 Pa
p2 = F2 A2 = 500 4 m2 = 125 Pa
50 kg mass on an area of 4 m2 exerts more pressure.
5. Rohan placed an iron cuboid of dimensions 4 cm × 7 cm × 10 cm on a tray containing fine sand. He placed the cuboid in such a way that it was made to lie on the sand with its faces of dimensions
(a) 4 cm × 7 cm
(b) 7 cm × 10 cm
(c) and 4 cm × 10 cm
If the density of iron is nearly 8 gcm–3 and g = 10 ms–2, find the minimum and maximum pressure as calculated by Rohan.
Ans. Given, density of iron = 8 g cm–3 and g = 10 ms–2
Mass of cuboid = Volume × Density
= (4 cm × 7 cm × 10 cm) × 8 g cm–3 [∴ Volume of cuboid = l × b × h]
= 2240 g or 2.24 kg
Weight of cuboid = mg = 2.24 × 10 = 22.4N [∴ g = 10 ms–2]
Pressure will be minimum when area of the face of cuboid kept on sand is maximum, i.e. in the case of face with 7 cm × 10 cm.
Area of the face = 7 cm × 10 cm = 7 100 m × 10 100 m = 0.07 m × 0.1 m = 0.007 m2
Minimum pressure = Force Area = 22.4 0.007 = 3200 Nm–2
Pressure will be maximum when area of face of cuboid kept on the sand is minimum, i.e. in the case of face with 4 cm × 7 cm.
Area of the face = 4 cm × 7 cm = 4 100 m × 7 100 m = 0.0028 m2
Maximum pressure = Force Area = 22.4 0.0028 = 8000 Nm–2
6. A coin is allowed to fall freely from a certain height. It reaches the ground after 0.5 s. From what height was it thrown from?
Ans. Given,
u = 0
g = 10 ms–2
t = 0.5 s
From 2nd equation of motion, s = ut + 1 2 at2
h = ut + 1 2 gt2
h = 1 2 gt2 = 0.5 × 10 × (0.5)2 = 1.25 m (as, u = 0)
7. A spacecraft carried a bucket containing soil weighing 60 N from the earth to the surface of the moon. When he weighed the soil there on the surface of the moon, he found that it was only 10 N. Calculate how much mass of the soil was lost and where did the rest of the soil go?
Ans. Given that, the weight of the soil on the earth is 60 N we know, gearth = 10 ms–2
Mass on the earth, m1 = 60 10 = 6 kg m = w gearth
Weight of the soil on moon = 10 N
gmoon = gearth 6 = 10 6 ms–2
Mass on the moon, m2 = 10 × 6 10 m = w gmoon
= 6 kg
Since, m1 = m2, hence there will be no loss in mass of the soil on the surface of the moon and decrease in weight is due to difference in the gravity.
8. A force of 20 N is distributed uniformly on one surface of a cube of edge 10 cm. Find the pressure on this surface.
Ans. Force acting on one surface = Thrust = 20 N
Area of one surface of cube, A = a2
= 10 × 10 × 10–4 m2
= 100 × 10–4 m2
Therefore, Pressure on this surface,
p = F A = 20 100 × 10–4 = 2000 Pa
9. The value of ‘g‘ on earth’s surface is 9.8 ms–2. Suppose the earth suddenly shrinks to one-third of its present size without losing any mass. What is the value of ‘g‘ on the surface of shrunken earth?
(CBSE 2016)
Ans. If radius of earth be R and mass of earth be M, then
g = 9.8 = GM R2
If suddenly earth shrinks to one-third of its present size without any change in mass then Rʹ = R 3. Hence new value of g will be
10. Calculate the buoyant force on a block of volume 0.005 m³ fully submerged in water. (Density of water = 1000 kg/m³, g = 9.8 m/s²)
Ans. Given,
Volume, V = 0.005 m³
Density, ρ = 1000 kg/m³
g = 9.8 m/s²
Buoyant force = weight of displaced fluid
= m × g = ρ × V × g = 1000 × 0.005 × 9.8 = 49 N
Practice Questions
Multiple Choice Questions
1. Two objects of different masses falling freely near the surface of moon would (NCERT Exemplar)
(a) have same velocities at any instant (b) have different accelerations
(c) experience forces of same magnitude (d) undergo a change in their inertia
2. What is the buoyant force on an object immersed in a liquid?
(a) The weight of the liquid displaced by the object
(b) The mass of the object
(c) The weight of the object
(d) The density of the liquid
3. If the mass of one of two objects is doubled, how will the gravitational force between them change?
(a) It will double (b) It will remain the same
(c) It will increase by four times (d) It will decrease by half
4. What is the acceleration due to gravity near the surface of the Earth? (a) 10 m/s² (b) 9.8 m/s² (c) 11 m/s² (d) 9.0 m/s²
5. The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become (NCERT Exemplar) (a) F 4 (b) F 2 (c) F (d) 2F
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
6. Assertion (A): A person feels lighter in a swimming pool.
Reason (R): The buoyant force exerted by water is equal to the weight of the displaced water and acts in the upward direction.
7. Assertion (A): The value of acceleration due to gravity does not depend upon mass of the body.
Reason (R): Acceleration due to gravity is a constant quantity.
Very Short Answer Questions (30-50 words)
8. If the Earth attracts two objects with equal force, can we say that their masses must be equal? Is weight a force?
9. How does the time of fall a freely falling body depend on (i) mass of the body? (ii) acceleration due to gravity?
10. A feather takes much longer to fall than a stone through the same distance. Explain the reason.
11. Prove that if a body is thrown vertically upward, the time of ascent is equal to the time of descent.
12. (a) Why can one jump higher on the surface of the Moon than on the Earth?
(b) If the weight of an object is 98 N on the Earth, what would be its mass on the Moon?
Short Answer Questions (50-80 words)
13. State reasons for the following:
(a) An army tank weighing over a thousand tonnes rests on a continuous chain.
(b) A truck has much wider tyres.
(c) Cutting tools have sharp edges.
14. (a) Write the formula to find the magnitude of gravitational force between the earth and an object on the earth’s surface.
(b) Derive how the gravitational force F between the Earth and an object is affected in the following cases:
(i) When the mass of the object is increased by four times.
(ii) When the radius of the Earth is reduced to half, while keeping the masses of both the Earth and the object unchanged.
15. (a) What is upthrust? What are the quantities that can vary upthrust?
(b) How does it account for the floating of a body? When a partially immersed body is pressed down a little, what will happen to the upthrust?
16. (a) Seema buys few grams of gold at the poles as per the instruction of one of her friends. She hand over the same when she meets her at the equator. Will the friend agree with the weight of gold bought? If not, why?
(b) If the moon attracts the earth, then why does the earth not move towards the moon?
17. (a) Why does a bucket of water feel heavier when taken out of water?
(b) Lead has greater density than iron and both are denser than water. Is the buoyant force on a lead object greater than, less than or equal to the buoyant force on an iron object of the same volume?
Long Answer Questions (80-120 words)
18. (a) Derive expression for force of attraction between two bodies and then define gravitational constant. (b) If the mass of the earth is 6 × 1024 kg, then the radius of the earth is 6.4 × 106 m and the gravitational constant is 6.67 × 10–11 Nm2/kg2. Calculate the value of acceleration due to gravity on the surface of earth.
19. (a) How does a boat float in water?
(b) State two factors on which the magnitude of buoyant force acting on a body, immersed in a fluid depends.
(c) A block of wood can float while a similar sized block of concrete can be used as an anchor. Why?
20. The radius of the earth at the poles is 6357 km and the radius at the equator is 6378 km . Calculate the percentage change in the weight of a body when it is taken from the equator to the poles.
Additional Numericals
1. The ratio of gravitational force on Neptune to the gravitational force on earth is 9 : 8. Ramesh weighs 792 N on earth. Calculate his weight on Neptune. (CBSE 2016)
2. A boy of mass 40 kg is standing on loose sand. If the area of his feet is 0.04 m2, calculate the pressure exerted by the boy on the sand. [g = 10 ms–2]
(CBSE 2012)
3. The weight of a person on the Moon is about 1 6 times that on the Earth. He can lift a mass of 15 kg on the Earth. What will be the maximum mass which can be lifted by the same force applied by the person on the Moon?
(NCERT Exemplar)
4. A stone is dropped from a cliff. What will be its speed when it has fallen 100 m?
5. Calculate the gravitational force between two objects of masses 10 kg and 20 kg at separation of 5 m.
6. The mass of earth is 6 × 1024 kg and that of moon is 7.4 × 1022 kg. If the distance between the earth and the moon is 3.84 × 105 km, calculate the force exerted by earth on the moon.
7. Calculate the acceleration due to gravity on the surface of a neutral satellite of mass 7.4× 1022 kg and radius 1.74 × 106 m, G = 6.7 × 10–11 Nm2 kg–2
8. An apple is dropped from the top of a multistorey building. At the same instant an arrow is shot vertically upwards from the ground with velocity 11 m/s. The arrow hits the apple after 2 s. Calculate the height of the building. [Take, g = 10 ms–2]
9. The pressure of water on the ground floor is 4 × 104 Pa and on the first floor is 104 Pa. Find the height of the first floor.
[Density of water = 1,000 kg m–3 and = 10 ms–2]
10. You are given a solid iron ball A of mass 15 g and a hollow iron ball of volume 20 cm3 and mass 15 g. Both are placed on the surface of water. Which ball will sink? Give reason.
[Density of iron = 7.8 g/cm3]
11. Calculate the apparent weight of a 15 kg object submerged in water. What fraction of its volume will be submerged in water. Given: g = 9.8 m/s², volume of object = 0.02 m3, and density of water = 1 gcm–3.
12. Calculate the pressure exerted by a sharp knife edge if a force of 20 N acts on an area of 0.0001 m².
13. A block of mass 2 kg floats in water, half immersed. Calculate the upthrust acting on it. (Density of water = 1000 kg/m³, g = 10 m/s²)
14. A boat of mass 500 kg is floating in the river with half of its volume inside the water. Calculate the volume of water (in litres) displaced by the ship. (use g = 10 ms–2)
15. Calculate the force of gravitation between the earth and the sun. Given that the mass of the earth = 6 × 1024 kg and of the sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.
Brain Charge
1. Crossword Puzzle
Across
3. Force acting per unit area
5. The force that allows objects to float in fluids
7. SI unit of pressure
8. The force experienced by a body due to gravity
9. The mass per unit volume of a substance
Down
1. The principle that explains why ships and icebergs float
2. Unit of force
4. The force with which Earth attracts objects towards its center
6. The scientist who formulated the law of universal gravitation
2. Word Puzzle
Use the first alphabet of each answer to the puzzle and combine them to spell a word.
1. The universal constant used in the formula for gravitational force.
2. The distance between the centers of two objects in gravitational calculations.
3. The change in velocity per unit time, due to gravity on Earth.
4. A quantity that describes speed in a specific direction.
5. The tendency of an object to remain at rest or in uniform motion unless acted upon.
6. The duration of one complete orbit or revolution.
7. A unit used to measure weight in the Imperial system.
3.
Who am I?
1. I pull everything toward me, from an apple to the moon.
2. I always act perpendicular to the surface and increase with depth in a fluid.
3. The more I spread out a force, the lesser my value becomes.
4. I am what causes you to weigh less in water.
5. Archimedes discovered me when he stepped into a bath and noticed water rise.
6. I determine whether you sink or float—if I’m higher than your density, you float.
7. I am the quantity that tells how heavy you feel when submerged in a fluid.
Challenge Yourself
1. Two astronauts, each of mass 60 kg, travel to two different planets—Planet X and Planet Y.
• Planet X has a radius of 6000 km and a density of 4.0 × 10³ kg/m³
• Planet Y has a radius of 8000 km and a density of 3.0 × 10³ kg/m³
Assuming both planets are spherical and neglecting atmospheric effects, determine:
(a) On which planet is the acceleration due to gravity greater?
(b) What is the ratio of gravitational acceleration on Planet X to that on Planet Y?
2. A person weighs 600 N on the surface of the Earth. At what height above the Earth’s surface would their weight be
(a) 150 N
(b) 66.667 N
(c) 6 N? (Take Earth’s radius as R)
3. Is there a point in outer space, at a finite distance from earth, where you will not feel any gravitational force? Justify your answer.
4. A test tube floats in water with a small coin at its bottom. The mass of this test tube is equal to mass of seven coins and external volume is 16 cm3. It just sinks when the third coin is added. Calculate the mass of each coin. (Density of water is 1 gcm–3)
Answers
Practice Questions
Additional Numericals
Brain Charge
1. Crossword Puzzle
2. Word Puzzle Gravity
1. G – Gravitational constant
2. R – Radius
3. A – Acceleration
4. V – Velocity
5. I – Inertia
6. T – Time period
7. Y – Yard
Challenge Yourself
1. (a) same on both planets (b) 1:1
2. (a) R (b) 2R (c) 9R
3. Yes
4. 1.78 g
3. Who am I?
1. Gravitational force
2. Thrust
3. Pressure
4. Buoyant force
5. Principle of floatation
6. Density of fluid
7. Apparent weight
Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 1.5 Hour
Multiple Choice Questions
1. Which of the following describes Archimedes’ principle?
Scan me for Solutions
Max. Marks: 50
(10 × 1 = 10 Marks)
(a) An object in a fluid experiences an upward force equal to the weight of the fluid displaced
(b) The force on an object is proportional to the square of the distance
(c) The object’s weight is proportional to the mass
(d) The gravitational force between two objects is proportional to the product of their masses
2. In which condition would an object sink in a liquid?
(a) When the object’s density is less than the liquid’s density
(b) When the object’s density is equal to the liquid’s density
(c) When the object’s density is greater than the liquid’s density
(d) When the object is very large in size
3. The SI unit of pressure is:
(a) Newton (b) Pascal (c) Joule (d) Watt
4. What happens to an object’s weight when it is on the Moon?
(a) It remains the same as on Earth (b) It decreases by one-sixth (c) It increases (d) It becomes zero
5. Three cuboids of same density are shown in Fig. 9.11. which of them applies more pressure on the surface?
(a) A
(b) B
(c) C
(d) All apply same pressure.
6. A and B are two metallic pieces. They are fully immersed in water and the buoyant force due to water on both pieces is found to be same. From this we conclude that (a) Masses of pieces A and B are equal. (b) Weights of pieces A and B are equal. (c) Densities of two metallic pieces A and B are equal. (d) Volumes of metallic pieces A and B are equal.
7. An object has a mass of 163 kg . When the object is kept on a weighing scale, it exerts a force of 1.597 × 103 N. What is the value of acceleration due to gravity? (CBSE QB) (a) 9.8 ms2 (b) 9.8 ms–2 (c) 9 : 8 × 102 ms–2 (d) 9.8 × 10–2 ms2
8. The acceleration due to the force applied by m1 on m2 in the below given configuration (Fig. 9.12) can be represented by: (CBSE QB)
(a) G m1 r2
(b) G m1 × m2 r2
(c) G m2 r2
(d) G r2 Fig. 9.11
Fig. 9.12
9. An object weighs 10 N when measured on the surface of Earth. What should be its weight on Moon? (a) 1.67 N (b) 4.18 N (c) 2.20 N (d) 16.7 N
10. Find the value of g on a planet with mass (M) = 4 × 10²³ kg and radius (R) = 2 × 10⁶ m (take G = 6.67 × 10⁻¹¹) (a) 6.67 m/s² (b) 9.8 m/s² (c) 10.8 m/s² (d) 15.3 m/s²
Assertion-Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): Gravitational force acts between all objects in the universe, irrespective of their size.
Reason (R): Gravitational force depends on the product of masses and the square of the distance between them.
12. Assertion (A): If the distance between two bodies of masses m1 and m2 is increased by a factor of 6 , the gravitational force is reduced by 1/36.
Reason (R): The gravitational force is inversely proportional to the square of the distance between two bodies.
13. Assertion (A): Acceleration due to gravity increases with altitude.
Reason (R): Acceleration due to gravity is inversely proportional to the square of the distance from the Earth’s center.
14. Assertion (A): Buoyant force on a body immersed in a liquid acts in the downward direction.
Reason (R): The pressure at the bottom of an object submerged in a liquid is greater than at the top.
15. According to the principle of flotation, an object will float in a fluid if its density is less than that of the fluid. When an object is partially or fully immersed in a fluid, the fluid exerts a force on it. At any small portion of the surface of the object, the force by the fluid is perpendicular to the surface and is equal to the product of the pressure at the surface and surface area. The resultant of all these contact forces is called buoyant force (or upthrust).
Archimedes’ principle states that: “Any object partially or completely submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object.”
The buoyant force depends on the submerged volume of the object, whether it is partly or wholly immersed. The volume of the object immersed is the same as the volume of the liquid displaced. Therefore, buoyant force = Vρg, where
V = volume of fluid displaced, ρ = density of fluid, g = acceleration due to gravity or Buoyant force = Weight of fluid displaced.
(a) An object weighs 36 g in air and has a volume of 80 cm3. What will be its apparent weight when immersed in water?
(i) 25 g (ii) 28 g (iii) 30 g (iv) 35 g
(b) What is the density of the gasoline?
(i) 0.89 g/cm3 (ii) 0.85 g/cm3 (iii) 1.05 g/cm3 (iv) None of these
(c) The density of water is 1 g/cm3. Which of the following substances will float in water?
Substance Mass Volume
a 6 g 3 cm3
b 8 g 4 cm3
c 10 g 2 cm3
d 2 g 4 cm3
(i) Substance a
(iii) Substance c
(ii) Substance b
(iv) Substance d
(d) An object is immersed in two liquids A and B in succession. The extent to which the object sinks in liquid B is less than in liquid A. What conclusion can be derived from such an observation?
(i) Density of liquid A is more than liquid B
(ii) Density of liquid B is more than liquid A.
(iii) Upthrust exerted by liquid B is more than exerted by liquid A.
(iv) Both (ii) and (iii)
16. Gravitation is defined as the force of attraction between any two bodies in the universe. The Earth attracts all objects lying on or near its surface towards its center. The force with which the Earth pulls objects towards its center is called the gravitational force of the Earth or simply gravity.
The concept of gravitation was initially studied by Galileo Galilei in the late 16th and early 17th centuries. Later, in 1687, Sir Isaac Newton formulated Newton’s theory of gravity, also known as the inverse-square law of gravitation.
Gravitational force can be defined as the force that attracts any two objects with mass. It is called attractive because it always pulls masses together and never pushes them apart.
(a) A body of mass 200 kg weighs 490 N on Mars. What is the acceleration due to gravity on Mars?
(b) Is the value of acceleration due to gravity (g) the same at different places on Earth? Give reason for your answer.
(c) At what distance from the centre of the earth, the value of acceleration due to gravity equal to the value calculated in (a).(Take radius of earth as R)
Very Short Answer Questions (30-50 words)
(3 × 2 = 6 Marks)
17. The Earth attracts an apple. Does the apple also attract the Earth? If it does, why does the Earth not move towards the apple?
18. Why is it easier to walk on soft sand if we have flat shoes rather than shoes with sharp heels?
19. A ball is thrown vertically upwards such that it reaches a maximum height of 125 m. calculate the initial speed and total time taken to reach maximum height.
Short Answer Questions (50-80 words)
(4 × 3 = 9 Marks)
20. A piece of iron of density 7.8 g/cm3 and volume 100 cm3 is fully immersed in water (density = 1 g/cm3).
Calculate:
(a) weight of iron piece in air.
(b) upthrust of water on iron piece.
(c) apparent weight of iron piece in water (g = 10 m/s2).
21. (a) State Archimedes’ principle. Give two applications of this principle.
(b) A sphere of iron and another of wood having the same radius are held under water. Which of these experiences a greater buoyant force?
22. A stone of mass 2 kg is released from rest at a height of 80 m above the ground. At the same moment, a ball of mass 100 g is projected vertically upward from the ground with a speed of 40 m/s. Both objects travel along the same vertical line. At what time and at what height from the ground will the two objects meet? (use g = 10 m/s2)
23. (a) Differentiate between mass and weight.
(b) Calculate the ratio of the mass and the weight of an object on the Moon compared to its values on Earth.
(c) An object of 5 kg mass is dropped from a height of 5 m. What will be momentum of the ball before it strikes the ground? (use g = 10 m/s2)
24. (a) What height above the surface of the earth, the value of g becomes 64% of its value at the surface of the earth? (Take, the radius of the earth = 6400 km)
(b) A sphere of mass 40 kg is attracted by a second sphere of mass 15 kg. If their centres are 200 cm apart, then calculate the attractive force experienced by 40 kg sphere. ( take G = 6.6 × 10⁻¹¹ N m²/kg²)
25. (a) A bowl of water, full up to the brim, has 2 ice cubes floating in it. Will water overflow once ice melts? Justify your answer.
(b) A wooden cube of edge 2 m length is submerged in water. If the ratio of densities of water and wood is 3:2. Find the percentage volume of wooden cube outside water. (Take density of water as 1 g/cm3)
10 Work and Energy
This chapter introduces the key concepts of work, energy, and power, and explains how they relate to force and motion. It distinguishes between the everyday use of the term “work” and its scientific definition—where work is said to be done only when a force causes displacement.
Energy, the capacity to do work, is explored through its main forms—Kinetic (due to motion) and Potential energy (due to position)—along with the principle of energy transformation and conservation. Through practical examples such as lifting objects, falling bodies and operating machines, the concepts are linked to real-world applications, demonstrating their significance in technology, natural phenomena, and mechanical systems.
Work and Energy
Work
Work is done on an object when a force (F) is applied to it and it moves a certain distance (s) in the direction of the applied force.
It is a scaler quantity.
Formula: W = F × s
SI unit: Joule (J) or Nm
Condition for Work to be Done:
(i) A force must be exerted on an object.
(ii) The object must be displaced along the direction of force.
Positive, Negative and Zero Work
• Work done is +ve, when angle θ = 0° i.e. force and displacement are in the same direction.
W = +F · s
• Work done is −ve, when angle θ = 180° i.e. force and displacement are anti-parallel to each other.
W = −F · s
• Work done is zero, when angle θ = 90° i.e. force and displacement are at right angles to each other.
W = 0
Energy
The capacity of doing work is called Energy.
SI unit: Joule (J)
Power
Power is the rate at which work is done or energy is transferred, indicating how fast energy is used or work is performed.
SI unit: watt
Power (P) = W t
Forms of Energy
Kinetic Energy (KE)
• Energy possessed due to the motion of the body is called Kinetic Energy.
• Kinetic energy of a body moving with a certain velocity is equal to the work done on it. It increases with the speed of an object.
• Kinetic Energy = 1 2 mv2
Potential Energy (PE)
It is the energy stored in a body due to its position or configuration. Potential Energy = mgh
Mechanical Energy (ME)
• The sum of total kinetic energy and potential energy is called Mechanical Energy.
• ME = PE + KE
• ME = mgh + 1 2 mv2
Law of Conservation of Energy
• Energy can neither be created nor destroyed. It can only be transformed from one form to another.
• During the motion of a body, total energy is always conserved if there is no loss of energy by the system.
ME = KE + PE = Constant
Chapter at a Glance
• Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the applied force.
• The SI unit of work is joule: 1 Joule = 1 Newton × 1 metre or 1 J = 1 N × 1 m.
• Work can be:
Positive: when force and displacement are in the same direction. Negative: when force and displacement are in opposite directions. Zero: when displacement is zero or perpendicular to the force.
• Energy is defined as the capacity to do work. The SI unit of energy is joule (J).
• Energy exists in nature in several forms such as kinetic energy, potential energy, heat energy, chemical energy, etc. The total of the kinetic and potential energies of an object is known as its mechanical energy.
• Types of mechanical energy:
Kinetic Energy (KE): It is the energy possessed by a body due to its motion. It depends on the mass of the object and the speed at which it moves.
Potential Energy (PE): It is the energy stored in a body due to its position or configuration.
• Law of Conservation of Energy states that energy can neither be created nor destroyed; it only transforms from one form to another, with the total energy remaining constant in an isolated system.
• Energy transformations occur in daily life, such as in machines, falling objects, running motors, and chemical reactions.
• Power is the rate at which work is done or energy is transferred, indicating how fast energy is used or work is performed. The SI unit of power is watt.
• The commercial unit of energy used by electricity companies is the kilowatt-hour (kWh), which represents a large amount of energy.
Formulae
1. Work (W) = F × d
2. Kinetic Energy (Ek) = 1 2 mv2
3. Potential Energy (E p) = mgh
4. Power (P) = W t
5. Commercial Unit of Energy: 1 kilowatt-hour (kWh) = 3.6 × 106 joules
NCERT Zone
Intext Questions
1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 10.1). Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Fig. 10.1
Ans. The work done is calculated by the formula: Work = Force × Displacement
Given, Force = 7 N
Displacement = 8 m
Work done (W) = 7 N × 8 m = 56 N m or 56 J.
2. When do we say that work is done?
Ans. In science, work is said to be done when a force acts on an object and the object is displaced in the direction of the force.
Thus, two conditions need to be satisfied for work to be done:
(a) A force must be exerted on an object.
(b) The object must be displaced along the direction of force. If either condition is missing, then no work is done.
Example: If you push a box and it moves in the direction you push it, work is done. But if you push a wall and it doesn’t move, no work is done—even though force was applied.
3. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Ans. The expression for work done (W) when a force (F) is acting on an object in the direction of its displacement (s) is given by:
Work done = Force × Displacement
W = F × s
4. Define 1 J of work.
Ans. 1 joule (J) of work is defined as the amount of work done when a force of 1 newton (N) displaces an object by 1 meter (m) in the direction of the force.
1 J = 1 N × 1 m
5. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Ans. Given: Force (F) = 140 N
Displacement (s) = 15 m
We know that, Work done (W) = F × s
= 140 N × 15 m
= 2100 J
Therefore, the work done by the bullocks is 2100 joules.
6. What is the kinetic energy of an object?
Ans. Kinetic energy is the energy possessed by a body due to its motion. It depends on the mass and speed of the object. A moving object has the ability to do work because of its kinetic energy.
7. Write an expression for the kinetic energy of an object.
Ans. For an object of mass m kg moving with velocity v, the expression for the kinetic energy is:
Kinetic Energy (Ek) = 1 2 mv2
8. The kinetic energy of an object of mass, m moving with a velocity of 5 m s−1 is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Ans. Given,
Initial velocity, v = 5 m/s
Initial kinetic energy, Ek = 25 J
We know that: Ek = �1 2� × m × v2
Substituting the values:
25 = �1 2� × m × (5)2
⇒ 25 = �1 2� × m × 25
⇒ mass, m = 2 kg
When velocity is doubled:
New velocity = 2 × 5 = 10 m/s
Ek = �1 2� × 2 × (10)2 = 1 × 100 = 100 J
When velocity is tripled:
New velocity = 3 × 5 = 15 m/s
Ek = �1 2� × 2 × (15)2 = 1 × 225 = 225 J
Therefore, the kinetic energy becomes 100 J when the velocity is doubled, and 225 J when the velocity is tripled.
9. What is power?
Ans. Power is defined as the rate at which work is done or the rate of energy transfer. It measures how fast or slow work is performed. It is measured in watts (W).
Power (P) = Work done (W) Time (t)
10. Define 1 watt of power.
Ans. 1 watt (W) of power is defined as the rate of work done or energy transferred when 1 joule (J) of work is performed or energy is transferred in 1 second (s).
1 W = 1 J/s.
11. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Ans. Given: Energy consumed (or Work done), W = 1000 J Time taken, t = 10 s
We know,
Power (P) = W t P = 1000 J 10 s = 100 W
The power of the lamp is 100 watts.
12. Define average power.
Ans. Average power is defined as the total work done or total energy transferred divided by the total time taken. It gives an overall measure of the rate at which work is done or energy is transferred over a specified time period.
Average Power = Total work done (W) Total time taken (t)
NCERT Exercises
1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
(a) Suma is swimming in a pond.
(b) A donkey is carrying a load on its back.
(c) A windmill is lifting water from a well.
(d) A green plant is carrying out photosynthesis.
(e) An engine is pulling a train.
(f) Food grains are getting dried in the Sun.
(g) A sailboat is moving due to wind energy.
(a) Suma is swimming in a pond. Yes
(b) A donkey is carrying a load on its back.
(c) A windmill is lifting water from a well.
(d) A green plant is carrying out photosynthesis.
No
Yes
No
(e) An engine is pulling a train. Yes
(f) Food grains are getting dried in the Sun.
(g) A sailboat is moving due to wind energy.
No
Yes
Justification
Suma applies force with her hands and legs, causing displacement of her body in the direction of the force.
The force (weight) acts vertically downward, while the donkey moves horizontally. Since displacement is not in the direction of the force, no work is done on the load.
The windmill applies a force to lift water, and the water is displaced upward in the direction of the force.
No mechanical force is applied and no displacement occurs; hence, no work is done in the mechanical sense.
The engine applies a force that causes the train to move in the direction of the force. Hence, work is done.
There is no mechanical force or displacement involved. Only heat energy is involved in drying.
Wind applies force on the sail, causing the boat to move in the direction of the force. Hence, work is done.
2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Ans. Work is defined as the product of force and displacement in the direction of the force. In this case, the object is displaced along a curved path, but its initial and final positions lie on the same horizontal level. The force of gravity acts vertically downward throughout the motion. Since the net vertical displacement is zero, the net work done by gravity is zero.
3. A battery lights a bulb. Describe the energy changes involved in the process.
Ans. When a battery lights a bulb, the chemical energy stored in the battery is first converted into electrical energy as the circuit is completed. This electrical energy flows through the bulb’s filament, which offers resistance and becomes hot. As a result, the electrical energy is converted into heat energy and light energy. Thus, the energy transformation involved is:
Chemical energy → Electrical energy → Light energy + Heat energy
4. Certain force acting on a 20 kg mass changes its velocity from 5 ms−1 to 2 ms−1. Calculate the work done by the force.
Ans. Given, Initial velocity (u) = 5 m/s
Final velocity (v) = 2 m/s
Mass (m) = 20 kg
K.E. = �1 2� × m × v2
Work done by the force (W) = Change in kinetic energy = Final K.E. − initial K.E.
W = �1 2 mv2 − 1 2 mu2�
= 1 2 m (v2 − u2) = 1 2 × 20 × (22 − 52)
= 1 2 × 20 × (4 − 25) = 10 × −21 = −210 J
The work done by the force is −210 J.
5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Ans. Work is defined as the product of force and displacement in the direction of the force. Here, the gravitational force acts vertically downward, whereas the displacement of the object is horizontal (from point A to B). Since the displacement is perpendicular to the direction of the force, the angle between force and displacement is 90°. The work done by the gravitational force is zero.
6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Ans. No, the decrease in potential energy of a freely falling object does not violate the law of conservation of energy.
As the object falls, its gravitational potential energy is converted into kinetic energy. Although the potential energy decreases, the kinetic energy increases by the same amount. Therefore, the total mechanical energy (potential energy + kinetic energy) remains constant throughout the motion. This adheres to the law of conservation of energy, which states that energy can neither be created nor destroyed but can only be transformed from one form to another, assuming no loss due to air resistance.
7. What are the various energy transformations that occur when you are riding a bicycle?
Ans. When riding a bicycle, chemical energy from the rider’s body (obtained from food) is converted into mechanical energy used for pedaling. This mechanical energy is transferred to the bicycle and converted into kinetic energy for motion. Some energy is also converted into heat due to friction between the tires and the road, and between the moving parts of the bicycle.
8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Ans. Yes, energy is transferred even if the rock does not move. Although no work is done on the rock (as there is no displacement), energy from the muscles is used and converted into heat due to muscle activity. Some of it may also be stored briefly as potential energy in the muscles.
9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Ans. Energy consumption in units is often measured in kilowatt-hours (kWh).
1 kilowatt-hour = 3.6 × 106 J
Total energy consumed = 250 × 3.6 × 106 = 900 × 106 J = 900 MJ
10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Ans. Given: m = 40 kg, h1 = 5 m, h2 = h1 2 = 2.5 m, g = 10 m/s2
Potential energy, PE = mgh = 40 × 10 × 5 = 2000 J
Potential energy at halfway = mgh2 = 40 × 10 × 2.5 = 1000 J
Kinetic energy at halfway = Total PE − PE at halfway = (2000 – 1000) J = 1000 J
11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Ans. The work done by gravity on a satellite moving in a circular orbit around the Earth is zero. This is because the force of gravity acts as a centripetal force, perpendicular to the direction of the satellite�s displacement. Since work is calculated as force times the displacement in the direction of force, no work is done when the direction of force and displacement are perpendicular.
12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
Ans. Yes, displacement can occur in the absence of force if the object continues moving due to inertia, as per Newton�s first law of motion, which states an object in motion remains in motion in a straight line unless acted upon by a force.
13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
Ans. In scientific terms, the person has not done any work on the bundle of hay. Work is defined as the product of force and displacement in the direction of the force. Although the person applies an upward force to support the hay, there is no displacement in the direction of the applied force (the hay remains stationary).
Therefore, work done is zero, even though the person feels tired because energy is spent internally.
14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Ans. Given, Power = 1500 W
Time = 10 h = 10 × 3600 s = 36000 s
The energy consumed by the electric heater can be calculated using the formula:
Energy = Power × Time = 1500 × 36000 = 54,000,000 J = 54 MJ
15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Ans. When a pendulum bob is drawn to one side, it gains gravitational potential energy at its highest point. As the bob is released, this potential energy is gradually converted into kinetic energy as it moves downward. At the lowest point of its swing, the kinetic energy is maximum, and the potential energy is minimum. As the pendulum swings back upward, kinetic energy is again converted into potential energy. This mutual conversion between kinetic and potential energy during oscillation demonstrates the law of conservation of energy, which states that energy can neither be created nor destroyed, only transformed from one form to another.
Eventually, due to air resistance and friction at the pivot, some mechanical energy is converted into heat and sound energy, and the pendulum comes to rest. This energy transformation does not violate the law of conservation of energy, as the total energy remains constant but changes form.
16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Ans. To bring a moving object to rest, work must be done to reduce its kinetic energy to zero. The amount of work required is equal to the initial kinetic energy of the object.
initial K.E. = 1 2 mv2
final K.E. = 0 (rest)
Work required = change in K.E. = 0 − 1 2 mv2 = − 1 2
This work done is negative which means that force applied is in the opposite direction of motion.
17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Ans. Given: Mass of the car, m = 1500 kg
Initial velocity, v = 60 km/h = �60 × 5 18� m/s = 16.67 m/s
Initial K.E. = 1 2 mv2
Final K.E. = 0 (rest)
= − 1 2 × 1500 × (16.67)2
= −750 × 278 ≈ −208500 J
Therefore, the work required to stop the car is approximately −2,08,500 joules.
18. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east as shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
Fig. 10.2
Ans. (a) Force is perpendicular to displacement, so work done is zero. (b) Force is in the same direction as displacement, so work done is positive. (c) Force is opposite to the displacement, so work done is negative.
19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
Ans. Yes, I agree with Soni. According to Newton�s First Law of Motion, an object remains at rest or in uniform motion if the net external force acting on it is zero. Even if multiple forces act, as long as they cancel out to no net force, the object remains at rest or continues moving uniformly, indicating zero acceleration.
20. Find the energy in joules consumed in 10 hours by four devices of power 500 W each.
Ans. Given,
Power of each device = 500 W
Number of devices = 4
Total power = 4 × 500 = 2000 W
Time = 10 hours = 10 × 3600 = 36000 seconds
Energy = Power × Time = 2000 × 36000 = 72,000,000 J
Therefore, the total energy consumed is 72,000,000 joules or 72 MJ.
21. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
Ans. When a freely falling object reaches the ground, its kinetic energy is transformed into another form of energy, such as heat, sound, and deformation of the ground or the object itself. The energy transfer depends on the nature of the object and surface. This transformation adheres to the law of conservation of energy, where energy merely changes form but isn�t lost.
Multiple Choice Questions
1. The sum of the kinetic and potential energies of an object is called its (a) Electrical energy (b) Chemical energy (c) Light energy (d) Mechanical energy
2. Choose the incorrect statement.
(a) Work has both magnitude and direction.
(b) The unit of work is Newton-metre (Nm) or joule (J).
(c) Work done is positive when the force is in the direction of displacement.
(d) Work done is negative when the force acts opposite to the direction of displacement.
3. One kilowatt-hour is equal to:
(a) 3600 J (b) 3600 kJ
(c) 3.6 × 105 J (d) 3.6 × 10⁶ kJ
4. A man is holding a bag without moving as shown in Fig. 10.3. The work done is:
(a) Positive
(c) Zero
(b) Negative
(d) Infinite
5. Which of the following quantities has the same unit as work?
(a) Power
(c) Energy
(b) Force
(d) Pressure
6. An object is moving with a uniform velocity along a particular direction. A retarding force of 5 N, is applied in the direction as shown. (CBSE QB)
N Direction of Motion Fig. 10.4
The object stops after a displacement of 5 m. What is the work done by the retarding force on the block?
(a) −5 J (b) −25 J (c) 5 J (d) 25 J
7. Two bodies of equal weight are kept at heights of h and 1.5h, respectively. The ratio of their P.E is (a) 3:2 (b) 2:3 (c) 1:1 (d) 4:3
8. When a body falls freely towards the Earth, then its total energy (NCERT Exemplar) (a) increases (b) decreases (c) remains constant (d) first increases and then decreases
9. In case of negative work, the angle between the force and displacement is (NCERT Exemplar) (a) 0° (b) 45° (c) 90° (d) 180°
10. A car weighing 500 kg is moving with a uniform velocity of 15 m/s. What is the kinetic energy possessed by the car? (CBSE QB)
(a) 56.250 kJ
(b) 112.500 kJ
(c) 56250 kJ
11. Which of the following situations involves negative work?
(a) A man lifting a box upwards
(b) A car accelerating on a highway
(c) Friction slowing down a sliding object
(d) A ball falling freely under gravity
(d) 112500 kJ
12. The kinetic energy of an object weighing 10 kg moving with a velocity of 5 m/s is 125 J. What is the maximum amount of work that can be done by the object? (CBSE QB)
(a) 5 J (b) 10 J (c) 125 J (d) 1250 J
13. A body is thrown vertically upward. At the highest point:
(a) KE is zero, PE is maximum
(c) Both KE and PE are zero
(b) PE is zero, KE is maximum
(d) KE = PE
14. Water stored in a dam possesses (NCERT Exemplar)
(a) no energy
(c) kinetic energy
(b) electrical energy
(d) potential energy
Fig.
15. If kinetic energy of an object becomes four times, then its velocity becomes:
(a) Twice (b) Four times (c) Half (d) Same
16. The child is doing work on the toy cart because:
(a) Force is applied but no motion
(b) Force and displacement are in opposite directions
(c) Force causes displacement in the direction of the force
(d) The toy cart is at rest
17. What happens to the potential energy of an object when it is lifted to a greater height?
(a) Decreases (b) Increases (c) Remains the same (d) Becomes zero
18. A boy of 40 kg runs up a staircase 5 m high in 10 s. What is his power? (g = 10 m/s2)
(a) 100 W (b) 150 W
(c) 200 W (d) 300 W
19. A device consumes 1500 W of power in 2 minutes. How much energy does it consume?
(a) 3000 J (b) 9000 J (c) 18000 J (d) 180000 J
20. At which point is the kinetic energy of the pendulum maximum?
(a) At the extreme left
(b) At the extreme right (c) At the center of the swing
(d) Same throughout the swing
21. Potential energy is given by the formula:
(a) mv2
(b) mgh
(c) 1 2 mv2 (d) F × s
22. A 2 kg ball is moving with a velocity of 3 m/s slows down to 1 m/s. What will be the change in kinetic energy?
(a) 9 J (b) 1 J
(c) 8 J (d) 10 J
23. A motor lifts 100 kg of water through a vertical height of 10 m in 5 seconds. What should be the minimum power of the motor? (Take g = 9.8 m/s2)
(a) 100 W (b) 500 W
(c) 1 kW
(d) 2 kW
24. Two bodies of equal mass are moving with velocities of 4 m/s and 2 m/s respectively. What is the ratio of their kinetic energies?
(a) 1:2 (b) 2:1 (c) 4:1 (d) 1:4
25. Which of the following correctly states the Law of Conservation of Energy?
(a) Energy can be created and destroyed.
(b) Energy can only be created but not destroyed.
(c) Energy can only be destroyed but not created.
(d) Energy can neither be created nor destroyed.
Answers
Fig. 10.5
Fig. 10.6
Constructed Response Questions
Very Short Answer Questions (30-50 words)
1. (a) What are the two essential conditions required for work to be done on an object?
(b) What is positive work? Explain with an example.
Ans. (a) The two essential conditions for work to be done on an object are:
(i) A force must be applied on the object.
(ii) The object must be displaced in the direction of the applied force.
(b) Work is said to be positive when the applied force and the displacement of the object are in the same direction.
Example: When a person pulls a lawn roller forward, the force applied and the movement of the roller are both in the same direction, so the work done is positive.
2. (a) What is mechanical energy? Give its two forms.
(b) Name the type of energy possessed by:
(i) stretched bow
(ii) flying arrow
Ans. (a) Mechanical energy is the energy that is available to do mechanical work. It is the sum of kinetic and potential energies of the object.
(b) (i) Stretched bow: Potential energy (ii) Flying arrow: Kinetic energy
3. What is the commercial unit of energy? Can you define it? How is it related to joules?
Ans. The commercial unit of energy is kilowatt-hour (kWh).
Kilowatt-hour (kWh) is defined as the energy used by a device of power 1000 W in 1 hour.
1 kWh = 1 kW × 1 h = 1000 W × 3600 s = 3.6 × 106 joules
4. (a) State the law of conservation of energy.
(b) How is work related to energy?
Ans. (a) Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy remains constant.
(b) Work and energy are interconvertible; when work is done, energy is transferred or transformed. Work done on an object equals the energy gained/lost by it.
5. Give one example each of potential energy:
(a) due to position (b) due to shape
Ans. (a) Potential energy due to position: Water stored at a height in a dam has potential energy because of its elevated position. This energy can be used to generate electricity when released.
(b) Potential energy due to shape: A compressed or stretched spring in a toy car stores potential energy. When the spring is released, this energy is converted into kinetic energy, causing the car to move.
6. A rocket is moving up with a velocity v. If the rocket engine is fired for few seconds, its velocity is tripled, what will be the ratio of the two kinetic energies? (NCERT Exemplar)
Ans. We know that kinetic energy (K.E.) is given by the formula:
K.E. = �1 2� × m × v2
Let the initial velocity be v.
Initial K.E. = �1 2� × m × v2
If the velocity is tripled, then the new velocity = 3v
Therefore, the ratio of new K.E. to original K.E. = 9:1
7. Define 1 watt of power. A machine does 500 J of work in 10 seconds. What is its power?
Ans. An agent is said to have power of 1 watt when it does 1 joule of work in 1 second.
Given, Work done = 500 J
Time taken = 10 s
Power = Work done Time = 500 10 = 50 W
8. (a) Can any object have mechanical energy even if its momentum is zero? Explain.
(b) Can any object have momentum even if its mechanical energy is zero? Explain. (NCERT Exemplar)
Ans. (a) Yes, mechanical energy comprises of both potential energy and kinetic energy. Zero momentum means that velocity is zero. Hence, there is no kinetic energy but the object may possess potential energy.
(b) No. Zero mechanical energy means that there is no potential energy and no kinetic energy. Therefore, if kinetic energy is zero, velocity becomes zero and hence, there will be no momentum.
9. In a pendulum, a bob of mass 2 kg is raised to a height of 0.5 m. Using energy conservation, calculate its speed at the lowest point.
Ans. Potential energy = mgh = 2 × 9.8 × 0.5 = 9.8 J
Using energy conservation,
KE = PE ⇒
1
2 × 2 × v2 = 9.8 J
⇒ v2 = 9.8
v = 9.8 ≈ 3.13 m/s
Hence, the speed at the lowest point is 3.13 m/s.
10. (a) State the SI unit of energy.
(b) A person carrying 10 bricks each of mass 2.5 kg on his head moves to a height 20 m in 50 s.
Calculate power of the person spent in carrying bricks. (Given, g = 10 m/s2 )
Ans. (a) The SI unit of energy is Joule (J). 1 Joule is the energy required to do 1 J of work.
(b) Given, mass of one brick = 2.5 kg
mass of 10 bricks = 2.5 × 10 = 25 kg
height, h = 20 m
time, t = 50 s
work done = change in potential energy = mgh
Therefore, Power, P = Work done Time = 25 × 10 × 20 50 = 100 W
11. (a) How do plants prepare their own food?
(b) Why do we rub our hands together during winters?
Ans. (a) Plants prepare their own food through the process of photosynthesis, where they convert solar energy into chemical energy in the form of glucose, using carbon dioxide and water in the presence of sunlight and chlorophyll.
(b) When we rub our hands together in winter, mechanical energy is converted into heat energy due to friction. This heat helps to warm our hands and protect us from the cold.
12. A 10 kg ball is thrown upwards with a speed of 5 m/s.
(a) Find its potential energy when it, reaches the highest point.
At the highest point, the kinetic energy becomes zero, and hence, the entire kinetic energy of 125 J is converted into potential energy. So, the potential energy at the highest point is 125 J.
(b) Suppose the ball reaches a maximum height h. Its potential energy will be mgh.
Potential energy at max. height = mgh = 125 J
⇒ h = 125 mg = 125 10 × 10
⇒ h = 1.25 m
13. If a stone of mass 3 kg is thrown up with a speed of 30 m/s. Calculate its maximum potential energy.
Ans. Mass of stone, m = 30 kg
Speed with which it is thrown, u = 30 m/s
final speed, v = 0
From 3rd equation of motion, v2 – u2 = 2as
v2 = u2 – 2gh
gh = u2 2 = 302 2 = 900 2 = 450
h = 450 g m
Maximum potential energy, PE = mgh = 3 × g × 450 g = 1350 J
14. A girl sits and stands repeatedly for 6 min. Draw a graph to show the variation of potential energy of her body with time.
Ans. From the graph shown above, we can take the sitting position of the girl as the position of zero potential energy. Let m be the mass of the girl and h be the position of centre of gravity while standing above the sitting position. The PE while standing is +mgh and while sitting is zero
15. What types of energy transformation takes place in the following:
(a) Electric heater
(b) Solar battery
(c) Steam engine
(d) Hydroelectric power station
Fig. 10.7
Ans. (a) Electric heater: Electric energy into heat energy.
(b) Solar battery: Solar energy into electric energy.
(c) Steam engine: Heat energy to mechanical energy.
(d) Hydroelectric power station: Mechanical energy into electric energy.
Short Answer Questions (50-80 words)
1. (a) Explain in detail how the law of conservation of energy applies to a roller coaster ride.
(b) If a body has zero velocity, can it still have energy? Explain.
Ans. (a) In a roller coaster, energy is continuously converted between potential and kinetic forms. At the top of a hill, it has maximum potential energy and minimal kinetic energy. As it descends, this potential energy is transformed into kinetic energy, increasing speed. At the lowest point, kinetic energy is maximum. As it climbs again, kinetic energy converts back into potential energy. Despite transformations, the total mechanical energy remains constant, showcasing the law of conservation of energy.
(b) Yes, if it is elevated or under tension, it can have potential energy. For instance, a stationary object at height has no velocity but possesses potential energy due to its position in a gravitational field.
2. Tabulate the difference between Energy and Power.
Energy
Power
Energy is the capacity to do work. Power is the rate at which work is done.
It is calculated as: Work = Force × Displacement. It is calculated as: Power = Work Time
The SI unit of energy is joule (J).
The SI unit of power is watt (W).
It indicates how much work can be done. It indicates how quickly the work is done.
Example: A battery stores 1000 joules of energy. Example: A 100 W bulb consumes 100 joules every second.
3. (a) A coolie carries luggage along a long platform at a railway station. Has he done any work on the luggage? Why?
(b) In a game of tug of war, one team is slowly giving way to the other. Which team is doing positive work and which is doing negative? Give reason.
Ans. (a) No, the coolie has not done any work on the luggage in the scientific sense. The force applied by the coolie is vertically upward (to support the weight of the luggage), while the displacement is along the horizontal platform. Since the angle between the force and displacement is 90°, the work done is zero.
(b) In tug of war, the winning team does positive work because the force applied and their displacement are in the same direction. The losing team does negative work because they are applying force opposite to the direction of displacement as they are being pulled.
Fig. 10.8
4. What is the amount of work done by an electron in the following situations?
(a) When it is revolving in a circular orbit of radius r around a nucleus.
(b) When it is moving with half the speed of light in empty space, free of all forces.
(c) When it is accelerated through a potential difference of 100 V. (Given w = q × V)
Ans. (a) In circular motion, the force acting on the electron (centripetal force) is always perpendicular to its displacement. Since force and motion are perpendicular to each other at all times, the work done by the electron is zero.
(b) When the electron moves in free space with uniform velocity (half the speed of light), there is no external force acting on it. Since work is done only when a force causes displacement, hence work done will be 0.
(c) When an electron is accelerated through a potential difference, work is done on it. The work done is given by:
W = q × V
where q = 1.6 × 10⁻¹⁹ C (charge of electron), V = 100 V
W = 1.6 × 10⁻¹⁹ × 100 = 1.6 × 10⁻¹⁷ J
Thus, the work done is 1.6 × 10⁻¹⁷ joules.
5. Water is falling on the blades of a turbine at the rate of 6 × 103 kg per minute. The height of the fall is 10 m. Calculate the power given to the turbine. [Take g = 10 m/s2]
Ans. Given, Mass of water, m = 6 × 103 kg
Time, t = 1 minute = 60 s
Height of fall, h = 10 m
Acceleration due to gravity, g = 10 m/s2
Potential energy lost by water per minute = mgh = 6 × 103 × 10 × 10 = 6 × 105 J
Power given = Energy lost Time = (6 × 105) 60 = 10 × 103 W = 10 kW
Therefore, the power given to the turbine is 10 kW.
6. Briefly describing the gravitational potential energy, deduce an expression for the gravitational potential energy of a body of mass m placed at a height h, above the ground.
Ans. When an object is raised through a certain height above the ground, its energy increases. This is because the work is done on it, against gravity. The energy present in such an object is called gravitational potential energy. Thus, the gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity.
Consider a body of mass m lying at point P on the Earth’s surface, where its potential measurement of potential energy is taken as zero.
10.9
As weight, mg acts vertically downwards, so to lift the body to another position Q at a height h, we have to apply a minimum force which is equal to mg in the upward direction. Thus, work is done on the body against the force of gravity.
We know that,
force = mg and vertical displacement = h
Work done = Force × displacement = mg × h
This work done on the body is equal to the gain in energy of the body. This is the potential energy of the body.
Therefore, Potential energy, P.E. = mgh.
7. Two boys A and B weighing 60 kg and 40 kg respectively, climb a staircase each carrying a load of 20 kg on their heads. The staircase has 10 steps, each of height 50 cm. If A takes 20 s to climb and B takes 10 s, then:
(a) Who possesses greater power?
(b) Find the ratio of their powers.
Ans. (a) Given,
Mass of A, mA = 60 kg
Mass of B, mB = 40 kg
Mass of luggage, m = 20 kg
Height of staircase, h = 0.5 × 10 = 5 m
Work done by boy A to climb staircase = (mA + m)gh = (60 + 20) × 10 × 5 = 4000 J
So, Power of A, PA = Work done by A Time = 4000 20 = 200 W
Similarly, work done by boy B = (mB + m)gh = (40 + 20) × 10 × 5 = 3000 J
Power of B, PB = Work done by B Time = 3000 10 = 300 W
Therefore, B possesses greater power than A.
(b) Ratio of their powers = PA PB = 200:300 = 2:3
8. The weight of a person on a planet A is about half that on the Earth. He can jump up to 0.4 m on the Earth. How high can he jump on planet A? (NCERT Exemplar)
Ans. Given,
height jumped on Earth, h1 = 0.4 m
weight on Earth, WE = mg
Potential energy on Earth = mgh1 = mg × 0.4
Weight on planet A, WA = WE 2 ⇒ ga = g 2
Let maximum height jumped on planet A be h2
Potential energy on planet A = m × �g 2� × h2
Assuming that he jumps with same K.E. on both planets, the person will have same P.E. on both planets at highest point.
Potential energy on Earth = Potential energy on planet A
mg × 0.4 = �mg 2 � × h2
⇒ h2 = 0.4 × 2 = 0.8 m
He can jump up to 0.8 m on planet A.
9. A girl is drawing water from a well of 10 m depth in 25 s, as shown in Fig. 10.10. If the mass of the bucket with water is 3 kg, What will be the magnitude of power used by her. (Take g = 10 m/s2)
Ans. Given, Mass of bucket, m = 3 kg
Height, h = 10 m
Time taken, t = 25 s
Work done by woman = change in P.E. = mgh = 3 × 10 × 10 = 300 J
Power = Work Time = 300 25 = 12 W
The power used by the girl is 12 W.
Fig. 10.10 h = 10 m m = 3 kg h = 0 reference level
10. A motorcycle and a car are moving on a levelled horizontal road with equal kinetic energy. Both the vehicles are brought to rest by equal retarding forces. Which of them will cover shorter distance to stop?
Ans. Let the retarding force on both be F and the stopping distance for the motorcycle be x and for the car be y.
Work done on motorcycle = W m = F × x
Work done on motorcycle = W c = F × y
Since both vehicles have equal kinetic energy
∴ KE m = KE c = 1 2 mv2
From conservation of energy, Kinetic energy lost by vehicle should be equal to work done on it.
KE m = KE c = W m = W c
⇒ F × x = F × y
⇒ x = y
Therefore, both the vehicles will cover equal distance before coming to rest.
11. (a) Define Power. Give its SI unit.
(b) An elevator with a load of 700 kg is moving down with a velocity of 0.5 m/s. What is the power produced by the motor in horsepower? (1 HP = 746 W)
Ans. (a) Power is defined as the rate at which work is done or energy is transferred. It tells us how fast or slow work is being performed.
Power = Work done Time taken
The SI unit of power is watt (W).
(b) Power = work done time = F × s t
= F × s t = mg × v
= 700 × 10 × 0.5 = 3500 W
∴ 746 W = 1 HP
∴ 3500 W = 3500 746 HP = 4.7 HP
Long Answer Questions (80–120 words)
1. (a) Define potential energy.
(b) Give an example where potential energy is acquired by a body due to change in its shape.
(c) A skier of mass 50 kg stands at A, at the top of a ski jump. He takes off from A for his jump to B. Calculate the change in his gravitational potential energy between A and B.
Ans. (a) Potential Energy is defined as the energy possessed by a body by virtue of its position or shape.
(b) In a toy car, the wound spring possesses potential energy. When spring is released, its potential energy changes into kinetic energy due to which the toy car moves.
(c) Given, m = 50 kg, h1 = 75 m, h2 = 60 m
At point A, PE1 = mgh1 = 50 × 10 × 75 = 37500 J
At point B, PE2 = mgh2 = 50 × 10 × 60 = 30000 J
Change in PE = PE1− PE2 = 37500 − 30000 = 7500 J
2. A vehicle of 1 tonne travelling with a speed of 60 ms−1 notices a cow on the road 9 m ahead and applies brakes. It stops just in front of the cow.
(a) Find out the kinetic energy (KE) of the vehicle before applying brakes.
(b) Calculate the retarding force provided by the brakes.
(c) How much time did it take to stop after the brakes were applied?
(d) What is the work done by the braking force? (NCERT Exemplar)
Ans. Given,
Mass of the vehicle, m = 1 tonne = 1000 kg
Initial speed, u = 60 ms−1
Final velocity, v = 0
Displacement, s = 9 m
(a) Kinetic Energy before applying brakes:
KE = 1 2 mu2
= 1 2 × 1000 × (60)2
= 500 × 3600 = 1800000 J
(b) Using the third equation of motion:
v2 = u2 + 2as
0 = (60)2 + 2 × a × 9
⇒ a = −200 ms⁻2
Retarding force, F = m × a = 1000 × (−200) = −200000 N
(c) Using the second equation of motion:
s = ut + 1 2 at2
9 = 60t + 1 2(−200)t2
9 = 60t − 100t2
100t2 − 60t + 9 = 0
(10t − 3)2 = 0
⇒ t = 0.3 s
(d) Work done by the braking force:
Work = Force × Displacement = −200000 × 9 = −1800000 J
3. (a) What remains unchanged in the oscillation of a pendulum if it is released from point A?
(b) Take the lowest position of pendulum (B) during the motion as reference point for potential energy, the height of the pendulum from reference point at any time t as h and its speed at that time as v. Derive an expression relating v and h.
(c) If the mass of pendulum is 5 kg and the difference in height of point A and point B is 0.5 m. What will be the energy of the pendulum at the point A?
(g = 10 m/s²)
(d) Calculate the total energy of the pendulum when it is at a height of 0.25 m from point B.
Ans. (a) The total mechanical energy remains constant during the oscillation of a pendulum.
Mechanical energy = K.E. + P.E. = constant
⇒ mgh + 1 2 mu2 = constant
(b) At Point A,
P.E.= mgh and K.E. = 0
Total energy = P.E. + K.E. = mgh
At Point B,
P.E.= 0 and K.E. = 1 2 mv2
Total Energy = P.E. + K.E. = 1 2 mv2
4.
From conservation of energy,
mgh = 1 2 mv2
⇒ v = 2gh
(c) P.E. point A = mgh = 5 × 10 × 0.5 = 25 J
K.E. at point A = 0 (at rest)
Total mechanical energy = P.E. + K.E. = 25 J
(d) Since energy is conserved, therefore total mechanical energy will always be 25 J.
(a) A shell is fired from a cannon vertically upwards from a point X. On reaching the highest point Y, it falls down. Give a brief explanation about the transformation of energy from X to Y and Y to X.
(b) If this shell (m = 2 kg) is taken to the top of a tower of height 40 m and is dropped at t = 0. What will be its kinetic and potential energy after:
(i) 1st second
(ii) 2nd second
(iii) 3rd second (Take g = 10 m/s² and ground as reference for Potential energy)
Ans. (a) While going vertically upward from the ground, kinetic energy is gradually converted to potential energy. At the highest point (Y), potential energy is maximum and kinetic energy is zero. While falling down from Y to X, the potential energy is converted back into kinetic energy.
(b) Given, mass of shell, m = 2 kg
height of tower = 40 m
g = 10 m/s2
Total energy of the shell: T.E. = mgh = 2 × 10 × 40 = 800 J
(i) After 1st second:
using 1st equation of motion, v = u + at
v = u + gt = 0 + 10 × 1
⇒ v = 10 m/s
K.E. = 1 2 × 2 × (10)² = 100 J
P.E. = TE − KE = 800 − 100 = 700 J
(ii) After 2nd second:
v = u + gt = 0 + 10 × 2
⇒ v = 20 m/s
K.E. = 1 2 × 2 × (20)² = 400 J
P.E. = TE − KE = 800 − 400 = 400 J
(iii) at t = 3 sec, displacement will be using 3rd equation of motion, s = ut + 1 2 at2
∴ s = 1 2 gt2 (∴ u = 0 and a = g)
⇒ s = 1 2 × 10 × 32
⇒ s = 45 m
Since height of tower is only 40 m, the displacement of shell cannot be more than 40 m. This means that shell was not in air for full 3 seconds and it hit the ground between 2 – 3 secs.
So, after 3 s, the shell has hit the ground and its total energy is now zero.
KE = 0 (at rest)
PE = 0 (on ground)
Fig. 10.13
5. (a) A body thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
(b) You slowly lift a heavily packed carton of mass m vertically upward through a height h. What is the work done: (i) by you on the carton (ii) by the force of gravity on the carton
(c) Anil is doing work at a rapid rate but works only for one hour. Ashok does work at a slower rate but continues to work for six hours. Who has greater power? Who has more energy?
Ans. (a) When a body is thrown at an angle to the ground, it follows a curved path (called projectile motion). The vertical displacement between the initial and final points is zero. Since gravity acts vertically and the net displacement in that direction is zero.
Therefore, work done by gravity = 0
This is because the positive work done by gravity during descent is cancelled by the negative work done during ascent.
(b) (i) The force applied is vertically upward and the carton is displaced in the same direction. Hence, Work done by you = gain in PE = +mgh.
The work done by you is positive.
(ii) Gravitational force (mg) acts downward, while displacement (h) is upward. Therefore, Work done by gravity = F × s = −mgh.
The work done by gravity is negative.
(c) Anil is more powerful because power is the rate of doing work, and he completes more work per unit time. Ashok uses more energy because he works for a longer duration, and thus does more total work overall.
6. Fig. 10.14 shows a conveyor belt transporting a package to a raised platform. The belt is driven by a motor.
(a) State three types of energy, other than gravitational potential energy, into which the electrical energy supplied to the motor is converted.
(b) The mass of the package is 36 kg. Calculate the increase in the gravitational potential energy (G.P.E.) of the package when it is raised through a vertical height of 2.4 m.
(c) The package is raised through the vertical height of 2.4 m in 4.4 s. Calculate the power needed to raise the package.
(d) Assume that the power available to raise the package is constant. A package of mass greater than 36 kg is raised through the same height. Suggest and explain the effect of this increase in mass on the operation of the belt.
Ans. (a) (i) Kinetic energy of belt or the package
(ii) Heat energy in moving parts of motor
(iii) Sound energy in moving parts of motor.
(b) Given, m = 36 kg, h = 2.4 m, g = 10 m/s2 Gravitational potential energy (G.P.E.) = mgh = 36 × 10 × 2.4 = 864 J
(c) Power = work done time = mgh t = 36 × 10 × 2.4 4.4 = 864 4.4 = 2160 11 = 196.36 W
(d) Given, mass is increased and power is constant.
Power = work done time = mgh t = mg × h t = mg × v
Motor
Conveyor belt Package
Fig. 10.14
Since mass has increased, the total required PE has also increased. For constant motor power, increase in mass will result into reduction in speed of conveyer belt and the motor will require more time to move the box to the same height.
Competency Based Questions
Multiple Choice Questions (Choose the most appropriate option/options)
1. A ball is thrown up with some speed v and it reaches a maximum height of is h m before falling back. What will be its kinetic energy at a height of h 2 from the ground?
(a) Maximum
(c) Half of the initial value.
(b) Zero
(d) Quarter of the initial value.
2. If the mass of an object is doubled and its speed is halved, what will happen to the kinetic energy of the object?
(a) It becomes half.
(c) It becomes one-eighth.
(b) It becomes twice.
(d) It becomes eight times.
3. What is the work done by gravity on a satellite revolving around the Earth in a circular orbit?
(a) Positive (b) Negative
(c) Zero (d) Data insufficient.
4. If an object placed at a height of 1 km above the Earth’s surface has potential energy of 600 J. What will be the potential energy of the object if it is placed at the same height but from the Moon’s surface? (Given: Gravity on the surface of the Moon is one-sixth than that of the Earth’s surface.)
(a) 100 J (b) 300 J
(c) 600 J (d) 3600 J
5. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process, the potential energy of the car. (NCERT Exemplar)
(a) does not change.
(c) becomes 4 times that of initial.
(b) becomes twice to that of initial.
(d) becomes 16 times that of initial.
6. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same. (NCERT Exemplar)
(a) acceleration. (b) momenta.
(c) potential energy. (d) kinetic energy.
7. The spring of a spring-loaded toy car is compressed and then released. Which energy transformation best describes this process?
(a) Chemical to mechanical
(b) Electrical to kinetic (c) Potential to kinetic (d) Kinetic to potential
8. A ball rolls down a hill from a height of 50 m and then travels for 50 m on a horizontal road before stopping. What energy conversion takes place during horizontal motion?
(a) Potential to kinetic
(b) Kinetic to potential
(c) Potential to heat
(d) Kinetic to heat
9. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be (g = 10 ms−2) (NCERT Exemplar)
(a) 6 kJ
(b) 6 J
(c) 0.6 J (d) 0
10. An object of mass m has a momentum p and kinetic energy K. Then (a) K = mp
(c) K = p2 m
(b) K = p2 2m
(d) K = m2 2p
11. Two objects, A and B, with masses 2m and m respectively, are released from the same height h. Which of the following statements correctly gives the relation between their momentum just before hitting the ground?
10.16
(a) Momentum of A = Momentum of B. (b) Momentum of A = 2 × Momentum of B. (c) 2 × Momentum of A = Momentum of B. (d) Momentum of A < Momentum of B.
12. Two particles of masses 1 g and 4 g have equal kinetic energies. What is the ratio between their velocities?
(a) 1:4 (b) 4:1
(c) 1:2 (d) 2:1
13. A boy pushes a lawn mower to a total distance of 125 m over the grass with a force of 60 N directed horizontally. How much work is done in joules by the boy? (CBSE QB)
(a) +7500 J (b) −7500 J
(c) +2.03 J (d) −2.03 J
14. A 50 kg skydiver jumps from a height of 20 m. What would be his kinetic and potential energies when he is halfway down? Assume g = 10 m/s2. (CBSE QB)
(a) KE = 5,000 J; PE = 5,000 J
(c) KE = 10,000 J; PE = 0 J
(b) KE = 0 J; PE = 10,000 J
(d) KE = 10,000 J; PE = 10,000 J
15. The block shown in Fig. 10.17 is initially at rest. It is acted upon by forces F1 = 20 N and F2 = 40 N. If the block moves 5 m due to these forces, the work done by force F1 is:
(a) 100 J
(c) 200 J
Assertion-Reason Based Questions
(b) −100 J
(d) −200 J
10.17
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): When there is no displacement of an object, the work done by a force is zero.
Reason (R): If displacement is in the opposite direction of the applied force, the work done is negative.
Fig.
Fig. 10.15
Fig.
2. Assertion (A): Potential energy is stored in a stretched bow.
Reason (R): Potential energy is stored in within the object and it depends on position and deformation of the object.
3. Assertion (A): A machine with greater efficiency transforms more energy into useful output.
Reason (R): Efficiency =Total work output / Total energy input.
4. Assertion (A): When a coolie walks with a load on his head, he does no work on the load.
Reason (R): Work is considered to be done only if the displacement is the direction of the force.
5. Assertion (A): Power of two machines are said to be equal if they do equal amount of work.
Reason (R): Power is the amount of work done in a unit of time.
6. Assertion (A): If the velocity of a body is doubled, its kinetic energy becomes four times greater.
Reason (R): Kinetic energy varies directly with the square of the velocity of an object.
7. Assertion (A): Work and energy have different SI units.
Reason (R): Work is expressed in joules, whereas energy is measured in kWh.
8. Assertion (A): An object falling freely under gravity experiences constant acceleration and increasing kinetic energy.
Reason (R): The speed of an object in free fall under gravity increases uniformly, leading to an increase in kinetic energy.
Case-Based/Source-Based/Passage-Based Questions
1. In physics, work is said to be done when a force applied on an object results in displacement of that object in the direction of the force. All three components—force, displacement, and direction—are crucial. If the force is zero, work is zero. If force is present but displacement is zero (e.g., pushing a wall), work is also zero. Likewise, if both force and displacement exist but the displacement is perpendicular to the force, the work done remains zero. Thus, work is defined as the product of force and displacement in the direction of the applied force:
W = F × s × cos θ
Where θ is the angle between the directions of force and displacement. It is also important to note that work does not depend on the time taken for the displacement. For example, if two porters lift the same box to the same height in different durations, the work done by each remains the same.
(Given: cos 60° = 0.5, cos 45° = 0.7)
(a) A force of 20 N moves an object 4 m in the same direction. The work done is:
(i) 60 J (ii) 70 J (iii) 80 J (iv) 90 J
(b) A 30 N force acts at an angle of 60° to the direction of displacement of 1 m. The work done is:
(i) 15 J (ii) 20 J (iii) 25 J (iv) 30 J
(c) For which of the following scenarios the total work done by gravity is non-zero?
(i) A coolie standing with a 2 kg load on the head for 5 minutes.
(ii) Moon revolving around the Earth.
(iii) A stone thrown downwards from a tall tower.
(iv) A stone thrown upwards from ground level.
(d) A porter pulls a block of mass 10 kg placed on a rough surface at constant speed of 2 m/s for 3 seconds. If 10 N force is applied by the porter on the rope, then work done by the surface is:
(i) Positive (ii) Negative (iii) Zero
(iv) Situation is not possible. When force is applied speed cannot be constant.
2. During a science fair, students set up a model where water stored in a tank at a height is released to rotate a turbine connected to a small electric generator. As the water flows down, it rotates the turbine blades, and electricity lights up a bulb connected to the generator. The group explains that this is how hydroelectric power plants work.
Answer the following based on the case study above:
(a) What type of energy is stored in the elevated water?
(b) Explain the energy transformations taking place from the water tank to the glowing bulb.
(c) If the height of the tank is doubled, what happens to the potential energy of water? Justify your answer.
(d) Why is it important to minimize friction in the turbine system in such models?
3. A moving object is characterised by its momentum (p = mv) and kinetic energy �Eₖ = 1 2 mv2�. Momentum is a vector quantity while kinetic energy is a scalar quantity.
(a) An object of mass 100 g is moving with a constant velocity of 2 m/s. What is its momentum? What is its kinetic energy?
(b) How is the kinetic energy of a moving object related to its linear momentum?
(c) A scooty and a car both are in motion with the same kinetic energy. Whose velocity is more and why?
(d) A bullet and a recoiled gun have equal momenta in mutually opposite directions. Which has greater kinetic energy and why?
Answers
Multiple Choice Questions
Assertion-Reason Based Questions
Case-Based/Source-Based/Passage-Based Questions
1. (a) (iii) W = F × s × cos θ = 20 × 4 × cos (0°) = 80 J
(b) (i) W = F × s × cos θ = 30 × 1 × cos (60°) = 30 × 1 × 0.5 = 15 J
(c) (iii) Vertical displacement is in direction of force. Therefore work done is positive.
(d) (ii) Work done by porter is positive and block does not experience any change in KE or PE, therefore the work done by friction has to be negative and of the same magnitude.
2. (a) The elevated water possesses gravitational potential energy.
(b) Energy transformations:
• Potential energy → Kinetic energy (as water flows)
• Kinetic energy → Mechanical energy (turbine rotation)
• Mechanical energy → Electrical energy (generator)
• Electrical energy → Light energy (bulb)
(c) Potential energy is directly proportional to height.
So, doubling the height doubles the potential energy (PE = mgh).
(d) Friction causes energy loss in form of heat and sound. Minimizing friction ensures more mechanical energy is transferred to the generator, improving energy output.
(c) Since kinetic energy is the same for both vehicles and mass of the car is much greater than that of the scooty, the velocity of the scooty must be greater to compensate for the lower mass �Eₖ = 1 2 mv²�.
(d) The bullet has a much smaller mass than the gun. Since both have the same momentum and Eₖ = p² 2m , the bullet has greater kinetic energy because of its smaller mass.
Numerical Questions
1. A bullet of mass 10 g is fired with velocity 500 m/s into a wooden block and gets embedded. Find the kinetic energy lost by the bullet.
Ans. Given,
m = 0.01 kg, u = 500 m/s, v = 0
Change in KE = 1 2 × m × (v2 u2) = 1 2 × 0.01 × −(500)2 = −0.005 × 250000 = −1250 J
Therefore, kinetic energy lost by bullet is 1250 J.
2. A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a roundabout of radius 100 m. He moves on the circular path for one and half cycles and then moves forward up to 2.0 km. Calculate the work done by him.
(NCERT Exemplar)
Ans. Given,
Frictional force, F = 5 N
Length of straight path = 1.5 km and 2.0 km
radius of circular path = 100 m or 0.1 km
Length of circular path travelled = number of cycles × 2πr = 3 2 × 2 × π × 0.1 = 0.94 km
Total distance = 1.5 km + 0.94 km + 2.0 km
= 1.5 + 0.94 + 2.0 = 4.44 km or 4440 m
Work done = Force × Distance = 5 × 4440 = 22200 J
3. A man lifts a 20 kg suitcase to a height of 1.5 m and then carries it 50 m on a level road. Calculate the total work done by gravity.
Ans. Given,
Mass of suitcase, m = 20 kg
Height, h = 1.5 m
horizontal displacement, s = 50 m
Work done in lifting the suitcase (W1) = change in P.E. = mgh = 20 × 9.8 × 1.5 = 294 J
Work done in carrying it on the level road (W2) = 0 (no vertical displacement)
Total work = W1 + W2 = 294 + 0 = 294 J
Fig. 10.19
4. A 2 kg ball falls freely from a height of 20 m. Ignoring air resistance, find the speed just before it hits the ground using energy conservation.
Ans. Given,
mass of ball, m = 2 kg
height, h = 20 m
At the top,
PE = mgh = 2 × 10 × 20 = 400 J
KE = 0
At the bottom,
PE = 0
KE = 1 2 mv2
using energy conservation, (PE + KE) at top = (PE + KE) at bottom
400 = 1 2 mv2
400 = 1 2 × 2 × v2
v2 = 400
v = 400 = 20 m/s
5. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how high can the ball bounce back? (g = 10 m/s²) (NCERT Exemplar)
Ans. Let the mass of the ball be m kg.
initial height, h = 10 m
Initial potential energy = mgh = m × 10 × 10 = 100m J
After losing 40% energy, energy left = 60% of 100m J = 60m J
Let new height on rebound be hb. Then at top point of bounce back,
P.E. = 60m
mghb = 60m
10hb = 60 ⇒ hb = 6 m
So, the ball can bounce back up to 6 m.
6. A crane lifts a load vertically through 15 m in 10 s. If the power expended is 2940 W, find the mass of the load.
Ans. Given, height, h = 15 m time, t = 10 s
power of crane, P = 2940 W using definition of power,
Work = Power (P) × Time (t) = 2940 × 10 = 29400 J
This work transforms into the increase in PE of the load.
∴ Potential energy = mgh = 29400 J
m = 29400
9.8 × 15 = 200 kg
7. How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of 1 m/s vertically?
(g = 10 m/s²) (NCERT Exemplar)
Ans. Let the mass of body be m kg, the body is lifted to a height h m in time t seconds and the power applied be P.
Power, P = Work done Time taken = Change in PE of the body Time taken
= mgh t = mg × h t �h t = v�
∴ P = mg × v
Given: P = 100 W, v = 1 m/s, g = 10 m/s²
m = P g × v = 100 10 × 1 = 10 kg
So, the man can lift 10 kg mass vertically at a speed of 1 m/s.
8. An object of mass 16 kg is situated at the top of a multi-storeyed building and its potential energy is 2800 J. What is the height of the building? [Take g = 10m s−2]
Ans. Given,
mass, m = 16 kg
potential energy, PE = 2800 J g = 10 ms−2
As potential energy, PE = mgh
∴ h = PE mg = 2800 16 × 10 = 17.5 m
9. What will be the output power (in watt) of a 55 kg athlete who accelerates from 0 to 12 m/s in 4 s?
Ans. Given,
mass, m = 55 kg
initial speed, u = 0
final speed, v = 12 m/s
From conservation of energy, Work done = change in kinetic energy.
W = 1 2 m × (v2 − u2)
= 1 2 m × (122 − 02) = 1 2 × 55 × 144 = 3960 J
Power output, P = Work Time = 3960 4 = 990 W
10. A lift is designed in such a way to carry a load of 5000 kg through 12 floors of a building, on an average height is 9 m per floor in 30 s. Find the power of the lift. (g = 10 m/s²)
Ans. Given,
Mass of load, m = 5000 kg
number of floors = 12
height of each floor = 9 m time of ascend = 30 s
Height of the building, h = 12 × 9 = 108 m
Potential energy at the top of building, PE = mgh = 5 × 103 × 10 × 108 = 540 × 104 J
This amount of energy is transferred in 30s with the help of lift.
Therefore, Power of the lift,
P = PE Time = 540 × 104 30 = 18 × 104 W
Practice Questions
Multiple Choice Questions
1. Two bodies of different masses have the same kinetic energy. Which body has greater momentum? (a) The one with lesser mass (b) The one with greater mass (c) Both have equal momentum (d) None of these
2. A person pushes a wall with a force of 100 N for 2 minutes. The work done is: (a) 12,000 J (b) 200 J (c) 0 J (d) Cannot be calculated
3. A crane lifts 500 kg of coal to a height of 20 m in 50 s. What power does it use? (g = 10 m/s²) (a) 1000 W (b) 2000 W (c) 3000 W (d) 4000 W
4. Which situation shows conversion of kinetic energy into potential energy?
(a) A body falling freely from height (b) A ball hitting the wall (c) A car climbing a hill (d) A compressed spring released
5. A 60 W bulb is used for 5 hours. How much energy does it consume in kilowatt-hour? (a) 0.03 kWh (b) 0.3 kWh (c) 0.6 kWh (d) 3.0 kWh
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
6. Assertion (A): A truck and a bus do the same amount of work in 40 s and 60 s respectively. Truck is more powerful than the bus.
Reason (R): Power = Work done Time taken
7. Assertion (A): An object can have energy even if it is not moving. Reason (R): Energy is always associated with motion and depends on velocity.
Very Short Answer Questions (30-50 words)
8. Differentiate between Kinetic energy and Potential energy. Give one example for each.
9. Why is power considered a better measure of performance than work?
10. If a stone is thrown upwards, what happens to its kinetic energy and potential energy?
11. A man of 50 kg jumps from a height of 3 m. What is the maximum PE that he will have?
12. An engine lifts 500 kg of coal vertically up to 15 m in 25 seconds. Find the power output in kilowatts. (g = 10 m/s²)
Short Answer Questions (50-80
words)
13. Derive the relation between kinetic energy and momentum of a body.
14. A child pulls a toy car through a distance of 10 m on a smooth horizontal floor. The string held in child’s hand makes an angle of 60° with the horizontal as shown in Fig. 10.20. If force applied by the child is 16 N, calculate the work done by the child in pulling the toy car.
15. A ball of mass 0.5 kg is thrown vertically upwards with velocity 20 m/s. Calculate (a) its initial kinetic energy, (b) maximum height reached, (c) potential energy at maximum height, and (d) total energy throughout the motion. (Take g = 10 m/s²)
16. Define potential energy with an example. Derive the expression for gravitational potential energy and state the factors on which it depends.
17. A man of mass 60 kg climbs a staircase of 20 steps, each of height 15 cm, in 25 seconds. Calculate:
(a) Total height climbed
(b) Work done against gravity
(c) Power generated by the man (Take g = 10 m/s²)
Long Answer Questions (80-120 words)
18. A bus of mass 800 kg is moving with a uniform speed of 20 m/s against an opposing force of 3000 N as shown in the Fig. 10.21.
(a) A fuel used by the engine to drive the bus. What form of energy is in the fuel?
(b) Find the kinetic energy of the bus.
(c) Find the energy used in 1s against the frictional force.
(d) Calculate the minimum power that the bus engine has to deliver to the wheels?
19. (a) Define kinetic energy. Write an expression for kinetic energy and its SI unit.
(b) A man of mass 75 kg lifts a load to a height of 5 m. The same load was carried by a young boy weighing 40 kg to the same height.
(i) Find the ratio of work done by the two.
(ii) Find the ratio of the potential energies possessed by the man and the boy if they are standing at a height of 50 m. (take g = 10 m/s²)
20. A car of mass 800 kg is moving at a speed of 20 m/s. The driver applies brakes to bring it to rest in a distance of 100 m.
(a) Calculate the kinetic energy of the car before applying brakes.
(b) What is the work done by the braking force?
(c) Calculate the retarding force applied by the brakes.
(d) How much time does the car take to stop?
(e) Calculate the power required to stop the car in this time if the work is done uniformly.
Additional Numericals
1. An object of mass 2 kg is dropped from a height of 10 m. Find the speed just before it hits the ground. (g = 10 m/s²)
2. An electric motor lifts a load of 500 N to a height of 20 m in 10 seconds. Calculate the power developed.
3. Compare the power at which each of the following is moving upwards against the force of gravity.
(Given g = 10 ms ²). (NCERT Exemplar)
(a) A butterfly of mass 1.0 g that flies upward at a rate of 0.5 ms−1.
(b) A 250 g squirrel climbing up a tree at a rate of 0.5 ms−1.
Fig. 10.21
4. What should be the mass of a bullet if its speed is 100 m/s and it possesses a kinetic energy of 0.1 kilojoule?
5. Two masses m and 2 m are dropped from a height h and 2 h. On reaching the ground, which will have more kinetic energy? Why?
6. A body weighing 1,000 kg accelerates uniformly from 10 m/s to 20 m/s. Find the amount of work done during this period.
7. A man weighing 500 N carried a load of 100 N up a flight of stairs 4 m high in 5 seconds. What is his power?
8. A body of mass 4 kg is thrown vertically upwards at a speed of 30 m/s. If g = 10 m/s2, calculate (a) maximum kinetic energy of the body (b) potential energy of the body at its maximum height
9. What amount of energy will be required to stop a truck of 1200 kg moving at a velocity of 25 km/h?
10. An object of mass 2 kg is dropped from a height of 1 m. What will be its kinetic energy as it reaches the ground? (Take, g = 9.8 ms−2)
11. In Fig. 10.22, a girl is carrying a bucket filled with water by applying a vertical force of 15 N. She moves a horizontal distance 10 m and then climbs up a vertical distance of 20 m. Calculate the total work done by the girl.
12. A girl weighing 50 kg runs up a hill raising herself vertically 10 m in 20 s. What is the power expended by girl?
13. An object of mass 10 kg is moving with a speed of 4 m/s. If its kinetic energy is halved, what will be the new speed? Show your calculations.
14. A car of mass 1200 kg is brought to rest from a speed of 72 km/h by applying brakes.
Calculate the work done by the brakes and the braking force if the car stops in 50 meters.
15. An object is thrown upward with 200 J of kinetic energy. Find the maximum height reached.
16. A cyclist increases speed from 10 m/s to 20 m/s.
(a) By what factor does kinetic energy increase?
(b) If mass = 50 kg, find the difference in kinetic energy.
17. An object of mass 20 kg is lifted vertically to a height of 5 m and then dropped.
Calculate the total mechanical energy at the highest point and just before hitting the ground. What does this indicate about energy conservation?
18. How fast should a man of mass 50 kg run so that his kinetic energy is 625 J?
19. A body of mass 3 kg is thrown vertically upwards with an initial velocity of 20 m/s. What will be its potential energy at the end of 2s? (Take g = 10 ms−2)
20. Express the power a car in kW when it is moving at a speed of 108 km/h under the action of 2000 N force.
20 m
10 m
Fig. 10.22
Brain Charge
1. Crossword Puzzle ACROSS
1. The product of force and displacement when both are in the same direction.
2. The SI unit of energy or work.
3. Type of energy in water at a height.
4. A term describing work done that adds energy to an object. DOWN
2. The energy possessed by a body due to its motion.
3. Type of work done when force acts opposite to the displacement.
4. A law that explains that total energy in a closed system remains constant.
5. The ability or capacity of a body to do work.
2. Word Puzzle
Use the first alphabet of each answer to the puzzle and combine them to spell a word.
1. It is the energy possessed by a body due to its position or configuration.
2. When a machine runs, the useful energy or work it produces is referred to by this term.
3. This quantity is considered zero if there is no displacement, even when a force is applied.
4. This physical quantity is never lost but only transformed from one form to another, such as from potential to kinetic.
5. This term refers to how quickly work is done or energy is transferred, and is synonymous with power.
Challenge Yourself
1. On a level road, a scooterist applies brakes to slow down from a speed of 54 kmh−1 to 36 kmh−1. What amount of work is done by the brakes? (Assuming, the mass of the empty scooter is 86 kg and that of the scooterist and petrol is 64 kg.)
2. (a) A body of mass m is raised to a vertical height h through two different paths A and B as shown in the Fig. 10.23. What will be the potential energy of the body in the two cases? Give reason for your answer.
(b) If two bodies X and Y having equal masses are dropped from the heights of 2 h and 5 h. Calculate the ratio of their potential energies when they have fallen through half of their initial height. Use g = 10 m/s2.
3. An engine can pump 30,000 litres of water to a vertical height of 45 metres in 10 minutes. Calculate the work done by the machine and its power.
[g = 9.8 m/s2; Density of water = 103 kg/m3, 1000 litre = 1 m3]
4. Can there be any change in the physical world without any change in energy? Justify.
Answers Scan me for detailed Solutions
Practice Questions
1. (b) 2. (c) 3. (b) 4. (c) 5. (b) 6. (a) 7. (c)
11. 1500 J
12. Work = 75000 J, Power = 3 kW
14. 80 J
15. (a) KE = 100 J (b) h = 20 m (c) PE = 100 J (d) Total mechanical energy = 100 J
17. (a) Height = 3 m (b) Work = 1800 J (c) Power = 72 W
18. (a) Chemical energy. (b) Kinetic energy of the bus = 160000 J (c) 60000 J or 60 k J (d) Minimum power = 60 kW
19. (b) (i) 1 (ii) 15 8
20. (a) 160000 J (b) −160000 J (c) −1600 N (d) 10 seconds (e) 16000 W or 16 kW
Additional Numericals
1. 14.14 m/s 2. 1000 W 3. (a) 5 × 10−3 W (b) 1.25 W 4. 0.02 kg
5. 2 m will have more kinetic energy 6. 150 kJ 7. 480 W
17. At top: P.E. = 980 J, K.E. = 0 At bottom: P.E. = 0, K.E. = 980 J
4800 N
Conclusion: Total mechanical energy = 980 J (constant)
18. 5 m/s 19. 600 J
Brain Charge
Crossword Puzzle
Word Puzzle
Solution: POWER
1. Potential
Challenge
1. −9375 N (ii) Potential energies of bodies X and Y is 2:5.
2. Work done = 1.323 × 107 joule, Power = 22 kW
SELF-ASSESSMENT
Time: 1.5 Hour
Max. Marks: 50
Multiple Choice Questions (10 × 1 = 10 Marks)
1. With the help of Fig. 10.24, write an expression for work in terms of force and displacement.
(a) W = −F × s
(b) W = F/s
(c) W = F × s
(d) W = −F/s
2. Two stones, A and B, are dropped from the same height from a tower. Stone A is lighter than stone B. Which stone will have more kinetic energy just before hitting the ground (neglecting air resistance)?
(a) Stone A (lighter one)
(b) Stone B (heavier one) (c) Both will have equal kinetic energy (d) Cannot be determined
3. When the kinetic energy of a body is increased by 800%, the momentum of the body is increased by: (a) 2000% (b) 20% (c) 200% (d) 100%
4. Why is no work done by gravitational force when a satellite moves in a circular orbit around Earth? (a) Because satellite is weightless (b) Force is tangential (c) Force is perpendicular to displacement (d) Earth pulls it inward
5. Two persons A and B do the same work, but A finishes it in half the time. Which of the following is correct?
(a) A has more energy
(b) A has more power (c) B has more power (d) Both have equal power
6. An object experiences a vertical force F = 50 N and moves a distance of 4 m in horizontal direction. What is the work done by force F?
(a) 0 J (b) 100 J (c) 200 J (d) 500 J
7. A 1.5 kg object moving at 20 m/s comes to rest in 5 seconds under constant force. What is the magnitude of this retarding force?
(a) 6 N (b) 3 N
(c) 12 N (d) 9 N
8. If two bodies of equal mass are moving with speeds of 3 m/s and 6 m/s, what is the ratio of their kinetic energies?
(a) 1:2 (b) 1:3 (c) 1:4 (d) 1:9
9. A motor is rated 2 kW. How much work will it do in 10 minutes?
10. Which of the following statements is/are true about work done?
(i) Work has magnitude but no direction.
(ii) If there is no displacement, no work is done.
(iii) Work done can be negative.
(iv) Work depends on the time taken to do work.
(a) (i), (ii), and (iii) only
(b) (i) and (iii) only
(c) (ii), (iii), and (iv) only (d) All of the above
Fig. 10.24
Assertion–Reason Based Questions
(4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): A 60-watt bulb consumes more energy than a 40-watt bulb when both are used for the same time.
Reason (R): Power is the rate of energy consumption.
12. Assertion (A): A moving hammer drives a nail into the wall.
Reason (R): Work done in driving the nail is positive.
13. Assertion (A): If momentum of a body is doubled, its kinetic energy increases by four-fold.
Reason (R): Kinetic energy is proportional to square of momentum.
14. Assertion (A): The total mechanical energy of a system remains constant even if energy is lost to friction.
Reason (R): In the presence of friction, mechanical energy is lost as heat.
15. Ordinarily the word work encompasses all sorts of mental and physical activities. But in scientific terminology, work is being done on an object when some force is applied on the object and the object moves through a certain distance in the direction of applied force. Usually, work is given by the product of force applied on the object and the displacement of the object in the direction of force. In a large number of cases motion of the object is at some angle to the applied force. When a gardener pulls a lawn roller by applying force on its handle, the roller moves horizontally on the lawn but the force is applied at some angle from horizontal. In such a case the work done is given by the product of displacement (s) and the projection of force along the direction of displacement (F cos θ).
Thus, Work W = F cos θ · s
(a) Annit applies a force to displace a large box from its position. He tries hard but is unable to displace the box. How much work is done by Amit and why?
(b) Radha covered a distance of 1.5 km from market to her home with a bag containing vegetables in her hand. What is the work done by Radha and why?
(c) A car is running on a road. Suddenly the driver applied brakes and the car came to a halt. What work is done by the braking force and why? What happens to the kinetic energy of the car?
16. Three balls are used in an experiment to study the effect of mass and velocity on kinetic energy.
• Ball A has a mass of 2 kg and moves with a velocity of 4 m/s.
• Ball B has the same mass (2 kg) but moves at a velocity of 8 m/s.
• Ball C has double the mass (4 kg) of Ball A but moves at the same velocity of 4 m/s.
(a) Using the work-energy theorem, derive the expression for the kinetic energy of a moving object.
(b) Without doing actual calculations, rank the three balls in order of increasing kinetic energy and justify your reasoning.
(c) Using the kinetic energy formula, demonstrate how doubling the velocity of an object affects its kinetic energy. Show mathematically why this relationship is not linear.
(d) Can two objects with different masses and velocities have the same kinetic energy? If yes, how?
Very Short Answer Questions (30-50 words)
(3 × 2 = 6 Marks)
17. A person pulls a block on a horizontal surface by applying a horizontal force of 5 N. Find the work done by this force in displacing the block through 2 m.
18. (a) State the law of conservation of energy. What happens when energy is transformed from one form to another?
(b) Under what condition is the law of conservation of energy valid? Is it a fundamental law of nature?
19. A labourer whose own mass is 50 kg carries a load of an additional 60 kg on his head to the top of a building 15 m high. Find the total work done by him. Also find the work done by him if he carries another additional block of mass 10 kg to the same height. [Take g = 10 m/s2]
Short Answer Questions (50-80 words)
(4 × 3 = 12 Marks)
20. Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it. (g = 10 ms−2)
(a) How much work is done by the men in lifting the box?
(b) How much work do they do in just holding it?
(c) Why do they get tired while holding it? (g = 10 ms−2)
21. (a) What is meant by the power of a machine? Name and define its SI unit.
(b) How is kilowatt different from kilowatt hour?
(c) From a 20 m high waterfall, nearly 25 metric tonnes of water falls per second. Calculate the equivalent power if all this energy is utilised.
22. (a) Define the work done by a constant force. Write its SI unit and define this unit.
(b) A 3000 kg truck moving at a speed of 72 km/h stops after covering some distance. The force applied by brakes is 24000 N. Compute the distance covered and work done by this force.
23. A bullet of mass 10 g is fired from a rifle with a speed of 800 ms−1. After passing through a mud wall 1 m thick, the speed of bullet drops to 200 ms−1. Calculate the work done by the mud wall on the bullet.
Long
Answer Questions
(80-120 words) (2 × 5 = 10 Marks)
24. (a) Define power and state its commercial unit.
(b) An electric geyser of 1500 W rating is used for 20 minutes daily. Calculate the energy consumed in kWh by the geyser in 30 days.
(c) A 100 W ceiling fan runs for 10 hours daily. Find the total energy consumed (in joules) by the fan in a day.
(d) If the cost of electricity is ₹6 per unit (1 unit = 1 kWh), find the total electricity cost for both the geyser and the fan in 30 days.
25. (a) What is gravitational potential energy? Derive the formula used to calculate it.
(b) Calculate the potential energy of a 3 kg object placed on a shelf 2 m high. (g = 10 m/s²)
(c) What will be the potential energy of the object if the reference level is shifted 1 meter higher? Explain your reasoning.
(d) Mention two real-life situations where potential energy gets converted into kinetic energy.
11 Sound
Sound is a form of energy that travels in waves and is generated by vibrations. It is perceived by our ears and interpreted by our brains. This makes communication possible and enriches our daily experiences. In this chapter, we explore the properties of sound waves, such as frequency, amplitude and speed. We will learn about how temperature influences the speed of sound. We will also discuss echo, reverberation, and SONAR, highlighting their practical applications in communication and technology worldwide and beyond.
Infrasonic Sounds
Sounds of frequency less than 20 Hz
Human Audibility Range
• Range of frequencies detected by human aar
• Sounds of frequency 20 Hz–20,000 Hz
Ultrasonic Sounds
Sounds of frequency greater than 20,000 Hz
Uses of Ultrasonic Sounds
• To clean parts located in places difficult to reach,
• To detect flaws in metal blocks without damaging them.
• To view parts of heart by echocardiography.
• To break small kidney stones.
• To examine foetus by ultrasonography.
Time period (T)
Minimum time required to complete one vibration SI unit: Seconds (s)
Loudness
Sound
Form of energy that gives us sensation of hearing
Propagation
• Vibrating objects push the molecules of the medium creating high density (compression) and low density (rarefaction) regions.
• Successive compression and rarefaction carry the energy from source to ears.
Medium
• Material/Substance through which sound travels
• Sound cannot travel through vacuum
Longitudinal Waves
• Particles of medium move parallel to the direction of propagation of wave
• It is a mechanical wave
Amplitude (A)
Maximum displacement of particles of the medium from their normal position. SI unit: Metres (m)
• Loudness is the perceived intensity or volume of a sound.
• It is primarily determined by the amplitude of the sound wave.
• The greater the amplitude, the louder the sound
• Loudness is commonly measured in decibels (dB).
Reflection
Reflection Laws of Reflection
(i) Directions in which sound is incident and is reflected make equal angles with the normal to the reflecting surface at the point of incidence.
(ii) All the three lie in the same plane.
Uses of Multiple Reflection
• Megaphones, horns, musical instruments like trumpets are designed to send sounds in particular direction.
• Stethoscope, a medical instrument, used for listening sounds within body.
• Ceilings of concert halls, conference halls and cinema halls are curved.
Reverberation
• Persistence of sound due to repeated reflection.
• It is reduced by covering roof and walls of auditorium with sound-absorbent materials like compressed fibre board, rough plaster or draperies.
Echo
• Repetition of sound due to reflection of original sound by a large and hard obstacle.
• To hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1 s and the minimum distance from a sound reflecting surface should be 17.2 metres (at 20°C).
Wavelength (λ)
Distance between two consecutive compressions or rarefactions. SI unit: Metres (m)
Quality/Timbre
• Distinguishes sounds of same pitch and loudness.
• Depends upon the waveform.
Pitch
Frequency (f/ )
Number of vibrations produced per second. SI unit: Hertz (Hz)
• Distinguishes between sounds of same loudness.
• Greater the frequency, greater is the pitch.
Speed (v)
Distance travelled by wave in 1 second. SI unit: Metres per second (m/s)
Factors Affecting Speed
• Property of medium through which it travels.
• Speed increases with increasing temperature.
• Speed decreases on going from solid to gaseous state.
• Speed increases with increase in humidity of air.
Chapter at a Glance
• Sound is produced by the vibrations of different objects.
• Sound travels through a material medium as longitudinal waves.
• It propagates as successive compressions and rarefactions in the medium.
• In the propagation of sound, it is the energy of the sound that travels and not the particles of the medium.
• A complete oscillation involves a change in density from a maximum value to a minimum value and back to the maximum value.
• The distance between two consecutive compressions or two consecutive rarefactions is called the wavelength (λ).
• The time taken by a wave for one complete oscillation of the density or pressure of the medium is called the time period (T).
• The number of complete oscillations per unit time is called the frequency (ν), where ν = 1 T
• The speed (v), frequency (ν), and wavelength (λ) of any wave (including sound waves) are related by the equation v = λ · ν or v = λ · f
• The speed of sound primarily depends on the nature and temperature of the transmitting medium.
• The speed of sound increases with increase in temperature and humidity.
• The law of reflection of sound states that the directions in which the sound is incident and reflected make equal angles with the normal to the reflecting surface at the point of incidence. The three lie in the same plane.
• To hear a distinct echo, the time interval between the original sound and the reflected one must be at least 0.1 s and the minimum distance between source and obstacle should be 17.2 m.
• The persistence of sound in an auditorium due to repeated reflections is called reverberation.
• Sound properties such as pitch, loudness and quality are determined by the corresponding wave properties.
• Loudness is a physiological response of the ear to the amplitude of the sound wave. It depends on energy of the sound wave.
• The amount of sound energy passing through a unit area per second is called the intensity of sound.
• The audible range of hearing for the average human being is 20 Hz to 20 kHz.
• Sound waves with frequencies below 20 Hz are called infrasonic, and those above 20 kHz are called ultrasonic.
• Ultrasound waves have significant medical and industrial applications, including imaging internal organs, detecting flaws in materials and cleaning complex objects.
Formulae
1. Wave Speed
2. Frequency and Time Period Relationship: ν = 1 T
3. Distance between source and observer = v × t
4. Distance between source and reflecting surface during echo (d) = () 2 vt ×
Fig. 11.1
NCERT Zone
Intext Questions
1. How does the sound produced by a vibrating object in a medium reach your ear?
Ans. The vibrating object creates compressions and rarefactions in the medium. These disturbances travel through the medium as sound waves, carrying energy to the ear, where they are detected.
2. Explain how sound is produced by your school bell.
Ans. When the school bell vibrates, it sets the surrounding air molecules into oscillatory motion. These oscillations create compressions and rarefactions in air, which propagate as sound waves to the listener’s ear.
3. Why are sound waves called mechanical waves?
Ans. Sound waves are called mechanical waves because they require a material medium (solid, liquid, or gas) to propagate and transfer energy through the vibration of particles in the medium.
4. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Ans. No, because the moon does not have an atmosphere or any medium for sound to travel through.
5. Which wave property determines (a) loudness, (b) pitch?
Ans. (a) Loudness is determined by the amplitude of the wave. A larger amplitude means a louder sound. (b) Pitch is determined by the frequency of the wave. A higher frequency means a higher pitch.
6. Guess which has a higher pitch: guitar or car horn?
Ans. A guitar generally has a higher pitch than a car horn because it produces sounds with higher frequencies.
7. What are wavelength, frequency, time period and amplitude of a sound wave?
Ans. • Wavelength (λ): The distance between two consecutive compressions or rarefactions.
• Frequency (ν): The number of oscillations per unit time.
• Time Period (T): The time taken for one complete oscillation.
• Amplitude (A): The maximum disturbance in the medium on either side of the mean value.
8. How are the wavelength and frequency of a sound wave related to its speed?
Ans. The speed of a sound wave (v) is the product of its wavelength (λ) and frequency (ν): v = λ ν
9. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Ans. Given,
Frequency(ν) = 220 Hz
Speed(v) = 440 m/s λ ν ===
10. A person is listening to a tone of 500 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source?
Ans. The time interval between successive compressions from the source is also called time period. The time period (T ) is the reciprocal of frequency: ν = 1 Time period () Frequency() T
Since, given frequency is 500 Hz.
Therefore,
T = ν == 1 0.002 sec 5 1 Frequency() 00 Hz
11. Distinguish between loudness and intensity of sound.
Ans. Both loudness and intensity are related to the energy of a sound wave and are often used interchangeably but they are different.
Intensity is the amount of sound energy passing through a unit area per second. It is absolute and measurable whereas loudness is a measure of the response of the ear to the sound. It is relative not absolute.
Though two sounds may have same intensity, one of them may appear to be louder than other.
12. In which of the three media—air, water, or iron—does sound travel the fastest at a particular temperature?
Ans. The speed of a sound wave is directly proportional to the density of the medium. Therefore, sound travels fastest in iron, followed by water and then air.
13. An echo is heard in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Ans. Given,
Speed of sound (v) = 342 m/s
Travel time of sound (t) = 3 s.
Let the distance between source and the reflecting surface be d metres. Since, sound travels from source to reflecting surface and back, it actually travels twice the distance between source and reflecting surface. i.e. distance travelled by sound = 2 × distance between source and reflecting surface = 2d total distance travelled by the sound (2d) = Speed of sound × Travel time of sound 2d = v × t = 342 × 3 m = 1026 m.
Therefore, the distance to the reflecting surface d == 1026 m513 m. 2
14. Why are the ceilings of the concert halls curved?
Ans. The ceilings of concert halls are curved to spread sound evenly across the hall. The curved shape helps sound waves bounce in all directions, so everyone in the hall can hear clearly. This makes the sound balanced and improves the listening experience
15. What is the audible range of the average human ear?
Ans. The audible range for humans is 20 Hz to 20,000 Hz.
16. What is the range of frequencies associated with (a) infrasound, (b) ultrasound?
Ans. (a) Infrasound: Below 20 Hz.
(b) Ultrasound: Above 20,000 Hz.
NCERT Exercises
1. What is sound and how is it produced?
Ans. Sound is a form of energy that produces the sensation of hearing created by the vibration of objects. These vibrations create disturbances in the surrounding medium, which propagate as sound waves.
2. Describe, with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.
Ans. Compressions are regions of high pressure formed when particles are forced closer together, while rarefactions are regions of low pressure formed when those particles spread farther apart. These
alternating high and low pressure areas arise as a vibrating object moves forward (creating a compression) and then backward (creating a rarefaction).
3. Why is a sound wave called a longitudinal wave?
Ans. A sound wave is called longitudinal because the particles of the medium vibrate parallel to the direction of wave propagation.
4. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?
Ans. The timbre or quality of the sound helps distinguish one person’s voice from another.
5. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?
Ans. Light travels much faster than sound does (~9,00,000 times faster), so the flash of thunder is seen much before its sound is heard.
6. A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m/s.
Ans. Given,
Speed of sound in air = 344 m/s
Frequency of waves = 20 Hz and 20 kHz
we know, wavelength (λ) ν = Speed() Frequency() v Therefore,
Wavelength at 20 Hz: λ ν === 344 m/s 17.2 m 20 v
Wavelength at 20 kHz: λ ν == 344 m/s 20,000 Hz v = 0.0172 m.
7. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.
Ans. Speed of sound in air: 346 m/s.
Speed of sound in aluminium: 6,420 m/s. Now, distance travelled = speed (v) × time (t) Since sound travels same distance, length of the rod(d), in both cases, therefore d = tal · val = tair · vair
Ratio: airal alair 6420 346 tv tv ==≈ 18.5.
8. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Ans. Vibrations in one minute = frequency (in Hz) · Time (in sec) = 100 × 60 = 6000 times
9. Does sound follow the same laws of reflection as light does? Explain.
Ans. Yes, sound follows the same laws of reflection
(i) The angle of incidence equals the angle of reflection
(ii) The incident wave, reflected wave, and normal all lie in the same plane.
10. When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remain the same. Do you hear echo sound on a hotter day?
Ans. Yes. Since the speed of sound increases with temperature, the echo will be heard sooner but still it should satisfy the conditions for distinct hearing.
11. Give two practical applications of reflection of sound waves.
Ans. Two practical applications of reflection of sound waves are
• Megaphones and loudhailers: These devices use reflection of sound to direct sound waves in a specific direction, increasing their reach.
• Stethoscopes: They use multiple reflections of sound to amplify the heartbeat or respiratory sounds for diagnosis.
12. A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? (Given, g = 10 m/s2, speed of sound (v) = 340 m/s).
Ans. When stone is dropped from the tower, it will take some time to fall into the lake. When it hits the lake, Sound is produced which also will take some time to travel to the top of the tower.
Let t1 be the time the stone takes to fall into the lake and t2 be the time the sound takes to travel to the top of the tower. The total time between dropping of stone and sound reaching the top of the tower be
t = t1 + t2
Given,
Acceleration due to gravity (g) = 10 m/s2
speed of sound (v) = 340m/s
Height of the tower = 500 m
Using the second equation of motion, h = ut + 1 2 gt2
we get, t = 2h g
Therefore, time taken for the stone to fall: t1 = 10010 sec. 10 22500 h g === ×
And, the distance travelled by sound = velocity of sound × time
Therefore,
Time taken for the sound to travel back: t2 = distance5 1.47 sec 340 00 speed ==
Total time: t = t1 + t2 = 10 + 1.47 = 11.47 s.
Therefore, the splash will be heard 11.47 sec after dropping the stone.
13. A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Ans. Given,
Speed of sound (v) = 339 m/s
Wavelength (λ) = 1.5 cm or 0.015 m
Since, Speed of wave (v) = wavelength (λ) × frequency (ν )
Path of Sound
Path of Stone 500 m Water
Fig. 11.4 Stone
Since 22,600 Hz is above the audible range of 20 Hz to 20,000 Hz, it will not be audible.
14. What is reverberation? How can it be reduced?
Ans. Reverberation is the persistence of sound in a closed space due to repeated reflections from walls and other surfaces.
It can be reduced by
• using sound-absorbing materials such as carpets, curtains, and foam panels.
• designing walls and ceilings with uneven or porous surfaces.
15. What is loudness of sound? What factors does it depend on?
Ans. Loudness is the physiological response of the ear to the amplitude of a sound wave. It depends on
• the amplitude of the wave (greater amplitude means louder sound).
• the sensitivity of the listener’s ear.
• the distance from the sound source.
16. How is ultrasound used for cleaning?
Ans. Ultrasound is used to clean hard-to-reach places such as spiral tubes and intricate parts of machines. High-frequency ultrasonic waves are passed through a cleaning solution, loosening dirt, grease and other particles, which are then detached and washed away.
17. Explain how defects in a metal block can be detected using ultrasound.
Ans. Ultrasound waves are passed through the metal block. If there are defects like cracks or voids, the waves are reflected back from these locations. Detectors pick up the reflected waves, allowing identification of the defect’s location and size.
Multiple Choice Questions
1. Which of the following quantities determines the loudness of a sound wave? (a) Frequency (b) Velocity (c) Amplitude (d) Wavelength
2. The frequency of a wave is measured in (a) metres (m). (b) hertz (Hz). (c) seconds (s). (d) decibels (dB).
3. Which characteristic of a sound wave is related to its pitch? (a) Frequency (b) Velocity (c) Amplitude (d) Time period
4. The minimum distance required between the source and the reflecting surface to hear an echo in air (at around 25°C) is approximately (a) 8.6 m. (b) 9.8 m. (c) 17.2 m. (d) 34.4 m.
5. Which of the following ranges represents the audible frequency range for a human being? (a) 2 Hz – 20 Hz (b) 20 Hz – 20,000 Hz (c) 2 kHz – 20 kHz (d) 200 Hz – 20,000 Hz
6. In a longitudinal wave, the regions of lower density are called (a) crests. (b) troughs. (c) rarefactions. (d) compressions.
7. The time period (T) of a sound wave is related to its frequency ( f ) by: (a) T = f × 2 (b) T = 1 f (c) T = f 2 (d) T = 2 1 f
8. Which of the following statements is correct about the speed of sound?
(a) Speed of sound is highest in air and lowest in solids. (b) Speed of sound is highest in solids and lowest in gases.
(c) Speed of sound is the same in all media.
(d) Speed of sound in liquids is always lower than in gases.
9. The velocity of sound in air at 20°C is approximately (a) 220 m/s. (b) 280 m/s. (c) 340 m/s. (d) 400 m/s.
10. A wave has a frequency of 50 Hz and travels with a speed of 100 m/s in a medium. What is its wavelength?
(a) 0.5 m (b) 2 m
(c) 5 m (d) 50 m
11. The phenomenon in which sound gets reflected from a surface and is heard again is known as (a) refraction. (b) reflection. (c) an echo. (d) reverberation.
12. When sound waves travel from air to water, which of the following remains unchanged? (a) Frequency (b) Velocity (c) Wavelength (d) All of these
13. The quality of sound that helps us distinguish between two sounds of the same pitch and loudness is (a) pitch. (b) timbre. (c) intensity. (d) velocity.
14. If the amplitude of a sound wave is increased, its loudness will (a) remain the same. (b) increase. (c) decrease. (d) fluctuate.
15. The sound intensity level is measured in (a) Hertz (Hz). (b) Decibels (dB). (c) Joules (J). (d) Metres per second (m/s).
16. An object is vibrating with a time period of 0.01 s. The correct statement is:
(a) The frequency is 10 Hz, and it is within the audible range.
(b) The frequency is 100 Hz, and it is within the audible range.
(c) The frequency is 100 Hz, and it is beyond the audible range.
(d) The frequency is 1000 Hz, and it is within the audible range.
17. A person hears thunder 3 seconds after seeing the lightning. If the speed of sound is taken as 340 m/s, approximately how far away did the lightning strike?
(a) 113 m (b) 340 m (c) 1020 m (d) 2040 m
18. Why do we generally not hear the sound produced by the vibration of a tuning fork under water, even if it is vibrating?
(a) Because the frequency of the fork is below the human audible range
(b) Because water dampens the vibrations quickly
(c) Because water completely reflects the sound waves
(d) Because sound waves do not travel in liquids
19. Two sound waves, A and B, have the same amplitude but different frequencies. Which statement is true?
(a) They will have the same pitch but different loudness.
(b) They will have the same loudness but different pitch.
(c) They will differ in both loudness and pitch.
(d) They will differ neither in loudness nor in pitch.
20. A stationary bat emits a high-frequency sound and hears its echo 0.2 s later. Calculate the distance between the bat and the obstacle if the speed of sound is 340 m/s.
(a) 17 m (b) 34 m (c) 68 m (d) 340 m
21. A device generates ultrasonic waves at 50,000 Hz. If these waves travel in air at 340 m/s, what is the approximate wavelength?
(a) 6.8 mm (b) 6.8 cm (c) 6.8 m (d) 68 m
22. A displacement–time graph for a sound wave shows that 2 complete cycles (vibrations) occur in 4 milliseconds (ms). Which of the following statements is correct?
(a) The wave’s time period is 2 ms, so the frequency is 500 Hz.
(b) The wave’s time period is 4 ms, so the frequency is 250 Hz.
(c) The amplitude of the wave is 2 cm.
(d) The wave’s time period is 1 ms, so the frequency is 1000 Hz.
23. A displacement–distance graph for a sound wave shows that the distance between two consecutive compressions is 1.2 m. If the wave travels through air at a speed of 360 m/s, what is the frequency of the wave?
(a) 300 Hz (b) 400 Hz (c) 600 Hz (d) 720 Hz
24. A music concert in a large auditorium uses thick curtains and cushioned seats. The main reason for having these materials is
(a) to improve the decorative look of the auditorium.
(b) to provide maximum comfort for seating.
(c) to reduce reverberations (excessive reflections of sound).
(d) to enhance the reflections of sound.
25. A ship sends out an ultrasonic pulse (sonar) to determine the depth of the ocean floor. The pulse returns to the ship after 2 seconds. If the speed of sound in water is 1500 m/s, how deep is the sea floor?
1. What is an echo? What is the minimum distance required to hear a distinct echo.
Ans. An echo is created when sound waves reflects off a surface and travels back to the listener. The delay between the original sound and the reflection makes the echo distinct. For humans to hear a distinct echo, minimum distance between the source and the obstacle should be 17.2 m.
2. Compare the speed of sound in iron, air, and vacuum?
Ans. Sound travels at about 5100 m/s in iron and 344 m/s in air, as the strong particle bonding in solids allows faster wave transmission. In a vacuum, sound cannot travel because there are no particles to transmit the vibrations needed for sound propagation.
3. Why does Anand hear the sound of the train before Akash?
Ans. Sound moves faster in solids like steel than in air because particles in solids are closely packed. Therefore, sound travelling through railway track reaches the listener’s ears faster than the sound travelling through air.
Fig. 11.5
4. What are mechanical waves, and why are sound waves called mechanical waves?
Ans. Mechanical waves are waves that require a medium (solid, liquid, or gas) to travel, as they propagate by the vibration of particles. Sound waves are called mechanical waves because they rely on the movement of air (or other particles) to carry energy from one point to another.
5. (a) Why are the roofs and walls of auditoriums covered with sound-absorbent materials?
(b) Why do bats use ultrasonic waves while flying in the dark?
Ans. (a) The walls and roofs of auditoriums are covered with sound-absorbent materials to reduce reverberation.
(b) Bats use ultrasonic waves in the dark because high-frequency signals bounce off objects, enabling precise echolocation for safe nighttime navigation.
6. The given graph (Fig. 11.6) shows the displacement versus time relation for a disturbance travelling with a velocity of 1500 m/s. Calculate the wavelength of the disturbance.
Ans. From the graph, we know that two wavelengths cross the point in duration of 1 – 5 μs. Therefore, the ime period,
Now, wavelength λ ν = 1 and ; v T
1500 ×
×
Displacement
(m) Time (in ms) 0 1 2 3 4 5 Time
Fig. 11.6
7. A sound wave of frequency 500 Hz travels with a speed of 340 m/s. Calculate its wavelength and time period.
Ans. Frequency (ν ) = 500 Hz
Speed of sound (v) = 30 m/s
Wavelength λ
8. Why is the sound of lightning heard after the flash is seen?
Ans. Light travels at an incredible speed of 300,000,000 m/s, which is far higher than the speed of sound, which moves at just 330 m/s. As a result, lightning is seen almost instantly, while the slower-traveling thunder reaches the observer a few moments later.
9. What is ultrasonography, and how is it applied for medical use?
Ans. Ultrasonography involves directing ultrasonic waves into the body. These waves reflect from tissues of varying densities, and the reflected signals are used to create detailed images of internal organs on a screen.
10. Why do we hear the sound produced by the humming bees while the sound of vibrations of a pendulum is not heard? (NCERT Exemplar)
Ans. The sound produced by humming bees has a frequency within the audible range of human hearing (20 Hz to 20,000 Hz), typically in the range of hundreds of hertz. In contrast, the vibrations of a pendulum occur at a much lower frequency, often below the audible range, and thus cannot be heard.
11. How does humidity affect the speed of sound in air?
Ans. The average molecular weight of air is greater than water vapour. In humid conditions, the percentage of water vapour in air increases, which reduces air’s overall density. This allows sound waves to travel more efficiently, increasing their speed compared to dry air, where particle density is higher.
12. What are infrasound waves, and how are they useful?
Ans. Infrasound waves have frequencies below 20 Hz, making them inaudible to humans. They are useful for monitoring natural events such as earthquakes, volcanic eruptions, and for studying animal communication over long distances.
13. A thunderstorm is heard 8 seconds after the lightning flash. Calculate its distance if sound travels at 340 m/s.
Ans. Assuming that lightning flash reaches the observer instantly, the time delay can be considered only for sound as the time sound takes to travel from thunderstorm to observer. The distance can be calculated using
Distance (d) = Speed (v) × Time (t)
Given,
Speed of sound (v) = 340 m/s
Time lag (t) = 8 sec
d = v × t = 340 × 8 = 2720 m or 2.72 km.
14. How does a megaphone amplify sound?
Ans. A megaphone’s conical shape directs sound waves and prevents them from dispersing. Through multiple internal reflections, the waves are focused into a single direction, amplifying the sound for the listener.
15. For hearing the loudest ticking sound heard by the ear, find the angle x in Fig. 11.7. (NCERT Exemplar)
Ans. The angle x corresponds to the angle which the sound makes with the normal. According to the second law of reflection of sound, we know that angle between incident ray and normal (angle of incidence) is equal to the angle between normal and reflected ray(angle of reflection).
Angle of incidence = 90° - 50° = 40°
Angle of reflection = x
Therefore, x = 40°
Short Answer Questions (50-80 words)
1. What are the characteristics of a longitudinal wave?
11.7
Ans. Longitudinal waves have particles vibrating parallel to the wave’s direction of propagation. They consist of compressions (high-pressure areas) and rarefactions (low-pressure areas). Examples include sound waves in air and waves traveling through liquids or solids. These waves require a medium for propagation, as they cannot travel through a vacuum.
2. How does sound differ in an empty room and a furnished room?
Ans. In an empty room, hard surfaces reflect sound, causing echoes and reverberations. In a furnished room, soft materials like carpets, curtains, and furniture absorb sound, reducing reflections and making the sound less harsh. Furnishings also improve sound clarity and reduce noise by damping vibrations.
3. What are ultrasonic waves, and how are they used in industries?
Ans. Ultrasonic waves have frequencies above 20 kHz, which is beyond human hearing. In industries, they are used for detecting cracks in metal, precision cleaning, and non-destructive testing. The highfrequency vibrations also help in welding thermoplastics, ensuring durability. Their ability to penetrate materials and reflect off imperfections makes them ideal for quality control and maintenance.
Fig.
Ear
50° Normal Clock Wall
Table
4. Explain the difference between longitudinal and transverse wave.
Ans. Longitudinal Waves Transverse Waves
Particle Motion particles vibrate parallel to the wave’s direction of propagation particles vibrate perpendicular to the wave’s direction
Wave type form compressions and rarefactions form crests and troughs
Medium require a medium (solid, liquid, gas) Can travel in a medium or vacuum example Sound waves Ripples in a pond
5. Why can we hear sound through water but not through space?
Ans. Sound requires a medium, such as water or air, to propagate because it travels via vibrations between particles. Water has closely packed molecules, making it an efficient medium for sound. Space, being a vacuum, lacks particles to transmit vibrations, so sound cannot propagate through it. This explains why astronauts cannot hear each other without communication devices.
6. Why does a loudspeaker’s diaphragm move back and forth when producing sound?
Ans. The diaphragm of a loudspeaker vibrates due to an alternating current passing through its coil. These vibrations create compressions and rarefactions in the air, producing sound waves. The amplitude of the vibrations determines the sound’s loudness, while the frequency controls its pitch. This mechanical motion converts electrical signals into audible sound.
7. What is the principle behind breaking kidney stones with ultrasound?
Ans. Ultrasound waves target kidney stones with high-energy vibrations. These waves generate pressure that fractures the stones into smaller pieces, which can pass through the urinary tract naturally. This noninvasive method is known as lithotripsy. It eliminates the need for surgery, offering a safer and more convenient treatment option.
8. In a ripple tank, 10 full ripples are produced in one second. If the distance between a crest and next trough is 20 cm, find
(a) wavelength, (b) frequency, and (c) velocity of the wave.
Ans. (a) Wavelength = twice the distance between consecutive crest and trough
λ = 2 × 20 = 40 cm or 0.4 m
(b) Frequency = number of ripples produced in one second = 10 Hz
(c) Velocity of wave = wavelength frequency
v = λ · ν = 0.4 × 10 = 4 m/s
9. A boy shouts in a small valley. The sound wave hits a rock face 25 m away and returns. If the speed of sound in air is 340 m/s, will he hear a distinct echo? If yes, how long does the boy wait to hear the echo? (based on the 0.1 s rule)?
Since, reflected sound takes 0.147 s to reach the boy, which is greater than 0.1s, the echo should be just about distinct.
10. Plot the following:
(a) A longitudinal wave in air on a density-distance graph.
(b) A transverse wave on a displacement-distance graph.
25 m
Fig. 11.8
Ans. (a) A longitudinal wave in air
Density or Pressure variations
Amplitude
C = Compression: Points of increased density or pressure
(b) A transverse wave
Average density or pressure
Direction of propagation R = Rarefaction: Points of decreased density or pressure
11. Waves of frequency 200 Hz are produced in a string as shown in the Fig. 11.10. Give its
(a) amplitude
(b) wavelength
(c) velocity
(d) nature
Ans. (a) Amplitude = 8 cm
(b) Wavelength = distance between two consecutive crests = 20 cm or 0.2 m
(c) Velocity (v) = λ × ν = 0.2 × 200 = 40 m/s
(d) Since the wave is travelling on a string, it must be a transverse wave.
12. A man produces a sound wave of frequency of 2 kHz and wavelength of 50 cm in front of a cliff. If he hears the echo after 7 sec. Calculate the distance between cliff and man.
Ans. Given,
Wavelength (λ) = 50 cm or 0.5 m
Frequency (ν ) = 2 kHz or 2000 Hz
Velocity (v) = λ × ν = (0.5) × 2000 = 1000 m/s
Distance between man and cliff ×× === 10007 350 m 22 vt
13. What is the audible range of the human ear? Define ultrasound and infrasound, and list two uses of ultrasound. (NCERT Exemplar)
Ans. Humans can typically hear sounds in the frequency range of 20 Hz to 20,000 Hz. Ultrasound refers to sound waves with frequencies above 20,000 Hz, while infrasound refers to frequencies below 20 Hz. Ultrasound is commonly used in medical imaging (ultrasonography) to visualize internal organs and in industrial cleaning to remove dirt from delicate objects.
14. Compare the two sound waves shown in the graph Fig. 11.11 based on their characteristics.
(a) Explain how the amplitude and loudness differ between the waves.
(b) Explain the difference in frequency and pitch of the waves.
Fig. 11.9
Fig. 11.10
(c) If both waves are travelling the same medium, what will the ratio of their wavelengths? (Take Vsound = 330 m/s)
Ans. (a) Amplitude of wave 2 is twice the amplitude of wave 1. Hence wave 2 is louder than wave 1.
(b) Since both waves have same time period, therefore they must have same frequency and hence same pitch.
(c) Since both waves are travelling in same medium they will have same speed and given that they have same frequency, they must have same wavelength. Hence the ratio of their wave lengths is 1.
Long Answer Question (80–120 words)
1. (a) How does a sound wave propagate?
(b) What is the audible range of sound for human beings?
(c) It is observed that some animals get disturbed before earthquakes. How?
Ans. (a) A sound wave propagates as a longitudinal wave in the form of compressions (regions of high pressure) and rarefactions (regions of low pressure). When an object vibrates, it disturbs the particles in the surrounding medium. These particles oscillate back and forth, transferring the sound energy from one particle to another until it reaches the listener.
(b) Humans can hear sounds in the frequency range of 20 Hz to 20,000 Hz. This is referred to as the audible range. Frequencies below 20 Hz are called infrasound, while those above 20,000 Hz are termed ultrasound.
(c) Animals such as elephants and dogs can detect infrasound waves generated before an earthquake. These low-frequency waves, which are beyond the human hearing range, create sensations in animals, causing them to react restlessly as an early warning.
2. (a) What is sound?
(b) What does a wave transfer—matter or energy?
(c) How does it transfer that from source to observer?
Ans. (a) Sound is a form of energy produced by vibrations, travelling as a mechanical wave through materials like air, water, or solids. These waves are characterised by alternating regions of compression and rarefaction in the medium.
(b) A wave transfers energy, not matter. While particles in the medium oscillate back and forth around their equilibrium positions, they do not move permanently with the wave.
(c) When the source vibrates, it pushes and pulls on adjacent particles, creating a disturbance that propagates outward. The particles themselves simply move back and forth, passing the energy along from one particle to the next, until the energy reaches the observer’s ear (or detecting device).
3. Fig. 11.12 shows a loudspeaker cone oscillating to produce sound waves.
(a) As the sound wave passes a point, it produces regions of higher and lower pressure. State the names of these regions.
(b) Describe how the movement of the loudspeaker cone produces these regions of higher and lower pressure.
Fig. 11.11
Fig. 11.12
(c) State the effect on the loudness and pitch of the sound from the loudspeaker when:
(i) The amplitude increases but the frequency of the sound stays the same.
(ii) The amplitude stays the same but the frequency increases.
Ans. (a) Regions of higher pressure are called compressions, and regions of lower pressure are called rarefactions.
(b) The loudspeaker cone vibrates to produce sound. When it moves forward, it compresses air particles, creating a compression. When it moves backward, it reduces air pressure, forming a rarefaction. These alternating compressions and rarefactions propagate as sound waves.
(c) Effect of changes in amplitude and frequency:
(i) If amplitude increases but the frequency stays the same: The sound becomes louder, but its pitch (frequency) remains unchanged.
(ii) If amplitude stays the same but frequency increases: The pitch of the sound becomes higher, but its loudness (amplitude) remains the same.
4. What is meant by reflection of sound? Also, describe an activity to study the reflection of sound.
Ans. (a) Reflection is the phenomenon in which a wave, after striking a surface, reverses its direction and continues to travel in the same medium. Sound waves, like other waves, also undergo reflection.
(b) To observe the reflection of sound, take a flat drawing board and place it on the floor. Position two metallic or cardboard tubes at an angle to each other, as shown in the Fig. 11.13. Place a clock near the opening of one tube and a screen between the tubes to block direct sound. The ticking sound waves pass through the first tube, reflect off the board, and enter the second tube. In the second tube, they can be heard by an ear positioned near its opening.
5. The units of certain parameters of a mechanical waves are given:
Name the corresponding parameter of a mechanical wave and define each one:
(a) (i) metre (ii) metre per second
(iii) hertz (iv) second
(b) Two sounds A and B are of different pitch. B appears to be heavier as compared to A. What can be said about their comparative frequencies.
Ans. (a) (i) Metre represents wavelength, the distance between two consecutive compressions or rarefactions in a wave.
(ii) Metre per second represents wave speed, the distance travelled by a wave in one second. (iii) Hertz represents frequency, the number of complete wave cycles passing a point per second. (iv) Second represents time period, the time taken for one complete wave cycle.
(b) Sound B has a higher frequency than sound A, as pitch is directly proportional to frequency. A higher frequency produces a higher pitch, which is perceived as heavier or sharper compared to lower-pitched sounds.
6. Two waves are shown in Fig. 11.14:
Fig. 11.13
Fig. 11.14
(a) Which of the two waves corresponds to a high-pitched sound?
(b) Name the characteristics of sound the graphs represent.
(c) Define the characteristic.
(d) Explain the reason why sound of school bell is heard over long distance while that of a blow by hand on a wooden desk remains limited to the room.
Ans. (a) Wave (i) corresponds to a high-pitched sound because it has a higher frequency compared to wave (ii) High frequency directly correlates with high pitch.
(b) The graphs represent two characteristics of sound: frequency and amplitude. Frequency determines the pitch of the sound, while amplitude determines its loudness.
(c) (i) Frequency: The number of oscillations or cycles per second, measured in hertz (Hz), determines the pitch of the sound. Higher frequency produces a higher pitch.
(ii) Amplitude: The maximum displacement of particles from their equilibrium position determines the loudness of the sound. Greater amplitude results in louder sounds.
(d) The school bell produces louder sound waves with higher energy and uniform vibrations, allowing them to travel further. A hand blow on a desk generates sound with lower energy, limiting its travel distance to the immediate surroundings.
7. Establish the relationship between speed of sound, its wavelength, and frequency. If velocity of sound in air is 340 m s–1, calculate (NCERT Exemplar)
(a) wavelength when frequency is 256 Hz.
(b) frequency when wavelength is 0.85 m.
Ans. The relationship between speed of sound (v), wavelength (λ), and frequency (ν) is given by:
Speed of sound (v) = wavelength (λ) × frequency (ν )
Using this relation, we get
(a) Wavelength when frequency is 256 Hz:
Given:
v = 340 m/s and ν = 256 Hz.
Using the formula: λ ν == 340 256 v ≈ 1.33 m.
Therefore, the wavelength is approximately 1.33 m.
(b) Frequency when wavelength is 0.85 m:
Given:
v = 340 m/s and λ = 0.85 m.
Using the formula: ν λ === 340 400 Hz. 0.85 v
Therefore, the frequency is approximately 400 Hz.
Competency Based Questions
Multiple Choice Questions
1. A key of a mechanical piano is struck gently and then struck again but much harder the second time. In the second case, (NCERT Exemplar)
(a) the sound will be louder but the pitch will not be different.
(b) the sound will be louder and the pitch will also be higher.
(c) the sound will be louder but the pitch will be lower.
(d) both loudness and pitch will remain unaffected.
2. In the curve (Fig. 11.15), half the wavelength is (NCERT Exemplar)
(a) AB
(b) BD
(c) DE
(d) AE
3. For waves 1 and 2, which of the following are correct?
(a) wave 1 is louder and has higher pitch.
(b) wave 2 is louder and has a higher pitch.
(c) wave 1 is louder but wave 2 has a higher pitch.
(d) wave 2 is louder but wave 1 has a higher pitch.
4. A tuning fork and a metal ball are arranged as shown in Fig. 11.17. The ball is now released from each position and observations are recorded. Which of the following observation is true?
(i) Amplitude is largest for position A.
(ii) Amplitude is largest for position C.
(iii) Frequency of vibration is highest for position A.
(iv) Frequency of vibration is highest for position C
(v) Frequency of vibrations is same for all three positions.
(a) only (i) and (iii) (b) only (ii) and (iv)
(c) only (i) and (v)
(d) only (ii) and (v)
5. Aman is standing 17 meters in front of a wall on a cold morning of January 26th and shouts, 'Jai Hind.' He is just able to hear a distinct echo after 0.1 seconds. If he repeats the same experiment on a hot afternoon of August 15th, he will:
(a) hear an echo after exactly 0.1 sec.
(b) hear an echo but it reaches him quicker.
(c) hear an echo but it takes a little longer to reach him.
(d) not hear an echo.
6. Intensity of a sound wave is defined as the total amount of energy in a sound wave. Which characteristics of sound does the intensity depends on?
(a) Speed of sound
(b) Wavelength
(c) Frequency (d) Amplitude
Fig. 11.16
Fig. 11.15
Fig. 11.17
7. Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting (NCERT Exemplar)
(a) intensity of sound only
(b) amplitude of sound only
(c) frequency of the sitar string with the frequency of other musical instruments
(d) loudness of sound
8. Johnny is standing at the centre of Eden Gardens Cricket Stadium in Kolkata. He claps his hands loudly and hears an echo after 0.5 seconds, which is due to the reflection of sound waves from the audience stands. What is the distance of the stands from the centre of the stadium? (Speed of sound = 340 m/s)
(a) 340 m
(b) 170 m
(c) 85 m
(d) 42.5 m
9. Amaan uses a sound wave generator to generate waves of wavelength 16 mm. If these waves travel in air at 340 m/s, then the sound will be .
(a) Infrasonic
(c) within audibility limits
(b) ultrasonic
(d) none of these
10. An engineer is designing a concert hall and needs to ensure that the sound from the stage is heard clearly at the back of the hall without electronic amplification. If the hall is designed to be a rectangular room with parallel walls, which of the following features should the engineer incorporate to reduce the issue of echo and enhance the acoustics?
(a) Sound absorbing panels on the side walls
(b) A flat ceiling with highly reflective surface materials
(c) Parallel, flat walls throughout with no modifications
(d) Hard, reflective surfaces near the stage area
Assertion–Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
1. Assertion (A): The speed of sound is higher in solids than in gases.
Reason (R): The particles are more closely packed in solids, which allows sound to travel faster.
2. Assertion (A): Humans can typically hear frequencies from 20 Hz to 20,000 Hz.
Reason (R): Sound waves below 20 Hz are called infrasound, and those above 20 kHz are called ultrasound.
3. Assertion (A): The amplitude of a sound wave determines its pitch.
Reason (R): The pitch of a sound wave depends upon the wave’s frequency.
4. Assertion (A): The time period (T) of a wave is the reciprocal of its frequency (ν ).
Reason (R): v = λ × ν
5. Assertion (A): An echo of sound can be distinctly heard only if the reflecting surface is at least 17.2 metres away from the listener (in normal temperature conditions).
Reason (R): To hear a distinct echo, the time interval between the original sound and the reflected sound must be at least 0.1 second.
6. Assertion (A): An auditorium is lined with soft materials such as thick curtains and carpets to reduce reverberations.
Reason (R): Soft and porous materials reflects more sound energy rather than absorbing it.
7. Assertion (A): You cannot hear a bomb explode on the moon.
Reason (R): Sound requires a medium (with particles) for its propagation because it is a mechanical wave.
Case-Based/Source-Based/Passage-Based Questions
1. Your teacher has asked you to perform a sound experiment inside an empty classroom to understand reverberation and echoes. You notice that when you clap once, the sound persists slightly before dying out. The teacher then places thick curtains on the walls and asks you to clap again to observe changes.
Activity Observations:
1. Before curtains: The sound of the clap lingers, making the room feel ‘echoey’.
2. After curtains: The sound dies out more quickly; the room sounds less ‘echoey’.
Questions:
(a) Explain why the sound persists longer in an empty classroom (without curtains).
(b) How do thick curtains or soft materials influence the reflections of sound?
(c) Briefly outline an activity or method to measure the time difference between your initial clap and its echo in a large hall.
(d) If the speed of sound in air is about 340 m/s, how far away must a reflecting surface (e.g., wall) be for you to hear a distinct echo?
2. A student records a sound wave on an oscilloscope, obtaining a displacement–time graph (Fig. 11.18). The graph below shows 2 complete cycles of the wave in 4 milliseconds (ms). The peak of the wave measures at +2 units and the lowest trough at −2 units.
Questions:
(a) How many complete waves (oscillations) are shown in the 4 ms interval?
(b) Determine the time period (T) of one complete oscillation.
(c) Calculate the frequency (ν ) of the wave in Hz.
(d) If the maximum positive displacement is +2 units (and negative is −2), which characteristic of the sound wave does this measure represent, and how does it affect loudness?
3. In an experiment, a displacement–distance graph (Fig. 11.19) is plotted for a sound wave traveling in air. The experimenter marks two consecutive compressions on the graph, which are 1.5 m apart.
Questions:
(a) In a longitudinal sound wave, what do these ‘peaks’ on the displacement–distance graph represent physically?
(b) If this 1.5 m represents the wavelength λ, and the wave is traveling at v = 330 m/s, find the frequency.
(c) Is the wave’s frequency audible to humans? Justify.
(d) If the temperature of air increases, how would this affect the speed of sound and the wavelength, assuming the frequency remains constant?
4. A ship uses sonar to determine the depth of the ocean. It emits an ultrasonic pulse that returns after 3 seconds. Assume the speed of sound in water is about 1500 m/s. The captain also wants to check if the depth is within safe limits to drop anchor.
Questions:
(a) What is the total distance travelled by the sound wave in these 3 seconds?
(b) Calculate the depth of the ocean at that spot.
(c) The captain is comfortable anchoring if the depth is less than 3000 m. Should the captain anchor here?
(d) Why do high-frequency (ultrasonic) waves work better for sonar than lower-frequency waves?
Answers
Multiple Choice Questions
1.
Assertion–Reason Based Questions
1.
Case-Based/Source-Based/Passage-Based Questions
1. (a) In an empty classroom, there are many hard surfaces (bare walls, floor, ceiling) that reflect the sound of the clap, causing the sound waves to bounce around. This leads to multiple reflections, which persist for a short while, creating reverberation (a prolonged echo-like effect).
(b) Thick curtains (or any soft, porous material) absorb a significant portion of the sound energy rather than reflecting it. As a result, there are fewer reflections, thus reducing the ‘echoey’ effect and shortening the time the sound can be heard.
(c) Activity to measure echo time:
(i) Stand at a known distance from a large flat wall (e.g., 10 m, 15 m, etc.).
(ii) Clap once and record with a timer or app when you hear the echo.
(iii) The difference between the clap and the echoed sound gives you the time for sound to travel to the wall and back.
(iv) From this time interval, you can calculate the distance if speed of sound is known, or vice versa.
(d) The human ear can distinguish two separate sounds if they are at least 0.1 second apart.
In 0.1 s, distance travelled by sound wave = 340 × 0.1 = 34 m (round trip).
Hence, one-way distance: 34 17 m. 2 =
Therefore, the wall/reflecting surface must be at least 17 m away for you to hear a distinct echo.
2. (a) The graph clearly shows 2 complete waves in the 4 ms interval.
(b) Time period = Total time () Number of waves T ==× 3 4 ms 2 ms or 210 s. 2 T
(c) Frequency ν=== × 3 11 () 500 Hz. (210s) T
(d) This maximum displacement is the amplitude of the wave.
Amplitude is directly related to the loudness: larger amplitude → louder sound.
3. (a) These ‘peaks’ represent compressions—regions of high pressure where particles of the medium are closely packed.
(b) λ = 1.5 m, v = 330 m/s
ν λ === 330 220 Hz. 1.5 v
(c) Human audible range is roughly 20 Hz to 20 kHz. 220 Hz is within that range, so it is audible.
(d) If temperature increases, the speed of sound v increases. Frequency (ν) is constant (if the source doesn’t change), so by v = f × λ, the wavelength λ must increase as well.
4. (a) Total time = 3 s, speed v = 1500 m/s.
Stotal = v × t = 1500 × 3 = 4500 m.
(b) Depth is one-way distance:
=== total 4500 Depth 2250 m. 22 S
(c) 2250 m is less than 3000 m, so it is safe to anchor.
(d) High-frequency (ultrasonic) waves diverge less, travel well in water, and offer better resolution for sonar. Lower-frequency waves spread out more and do not give as precise data for locating objects or measuring depth.
Numerical Questions
1. A sound wave traveling in air has a frequency of 200 Hz. If the speed of sound in air is taken as 340 m/s, find the wavelength (λ) of this sound.
Ans. Given
Frequency (ν) = 200Hz
Velocity of sound (v) = 340m/s.
Wavelength λ ν === 340 m/s () 1.7 m. 200 Hz v
2. A source produces sound waves at a frequency of 500 Hz. Calculate the time period T of these waves in milliseconds.
Ans. Given,
Frequency (ν ) = 500 Hz.
ν === 11 0.002 sec or 2 ms. 500 Hz T
3 At around 20oC, the speed of sound in air is about 340 m/s. What is the minimum distance from a reflecting wall required to hear a distinct echo? Assume the ear needs at least 0.1 s to distinguish between the original sound and the echo.
Ans. Given,
Speed of sound in air (v) = 340 m/s
For a distinct echo, time gap (t) should be greater than 0.1 s.
Total (round-trip) distance sound will travel = dround = v × t = 340 × 0.1 = 34 m.
Therefore, distance between wall and observer === 34 17 m. 22 dround
4. An sound wave has a wavelength of 0.008 m in air. If the speed of sound is 340 m/s, calculate the frequency. Can humans hear this sound?
Ans. Given,
Wavelength (λ) = 0.008 m
Speed of sound wave (v) = 340 m/s.
ν === 340 42500 Hz 0.008
λ
v The human audible range is roughly 20 Hz to 20000Hz.
Since, 42500 Hz is above 20000 Hz, so it is ultrasonic sound and will not audible to humans.
5. A cracker is burst at some distance. A person sees the flash instantly but hears the sound after 5 s. If the speed of sound is 340 m/s, how far away did the cracker explode?
Ans. Light arrives almost instantaneously compared to sound, so the delay is due to the sound travelling from cracker to person.
Given,
Speed of sound (v) = 340 m/s
Time delay (t) = 5 sec
Distance between cracker and person (d) = speed of sound (v) × time (t) d = 340 × 5 = 1700 m or 1.7 km.
6. A girl is sitting in the middle of a park of dimension 12 m × 12 m. On the left side of it there is a building adjoining the park, and on the right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound?
Explain your answer.
Ans. For the girl to hear an echo, the difference between time taken for the sound to travel from road to the girl and the sound coming after reflection from the building must be less than 0.1 sec.
Given,
the speed of sound in air (v) = 344 m/s
From Fig. 11.20, the distance travelled by original sound (do) = 6 m and the total distance travelled by reflected sound (dr) = 18 m
Time taken by sound to travel from one point to another distance Speed t =
Time gap between both sounds
Since, the time gap between both sounds is less than 0.1 sec, the girl will not hear an echo.
7. A submarine sends out a sonar pulse and detects its reflection from the ocean floor after 4 seconds. If the speed of sound in water is 1500 m/s, find the depth of the ocean floor below the submarine. Ans. Given,
Speed of sound in water (v) = 1500 m/s
Total time for sonar pulse to go down and come back (t) = 4 s.
Hence, one-way travel time oneway 4 () 2 s. 2 t ==
Distance (d ) = v × t oneway = 1500 × 2 = 3000 m.
11.21
8. A laboratory experiment produces sound waves with a frequency of 660 Hz. If these waves have a measured wavelength of 0.5 m, calculate the speed of sound in that medium. Ans. Given,
Frequency (ν ) = 660 Hz
Wavelength (λ ) = 0.5 m
Speed of sound (v) = frequency (ν ) × wavelength (λ )
v = 660 × 0.5 = 330 m/s.
9. A sound wave of frequency 500Hz travels from air into water. In air, its speed is 340 m/s, and in water, it is 1500 m/s.
(a) Calculate the wavelength in air.
(b) Calculate the wavelength in water.
(c) Which quantity remains unchanged (frequency, wavelength, or speed) as the wave crosses from air to water?
Ans. Given,
Frequency (ν ) = 500 Hz
Speed of sound in air = 340 m/s
Speed of sound in water = 1500 m/s
(a) Wavelength in air: λ ν == air air 34 5 0 00 Hz m/s v = 0.68 m.
(b) Wavelength in water: ν λ=== water water 1500 m/s 3 m 500 Hz v
(c) Frequency remains the same across media (the source does not change the vibration rate).
The speed and wavelength change, but frequency is constant.
10. A jet flying at an altitude of × 33100 m goes supersonic at point A (t = 0 s) and continues with a constant speed of 400 m/s after that.
After ‘t’ seconds, a stationary observer on ground hears the boom of the supersonic jet and immediately looks up to find that the jet has already passed him. If the distance AC is equal to distance CD, then find the time ‘t’. (Take the speed of sound to be 350 m/s)
Ans. Given,
Speed of sound (v) = 350 m/s
Altitude of plane or BC33100 m =×
Jet’s horizontal speed (vj) = 400 m/s.
Let the sonic boom be heard at time t s.
Now, in t s, the sound waves travels a distance ‘d’ along the hypotenuse AB, to reach the observer. In the same duration jet travels horizontally along line ACD to reach point D.
Using formula,
Distance = velocity × time
Distance travelled by sound wave AB, d = v s × t = 350 × t
Distance travelled by the jet along AD = vj × t
Distance ==× × = j Distance AD AC 200 22 vt t
From the Pythagoras theorem, (AB)2 = (AC)2 + (BC)2
×
2 2 j 2 s () 33100 2 vt vt
()2 22 (350)(200)33100 tt ×=×+×
On simplifying,
t2 × (3502 - 2002) = 33 × 1002
33100100 (350200)(350200)
Practice Questions
Multiple Choice Questions
1. Which of the following properties of sound is affected by change in temperature?
(a) Loudness (b) Pitch
(c) Quality (d) Speed
2. In which medium does sound travel the slowest?
(a) Nitrogen gas
(c) Iron
(b) Honey
(d) Water
3. Which of these units is used to measure the loudness of sound?
(a) Hertz
(c) Decibel
(b) Joule
(d) Metre
4. A sound wave travels at a speed of 330 m/s and has a frequency of 550 Hz at 20ºC. When temperature is increased to 40ºC, its wavelength will be
(a) 0.6 m
(c) less than 0.6 m
(b) more than 0.6 m
(d) cannot be predicted
5. If you are standing at the centre of a spherical room of radius 9 m and say “a” aloud, how many “a” will you hear?
(a) None, no sound will be heard.
(c) One, an elongated “aaaaa..” due to reverberation
Assertion-Reason Based Questions
(b) One, only your original voice.
(d) Multiple “a” due to multiple echoes
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
6. Assertion (A): Sound waves can travel in solids, liquids, and gases.
Reason (R): Sound waves are transverse waves.
7. Assertion (A): Increasing the amplitude of a sound wave increases its loudness.
Reason (R): Loudness depends directly on the energy carried by the wave.
Very Short Answer Questions (30-50 words)
8. Define the term Intensity and quality of a sound.
9. Why do sound waves travel faster in liquids than in gases?
10. State four practical application of ultrasonic waves.
11. Explain why sound cannot propagate in a vacuum.
12. A tuning fork produces two sound waves of different amplitudes. Predict which one will be louder and why.
Short Answer Questions (50-80 words)
13. Explain the difference between echo and reverberation with examples.
14. Determine the time taken by a sound wave to travel a distance of 1,720 m in a medium, given that its frequency is 200 Hz and its wavelength is 17.2 m.
15. Classify sound based on human hearing frequency ranges and provide examples for each category.
16. Aryan sees a lightning flash at a distance of 1.65 km. Calculate the time lag between lighting flash and thunder if the speed of sound in air is 330 m/s.
17. Explain how bats move in darkness without hitting any obstacle.
Long Answer Questions (80-120 words)
18. The graph, in Fig. 11.23, shows the depth of water in a harbour as a wave passes through.
(a) From the graph, measure the wavelength of the wave.
(b) Determine the amplitude of the wave.
(c) The speed of the waves is 1.25 m/s. Calculate how long it takes a wave to pass a given point.
19. Discuss the impact of temperature and humidity on the speed of sound. Use examples to illustrate your answer.
20. Design a simple experiment to measure the speed of sound using a loudspeaker and a long measuring tape.
Brain Charge
1. The Puzzle
A group of scientists are locked in an echo chamber! To escape, they must answer the following riddles related to sound. Each correct answer will unlock one part of the code. Combine all the answers to find the final escape code!
Riddle 1: I am the reason you can hear a clap from across the room, yet I cannot travel through empty space. What am I?
Riddle 2: I am the time it takes for one wave to pass a given point. What am I?
Riddle 3: I’m the loudest of all, yet I depend on how much energy is in the wave. What am I?
Riddle 4: When I hit a hard surface, I bounce back. If I travel a long enough distance, I may return as an echo. What am I?
Riddle 5: I decide whether a sound is high-pitched or low-pitched. Musicians often tune their instruments to match my value. What am I?
Final Challenge: Once you’ve solved all the riddles, arrange the first letters of each answer to spell out a word related to sound. That word is your escape code!
2. Sound Quest: Choose Your Adventure!
You are an intrepid sound wave, setting off on an exciting journey across different environments. As you travel, you encounter challenges that require you to make choices. Select the correct option for each scenario to continue your adventure!
(a) You begin your journey in a classroom full of students. Suddenly, the teacher claps her hands, and you’re sent bouncing off the walls. What happens to you?
(i) You get absorbed by the walls.
(ii) You travel in a straight line and disappear.
(iii) You bounce off the walls and create an echo.
(b) Now, you find yourself traveling through a tube filled with air. What property will change if the tube is made of steel instead of air?
(i) Your speed increases significantly.
(ii) Your frequency increases.
(iii) Your wavelength decreases.
(c) You enter a lake and find yourself moving faster than before. Why does this happen?
(i) Water molecules are more tightly packed than air molecules, allowing faster transmission.
(ii) Water has less density than air, speeding you up.
(iii) Water increases your frequency automatically.
(i) You’ll travel faster than ever.
(ii) You’ll stop moving completely.
(d) Uh-oh! You’ve been sucked into a vacuum chamber during a science experiment. What will happen to you?
(iii) You’ll slow down but still propagate.
(e) You join a group of sound waves produced by a tuning fork. If the fork vibrates at 256 Hz, what determines the pitch of the sound?
(i) The speed at which you’re moving.
(ii) The frequency of the vibrations.
(iii) The loudness of the sound.
(f) During your travels, you notice you’re getting louder! What has increased?
(i) Your frequency.
(ii) Your amplitude.
(iii) Your wavelength.
(g) Finally, you enter a large auditorium with walls covered in thick curtains. What happens to you?
(i) You bounce around the room and create a loud echo.
(ii) You are absorbed by the curtains, reducing your intensity.
(iii) You pass straight through the curtains with no change.
3. Two immortal aliens embark on a journey across the planets of our solar system to test whether they can hear each other’s voices. On which planets of our solar system will sound travel, allowing them to communicate?
Challenge Yourself
1. A tuning fork produces a sound wave of frequency 256 Hz. If the speed of sound in air is 340 m/s, what is the distance between two consecutive compressions in the sound wave? How will this distance change if temperature of the medium is raised by 20ºC.
2. Explain how the speed of sound changes with the temperature of the medium. If the speed of sound at 0°C is 331 m/s, calculate the speed of sound at 27°C.
(Hint: Use the formula v = 331 + 0.6T, where T is the temperature in °C.)
3. A sound wave propagates through a steel rod at a speed of 5,960 m/s. First, calculate how long it takes for the wave to cover a distance of 3 m in the steel rod. Then, using that same time interval, determine the distance the wave would travel if it were moving through air at a speed of 340 m/s.
4. A construction worker’s helmet slips and falls when he is 80 m above the ground. He hears the sound of the helmet hitting the ground 4.25 seconds after it slipped. Find the speed of sound in air. (Take g = 10 m/s2)
Practice Questions
1. (d)
(a)
18. (a) 2.5 m (b) 0.3 m (c) 2 sec
Brain Charge
1. M.T.A.R.F
2. (a) (iii) (b) (i) (c) (i) (d) (ii) (e) (ii) (f) (ii) (g) (ii)
3. Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune
Challenge Yourself
1. 1.33 m
2. 347.2 m/s
3. t = 0.5 ms; d = 0.171 m
4. 320 m/s
Answers Scan me for detailed Solutions Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 1.5 Hour
Multiple Choice Questions
Max. Marks: 50 Scan me for Solutions
(10 × 1 = 10 Marks)
1. When a tuning fork is struck with greater force, which of the following characteristics of the sound it produces will change?
(a) frequency (b) wavelength (c) amplitude (d) velocity
2. In a school auditorium, the sound produced on stage reaches the back of the hall clearly because of (a) absorption of sound (b) reflection of sound (c) diffraction of sound (d) refraction of sound
4. If a sound wave has a frequency of 500 Hz, what is its time period?
(a) 0.002 s (b) 2 s (c) 0.05 s (d) 0.005 s
5. The correct order of speed of sound in different media (from slowest to fastest) is:
(a) Air < Water < Steel (b) Air < Steel < Water (c) Steel < Water < Air (d) Water < Steel < Air
6. A ship in the ocean sends a sound signal downwards and receives the echo after 4 seconds. If the speed of sound in water is 1500 m/s, what is the depth of the ocean at that point?
(a) 3000 m (b) 1500 m (c) 6000 m (d) 750 m
7. The human ear is most sensitive to which approximate range of frequencies?
(a) 2 Hz to 20 kHz (b) 20 Hz to 20 kHz (c) 2 kHz to 20 kHz (d) 200 Hz to 20,000 Hz
8. Quality of sound is used to differentiate between two sound waves of . (a) different loudness.
(c) similar loudness and pitch.
(b) different pitch.
(d) varying loudness and pitch.
9. Which phenomenon is responsible for the reverberation in a large hall?
(a) Reflection of sound
(c) Diffraction of sound
(b) Refraction of sound
(d) Interference of sound
10. Identify the pair that incorrectly matches a characteristic of sound with its SI unit.
(a) Loudness – Decibel
(c) Intensity – Joule
(b) Frequency – Hertz
(d) Wavelength – Metre
Assertion-Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): Sound can be heard on Mars but not on the Moon. Reason (R): Sound waves need a material medium for propagation.
12. Assertion (A): An echo is heard only when the reflecting surface is at a suitable distance from the source.
Reason (R): The minimum distance for an echo to be heard distinctly is related to the speed of sound and the persistence of hearing.
13. Assertion (A): The loudness of a sound wave decreases as we move away from the source.
Reason (R): The amplitude of a sound wave diminishes with distance, due to the decrease in wavelength.
14. Assertion (A): The frequency of a tuning fork remains unchanged even if you strike it more forcefully.
Reason (R): Frequency = Speed of sound wave/Wavelength
Case-Based/Passage-Based/Source-Based Questions
(2 × 4 = 8 Marks)
15. A large hall is used for an event. The organiser notices that when people speak on the stage, they hear a prolonged sound even after they stop talking. The walls are made of hard concrete and no sound-absorbing material has been used.
(a) What term is used to describe the phenomenon where the sound persists due to repeated reflections?
(b) Suggest one method to reduce this unwanted effect in the hall.
(c) If a person in the hall claps and hears an echo 0.2 s later, how far away is the reflecting wall?
(Assume speed of sound = 340 m/s.)
(d) Explain why curtains or carpets might help to control this issue.
16. A school organises a musical concert in its auditorium. Various musical instruments like drums, guitars, and flutes are played. Students notice different pitches and loudness levels.
(a) State one factor affecting the pitch of the instruments.
(b) State one factor affecting the loudness of the musical notes.
(c) Why do different instruments produce distinct quality (timbre) of sound, even if they play the same note?
(d) During a drum performance, the sound waves travel through air at 340 m/s. If a certain beat has a wavelength of 1.7 m, calculate its frequency.
Very Short Answer Questions (30-50 words)
17. Compare the speed of sound in ice, fresh water, and sea water?
18. What causes a listener’s eardrums to rupture under extremely loud sounds?
19. Define reverberation. How is it different from a simple echo?
Short Answer Questions (50-80 words)
20. Which quantities characterise a sound wave? Explain each one and specify its SI unit.
(3 × 2 = 6 Marks)
(4 × 3 = 12 Marks)
21. Explain why the sound produced by a vibrating object in a medium travels in the form of a wave. What type of wave is a sound wave moving in air?
22. Why is the ceiling and wall behind the stage of good conference halls or concert halls made curved? (NCERT Exemplar)
23. A stationary boat sounds its horn at a frequency of 400 Hz. A bird and a whale, both at the same distance of 10 m from the boat, hear it. Will they detect it at the same moment? Determine how long it takes for the sound to reach each one, given that the speed of sound in air is 330 m/s and in seawater is 1,650 m/s.
Long Answer Questions (80-120 words)
(2 × 5 = 10 Marks)
24. Discuss the concept of audible, infrasonic, and ultrasonic ranges of sound. How do they differ, and what are some real-life uses for each category? Provide relevant examples to support your answer.
25. Represent graphically by two separate diagrams in each case (NCERT Exemplar)
(a) Two sound waves having the same amplitude but different frequencies?
(b) Two sound waves having the same frequency but different amplitudes.
(c) Two sound waves having different amplitudes and also different wavelengths.
12 Improvement in Food Resources
Food is essential for the survival of all living organisms. With a growing population, the demand for food is increasing rapidly. This chapter, Improvement in Food Resources, explores various strategies to enhance food production through improved agricultural practices, animal husbandry and sustainable resource management. It focuses on scientific methods to increase crop yields, protect crops from pests and diseases and improve the productivity of livestock. Also, the chapter emphasises the importance of adopting improved practices to achieve a balance between food production and environmental sustainability.
Improvement in Food Resources
Crop Variety Improvement
Developing high-yielding, disease-resistant and pest-tolerant crop varieties
Factors for Variety Improvement
• Higher yield.
• Improved quality.
• Biotic and abiotic resistance.
• Change in maturity duration.
• Wider adaptability
• Desirable agronomic features.
Cattle Farming
For milk and draught labour
Crop Production Improvement
Efficient use of resources like irrigation, fertilisers and proper farming practices
Improvement in Crop Yield
Plants
Resources of Food
Animals
Animal Husbandry
Poultry Farming
For meat and egg production
Practices
Fish Farming
Crop Protection Management (Protecting crops from pests, diseases and weeds using chemical, biological or cultural methods)
Disease Management
• Pest control
• Disease control
• Weed and insects control Factors for losses during storage of Grains:
Biotic: Insects, rodents, fungi, mites, bacteria, etc.
Abiotic: Moisture, temperature
Beekeeping
Capture and culture fish for animal protein
Rear and care for bees for commercial honey production
• Nutrient management: Manure-compost, vermicompost, green manure, fertilisers
• Irrigation: Wells, canals, river lift systems, tanks
o Fresh initiatives: rainwater harvesting, water-shed management
• Cropping patterns:
o Mixed cropping: growing of two or more crops simultaneously on the same piece of land
o Intercropping: growing two or more crops in definite row patterns
o Mixed cropping: growing of two or more crops simultaneously on the same piece of land
Chapter at a Glance
• Food is essential for human survival, growth and energy. The production of food must meet the needs of a growing population.
• Sustainable agricultural practices are required to ensure food security and environmental conservation.
• Kharif and rabi crops are grown in different seasons, depending on climatic conditions.
• The major groups of activities for improving crop yields can be classified as:
º Crop variety improvement
º Crop production improvement
º Crop protection management
• Improving crop variety through hybridisation and genetic engineering leads to higher yields and better resistance to pests and diseases.
• Cultivation practices and crop yield are related to weather, soil quality and availability of water.
• There are some basic factors for which variety improvement is done. They are: higher yield, improved quality, biotic and abiotic resistance, change in maturity duration, wider adaptability, desirable agronomic characteristics, etc.
• Crop production practices can be at different levels. They include ‘low cost’ production and ‘high cost’ production practices.
• Nutrient management focuses on providing crops with macronutrients like nitrogen, phosphorus and potassium, and micronutrients.
• There are several nutrients essential for crops. Of these, some are required in large quantities and are known as macronutrients whereas the rest of the nutrients are required in small quantities and are known as micronutrients.
• Soil fertility is maintained using organic manure, chemical fertilisers and bio-fertilisers.
• Manure and fertilisers are the main sources of nutrient supply to crops.
• Organic farming is a farming system with minimal or no use of chemicals, such as fertilisers, herbicides and pesticides and with a maximum input of organic manures, recycled farm wastes and bio-agents, with healthy cropping systems.
• Proper irrigation practices, including modern methods like drip and sprinkler irrigation, ensure efficient water use.
• Crop production practices, such as mixed cropping, intercropping and crop rotation, help maximise productivity and minimise risks.
• Mixed cropping is the growing of two or more crops simultaneously on the same piece of land.
• Growing two or more crops in definite row patterns is known as intercropping
• The growing of different crops on a piece of land in pre-planned succession is called crop rotation.
• Crops require protection from diseases caused by bacteria, fungi and viruses through preventive measures.
• Post-harvest storage techniques, such as cold storage and silos, prevent spoilage and maintain food quality.
• Food safety and quality are essential to ensure that food remains nutritious and free of harmful substances.
• Weed control is necessary to prevent competition for resources, achieved using manual, chemical and biological methods.
• Integrated pest management (IPM) combines natural and chemical methods to control pests and protect crops.
• Animal husbandry is crucial for enhancing food resources and providing milk, meat, eggs and other products.
• Dairy farming involves the rearing of high-yielding cattle breeds under hygienic conditions for increased milk production.
• Poultry farming focuses on raising chickens for eggs and meat, with proper feeding and disease management.
• Fisheries involve fish farming (pisciculture) and prawn farming, which are major sources of protein.
• Fish can be cultured in marine and inland ecosystems to increase production.
• Marine fish capture is done by fishing nets guided by echo-sounders and satellites. The composite fish culture system is commonly used for fish farming.
• Apiculture is the practice of beekeeping for honey and wax with benefits for pollination.
NCERT Zone
Intext Questions
1. What do we get from cereals, pulses, fruits and vegetables?
Ans. Cereals such as rice, potatoes, corn and barley contain carbohydrates, which provide energy.
Pulses such as gram, lentils, green gram and chickpeas contain proteins, which build our body.
Various vegetables and fruits provide vitamins and minerals that help prevent disease and boost immunity in the body.
2. How do biotic and abiotic factors affect crop production?
Ans. Factors responsible for the loss of grains, during storage and production are:
(a) Biotic factors like rodents, pests, insects, etc.
(b) Abiotic factors like temperature, humidity, moisture, etc.
A combination of both biotic and abiotic factors causes:
• infestation of insects
• weight loss
• poor germination ability
• degradation in quality
• discolouration
• poor market price
3. What are the desirable agronomic characteristics for crop improvements?
Ans. Desirable agronomic characteristics for crop improvements are:
(a) Tallness and profuse branching in fodder crops
(b) Dwarfness in cereals, to reduce nutrient consumption by these crops
4. What are macronutrients and why are they called macronutrients?
Ans. Macronutrients are essential elements that are utilised by plants in large quantities. Many macronutrients are required by the plants for the following functions:
• They are the constituents of protoplasm.
• Nitrogen (N), Phosphorus (P) and Sulphur (S) are present in proteins.
• Calcium (Ca) is present in the cell wall.
• Magnesium (Mg) is an important constituent of chlorophyll.
5. How do plants get nutrients?
Ans. Plants get nutrients from air, water and soil. There are sixteen nutrients essential for the growth of plants. Air supplies carbon and oxygen. Water supplies hydrogen and oxygen. Soil supplies the remaining thirteen nutrients.
6. Compare the use of manure and fertilisers in maintaining soil fertility.
Ans. Effects of using manure on soil quality:
1. Manure enriches the soil with nutrients.
2. It provides organic matter (humus) to the soil and thus restores the water-retention capacity of sandy soils and drainage in clayey soil.
3. The addition of manure reduces soil erosion.
4. It provides food for soil organisms, like soil-friendly bacteria.
Effects of using fertilisers on soil quality:
1. Continuous use of fertilisers leads to powdery, dry soil and the rate of soil erosion increases.
2. The use of fertilisers reduces organic matter in the soil, leading to decreased porosity of soil. As a result, plant roots do not get enough oxygen.
3. The nature of soil changes to acidic or basic.
7. Which of the following conditions will give the most benefits? Why?
(a) Farmers use high-quality seeds, do not adopt irrigation or use fertilisers.
(b) Farmers use ordinary seeds, adopt irrigation and use fertilisers.
(c) Farmers use quality seeds, adopt irrigation, use fertilisers and use crop protection measures.
Ans. (c) Farmers use quality seeds, adopt irrigation, use fertilisers and use crop protection measures.
The use of seeds of any quality alone is not sufficient; they must be properly irrigated, enriched with fertilisers and protected from biotic factors.
Hence, option (c) will give the most benefits.
8. Why should preventive measures and biological control methods be preferred for protecting crops?
Ans. Diseases in plants are caused by pathogens. To get rid of pathogens, some preventive measures and biological control methods are used as they are simple and economic and minimise pollution without affecting the soil quality.
9. What factors may be responsible for losses of grains during storage?
Ans. The factors responsible for losses of grains during storage are:
• Abiotic factors like moisture (present in food grains), humidity (in air) and temperature
• Biotic factors such as insects, rodents, birds, mites, bacteria and fungi
10. Which method is commonly used for improving cattle breeds and why?
Ans. Crossbreeding is a process in which indigenous varieties of cattle are crossed with exotic breeds to get high-yielding breeds. During crossbreeding, the desired characteristics are taken into consideration. The offspring should be high-yielding, have early maturity and be resistant to climatic conditions.
11. Discuss the implications of the following statement:
‘It is interesting to note that poultry is India’s most efficient converter of low-fibre foodstuff (which is unfit for human consumption) into highly nutritious animal protein food.’
Ans. The basic aim of poultry farming is to raise domestic fowl for egg production and chicken meat. These poultry birds are not only efficient converters of agricultural by-products, particularly cheaper fibrous wastes (which are unfit for human consumption but can be formulated into cheaper diets for poultry birds) into high-quality meat but also help in providing eggs, feathers and nutrient-rich manure. For these reasons, it is said that ‘poultry is India’s most efficient converter of low-fibre foodstuff into highly nutritious animal protein food’.
12. What management practices are common in dairy and poultry farming?
Ans. In dairy and poultry farming, some common management practices are as follows:
1. Shelter: Dairy animals and poultry birds require well-designed and hygienic shelters to ensure health and safety.
2. Feeding: To achieve a high yield of food products, dairy animals and poultry birds must be given proper feed.
3. Caring for animal health: Animals and birds must be protected from diseases caused by viruses, bacteria or fungi.
13. What are the differences between broilers and layers and in their management?
Ans.
Broilers Layers
1. A broiler is a poultry bird groomed to obtain meat. 1. A layer is an egg-laying poultry bird.
2. They need proper temperature control and hygienic conditions to ensure growth and reduce mortality.
3. The ration (daily food requirement) for broilers is protein-rich with adequate fat.
4. For example: Chickens are raised for meat.
14. How are fish obtained?
Ans. There are two ways of obtaining fish:
2. They need space and lighting during their growing and laying periods.
3. They need to be fed vitamins, minerals and micronutrients.
4. For example: Chickens are raised for the production of eggs.
• One is from natural resources, which is called capture fishing
• The other way is by fish farming, which is called culture fishery
15. What are the advantages of composite fish culture?
Ans. In composite fish culture, a combination of five or six fish species is used in a single fish pond. These species are selected so they do not compete for food among themselves and because they have different food habits.
Advantages of composite fish culture:
• Increased yield: Composite fish culture can produce 8–9 times more yield than monoculture.
• Economic: Composite fish culture is economically feasible and can be more profitable.
• Efficient use of food: Different fish species use the food available in the pond without competing with each other.
• Increased survival rate: Composite fish culture can enhance the survival of different species within the same area.
• Variety of fish: Composite fish culture allows one to produce different varieties of fish.
• Distinct zones: Fish live in distinct zones within the pond and have distinct feeding habits.
• No competition: Fish do not compete with each other for space and nutrition.
• Complete utilisation of food resources: Different fish species might have different food habits.
16. What are the desirable characteristics of bee varieties suitable for honey production?
Ans. The desirable characteristics are:
• Bee varieties should be able to collect a large amount of honey.
• The bees should stay in a given beehive for a long period.
• The bees should have a strong breeding capacity.
• Bee varieties should be disease-resistant.
17. What is pasturage and how is it related to honey production?
Ans. Pasturage refers to the availability of flowers for bees to collect nectar and pollen. In addition to adequate quantities of pasturage, the kinds of flowers available determine the taste of the honey.
NCERT Exercises
1. Explain any one method of crop production which ensures high yield.
Ans. One method of crop production which ensures high yield is plant breeding. It is the science involved in improving crop varieties by breeding plants. Plants from different areas/places with desired traits are picked and then hybridised or cross-bred to obtain a plant/crop of desired characteristics.
A high-yielding crop variety shows the following characteristics:
High yield, early maturation, less water for irrigation, better quality seeds produced, less fertilisers required, adapts itself to the environmental conditions
2. Why are manure and fertilisers used in fields?
Ans. Manure and fertilisers are used in fields to enhance soil fertility and increase crop yield. They provide essential nutrients like nitrogen (N), phosphorus (P), and potassium (K), which are necessary for plant growth.
Manure, which is made from decomposed organic matter, improves soil structure, water retention, and microbial activity. On the other hand, fertilisers are chemical or natural substances that supply specific nutrients quickly to the soil, promoting faster and healthier crop growth. Their combined use helps maintain soil health, improve productivity, and ensure sustainable agriculture.
3. What are the advantages of inter-cropping and crop rotation?
Ans. Advantages of inter-cropping:
1. It helps to maintain soil fertility.
2. It increases productivity per unit area.
3. It saves labour and time.
4. Both crops can be easily harvested and processed separately. Advantages of crop rotation:
1. It improves soil fertility.
2. It avoids the depletion of a particular nutrient from the soil.
3. It minimises pest infestation and diseases.
4. It helps in weed control.
5. It prevents change in the chemical nature of the soil.
4. What is genetic manipulation? How is it useful in agricultural practices?
Ans. Genetic manipulation is the process of incorporating desirable characteristics (genes) into crop varieties by hybridisation. Hybridisation involves crossing genetically dissimilar plants. This is done for producing varieties with desirable characteristics like profuse branching in fodder crops and high-yielding varieties in maize, wheat, etc.
Genetic manipulation is useful in developing varieties which show:
• Increased yield
• Better quality
• Shorter and early maturity period
• Better adaptability to adverse environmental conditions
• Desirable characteristics
5. How do storage grain losses occur?
Ans. The factors responsible for loss of grains during storage are:
• Abiotic factors like moisture (present in food grains), humidity (present in air) and temperature
• Biotic factors like insects, rodents, birds, mites and bacteria
6. How do good animal husbandry practices benefit farmers?
Ans. Good animal husbandry practices benefit farmers in the following ways:
• Improvement of domesticated animal breeds.
• Increasing food yields such as milk, eggs and meat.
• Proper management of domestic animals in terms of shelter, feeding, care and protection against diseases.
• Increased production and improved animal health lead to increased profit for farmers and improved economic conditions.
7. What are the benefits of cattle farming?
Ans. Cattle farming is beneficial in the following ways:
• High-yielding animals increase milk production.
• Good quality of meat, fibre and skin can be obtained.
• Good breed of draught animals can be obtained.
8. For increasing production, what is common in poultry, fisheries and bee-keeping?
Ans. To increase production in poultry, fisheries and beekeeping, we can use proper management techniques, such as
• cross-breeding
• proper feeding
• disease management
9. How do you differentiate between capture fishing, mariculture and aquaculture?
Ans. Capture fishing: Here, fishes are captured from natural resources such as ponds, seawater and estuaries. Mariculture: It is the culture of fish in marine water. Varieties such as prawns, oysters, bhetki and mullets are cultured for fishing.
Aquaculture: It is done both in fresh water and in marine water.
Multiple Choice Questions
1. Which one is an oil yielding plant among the following?
(a) Lentil and Hibiscus
(c) Cauliflower and cotton
2. Find out the wrong statement from the following.
(b) Sunflower and mustard
(d) Hibiscus and Parthenium
(a) White revolution is meant for increase in milk production.
(b) Blue revolution is meant for increase in fish production.
(c) Increasing food production without compromising with environmental quality is called as sustainable agriculture.
(d) Green revolution is meant for increasing milk, rice and wheat production.
3. Which one is not a source of carbohydrate?
(a) Rice
(b) Millets
(c) Sorghum
(NCERT Exemplar)
(d) Gram
4. Weeds affect crop plants by (NCERT Exemplar)
(a) killing plants in the field before they grow. (b) dominating the plants to grow.
(c) competing for various resources of crops (plants), causing low availability of nutrients. (d) all of the above.
5. Which one of the following species of honey bee is an Italian species? (NCERT Exemplar) (a) Apis dorsata (b) Apis florae (c) Apis cerana indica (d) Apis mellifera
6. Which of the following are Indian cattle? (NCERT Exemplar)
(i) Bos indicus (ii) Bos domestica (iii) Bos bubalis (iv) Bos vulgaris (a) (i) and (iii) (b) (i) and (ii) (c) (ii) and (iii) (d) (iii) and (iv)
7. Which fish is a surface feeder?
(a) Rohu (b) Mrigal (c) Common carp (d) Catla
8. Which one of the following nutrients is not available in fertilisers?
(a) Nitrogen (b) Phosphorus (c) Iron (d) Potassium
9. Which of the following is not an objective of crop improvement?
(a) High yield (b) Resistance to biotic and abiotic stresses (c) Reduction in crop duration (d) Elimination of weeds
10. Which crop is commonly grown in crop rotation to replenish soil nitrogen? (a) Wheat (b) Rice (c) Groundnut (d) Sugarcane
11. Which of the following is a limitation of using high-yielding variety (HYV) seeds?
(a) Resistance to pests
(b) Higher requirement for fertilisers and irrigation
(c) Shorter duration of crop maturity
(d) Increased tolerance to drought
12. Which component is NOT a part of organic farming? (a) Vermicomposting (b) Green manure (c) Chemical pesticides (d) Crop rotation
13. Which of the following is an example of a biotic factor affecting crop yield? (a) Flood (b) Soil pH (c) Pest attack (d) Temperature
14. Which micronutrient deficiency leads to chlorosis in plants?
(a) Iron (b) Zinc (c) Copper (d) Manganese
15. Which of the following is NOT a desirable characteristic of poultry breeds?
(a) Low maintenance (b) High feed consumption (c) Disease resistance (d) High egg production
16. Which of the following crops is most suitable for dryland agriculture?
(a) Rice (b) Cotton (c) Maize (d) Pearl millet
17. Which pest management method reduces environmental pollution the most?
(a) Use of chemical pesticides (b) Integrated Pest Management (IPM) (c) Use of genetically modified crops (d) Burning crop residues
18. Which fish species is widely used in composite fish culture due to its ability to feed at the bottom of the pond?
(a) Catla (b) Rohu (c) Mrigal (d) Hilsa
19. Which of the following is the main advantage of genetically modified (GM) crops?
(a) High water requirement
(c) Reduced use of chemical pesticides
(b) High susceptibility to pests
(d) Decreased crop yield
20. What is the purpose of growing cover crops in agricultural fields?
(a) To increase the cost of farming
(b) To provide shade to the main crop
(c) To protect the soil from erosion and improve fertility
(d) To increase the water retention capacity of the plants
Answers
Constructed Response Questions
Very Short Answer Questions (30-50 words)
1. Define mixed cropping and give two examples.
Ans. Mixed cropping is the practice of growing two or more crops simultaneously on the same field to reduce the risk of crop failure.
Examples: Wheat and mustard
2. What is intercropping? How does it differ from mixed cropping?
Ans. Intercropping is growing two or more crops in distinct rows in the same field. It differs from mixed cropping as crops are grown in rows, ensuring better utilisation of nutrients.
3. How does crop rotation help maintain soil fertility?
Ans. Crop rotation helps maintain soil fertility by alternating crops with different nutrient requirements, such as cereals followed by legumes.
Benefits of crop rotation
• Improves soil health
• Reduces the need for chemical fertilisers and pesticides
• Improves crop emergence, growth and health
• Reduces the risk of pests and disease
• Increases biodiversity on the farm
4. Name two methods of irrigation and mention one advantage of each.
Ans. • Sprinkler system: Suitable for sandy soil and uneven land
• Drip irrigation: Conserves water by delivering it directly to the roots
5. What is the significance of manure in agriculture?
Ans. Significance of manure:
• It improves soil structure, which helps the soil hold more water and nutrients.
• It improves the soil's water retention capacity, which reduces crop water stress and soil erosion.
• It encourages soil microbial activity, which improves the soil's trace mineral supply and plant nutrition.
• It is a key component of organic farming practices and promote sustainable agriculture.
6. Distinguish between organic farming and conventional farming.
Ans.
Organic Farming
1. Organic farming is the method of growing crops with natural resources and processes.
Conventional Farming
1. Conventional farming is a farming method that uses synthetic chemicals, heavy machinery and other inputs to grow crops and raise livestock.
2. Uses natural inputs like compost and biofertilisers. 2. Relies on chemical fertilisers and pesticides.
7. Differentiate between macronutrients and micronutrients, with examples.
Ans.
Macronutrients
1. Nutrients that are needed in large quantities are called macronutrients.
2. Examples: N, P, K, Mg, S, Ca, etc.
Micronutrients
1. Nutrients that are needed in small quantities are called micronutrients.
2. Examples: Fe, Zn, Cu, B, Mn, Cl etc.
8. Why are biofertilisers considered beneficial for farmers?
Ans. Biofertilisers contain living microorganisms that increase plant growth by providing nutrients to the soil. They enhance soil fertility by
• fixing nitrogen,
• solubilising phosphorus and
• improving soil health naturally.
9. Explain the role of vermicompost in farming.
Ans. Vermicompost is made from organic waste that is broken down by earthworms. It improves soil structure and provides essential nutrients like nitrogen and potassium.
10. What is green manure? Mention its importance in improving soil health.
Ans. Green manure is made by ploughing green plants (like legumes) into the soil. It enriches the soil with organic matter and nutrients.
11. What is the role of Integrated Pest Management (IPM) in agriculture?
Ans. IPM uses a combination of biological, cultural and chemical methods to manage pests while minimising environmental damage.
12. Differentiate between milch animals and draught animals with examples.
Ans. Milch animals: Used for milk production (e.g., cow, buffalo)
Draught animals: Used for labour, like ploughing (e.g., bullocks)
13. Distinguish between marine fisheries and inland fisheries.
Ans. Marine fisheries: Fish are caught from seas and oceans (e.g., tuna, mackerel).
Inland fisheries: Fish are reared in freshwater sources like ponds and rivers (e.g., rohu, catla).
14. What are the two main crop seasons in India? Give examples.
Ans. The two main crop seasons in India are:
• Kharif season: Crops are sown in the rainy season (June to September). Examples: Paddy, maize, soybean, groundnut, cotton
• Rabi season: Crops are sown in the winter season (October to March). Examples: Wheat, gram, pea, mustard, linseed
15. Why is organic farming considered environmentally friendly?
Ans. Organic farming avoids the use of chemical fertilisers, pesticides and genetically modified organisms. Instead, it uses natural inputs like compost, green manure and biopesticides. This reduces soil, water and air pollution, maintaining ecological balance.
16. What are the main elements of animal husbandry?
Ans. The main elements of animal husbandry are:
• Feeding animals properly
• Providing fresh water to animals
• Providing safe and hygienic shelter to animals
• Ensuring proper health of animals and protection against diseases
• Breeding animals properly
17. What are pathogens? Name any two plant diseases caused by pathogens.
Ans. The disease-causing microorganisms like bacteria, fungi and viruses are called pathogens. They reach the plants through water, air, soil and seeds.
Two plant diseases caused by pathogens are rust in wheat and blast in paddy/stem rot in pigeon peas.
18. Why is honey important? Give two uses of honey.
Ans. Honey is a dense sweet liquid that contains 20-40 percent sugar, 60-80 percent moisture, 0.22-0.3 percent minerals and 0.2-0.5 percent vitamins. Apart from that, it also contains certain enzymes and pollen.
The uses of honey are as follows:
• Honey has medicinal value, and is said to help in disorders related to digestion and liver ailments.
• As it contains iron and calcium, it also helps in growth.
• It is used as a source of sugar in various confectionery items.
19. Group the following and tabulate them as energy-yielding, protein-yielding, oil-yielding and fodder crop.
• Crop production management: Efficient use of resources like irrigation, fertilisers and proper farming practices
• Crop protection management: Protecting crops from pests, diseases and weeds using chemical, biological or cultural methods
2. What are the advantages of intercropping and crop rotation?
Ans. • Intercropping: Growing two or more crops simultaneously in the same field Advantages:
º Reduces the chances of pests and diseases
º Ensures optimum use of nutrients from the soil
• Crop rotation: Growing different crops in a sequential manner on the same land Advantages:
º Prevents depletion of specific nutrients from the soil
º Helps in pest and weed control
3. What are the three modes of irrigation? Mention their advantages.
Ans. The different modes of irrigation are:
• Canal irrigation: Water is brought to fields through canals from reservoirs or rivers.
º Advantage: Suitable for large agricultural areas
• Sprinkler irrigation: Water is sprayed over fields like rain using sprinklers.
º Advantage: Ideal for uneven terrains and water conservation
• Drip irrigation: Water is supplied directly to the plant roots through a network of pipes.
º Advantage: Conserves water and minimises weed growth
4. What mineral nutrients are supplied to plants by air, water and soil?
Ans. Air supplies two nutrients to plants, namely carbon and oxygen. Water supplies hydrogen and soil supplies thirteen nutrient elements to plants. Among these thirteen nutrients, six are macronutrients and seven are micronutrients. Macronutrients include nitrogen, phosphorus, potassium, calcium, magnesium and sulphur while micronutrients include iron, manganese, boron, zinc, copper, molybdenum and chlorine.
5. What are the three main advantages of beekeeping?
Ans. The following are the main advantages of beekeeping:
• In addition to honey, beekeeping also yields other products like wax, royal jelly and bee venom on a commercial scale.
• Beekeeping requires low investment, allowing farmers to engage in it alongside agriculture to generate additional income.
• It also helps in cross-pollination, as pollen is transferred from one flower to another by bees while collecting nectar.
6. Name two infectious diseases each for cows, poultry and fish.
Ans. • Cows – Anthrax and foot and mouth disease
• Poultry – Ranikhet and Salmonellosis
• Fish – Viral Haemorrhagic Septicaemia (VHS) and Infectious Pancreatic Necrosis (IPN)
7. Enumerate three advantages of mixed farming.
Ans. The following are the main advantages of mixed farming:
• Farmyard manure is made available from livestock, which is used in agricultural farms.
• Organic waste materials like straw, husk and chaff of grains and household kitchen waste are converted into human food through the agency of cattle, sheep, poultry and pigs, based on the farmer’s preference.
• It provides work to all family members throughout the year, thus providing subsidiary occupation without the need to employ special labour.
• By adopting an exact combination in mixed farming, income can be increased. For example, the number of animals can be increased (as per the food/crop available) to enhance milk production.
8. Answer the following questions.
(a) Give any three preventive measures for pest control.
(b) What preventive and control measures are used before grains are stored for future use?
Ans. (a) The three important preventive measures for pest control are:
• Employing crop rotation
• Use of pest-resistant varieties
• Employing optimum time of sowing crops
(b) Preventive and control measures used before grains are stored for future use include strict cleaning of the produce before storage, proper drying of the produce first in sunlight and then in the shade and fumigating by using chemicals that kill pests.
9. Write three harmful effects of weeds on crops.
Ans. • Weeds compete with crops for nutrients, water and sunlight, reducing yield.
• They harbour pests and diseases, which can spread to crops.
• They increase the cost of production as additional measures are needed for their removal.
10. Differentiate between manure and fertiliser.
Ans. Manure Fertiliser
1. Made from organic materials like plant and animal waste
1. Manufactured chemically in factories
2. Improves soil texture and fertility gradually 2. Provide immediate nutrients to plants
3. Eco-friendly and sustainable
3. Excessive use can harm the environment
11. Differentiate between mixed cropping and intercropping.
Ans.
Mixed Cropping
1. There is no definite pattern of rows.
Intercropping
1. Crops are grown in a definite pattern of rows like 1:1, 1:2 or 1:3.
2. It is undertaken to reduce the chances of crop failure. 2. It is undertaken to enhance the production of crops per unit area.
3. Mixed cropping cannot be done separately for crops. 3. In intercropping, crops can be harvested and threshed separately.
4. Seeds are mixed up before sowing. 4. Seeds are not mixed before sowing.
5. Application of fertilisers and spraying of pesticides for separate crops is not possible.
5. As per the need of the individual crop, fertilisers and pesticides can be applied easily.
12. Differentiate between chemical fertilisers and biofertilisers.
Ans. Chemical Fertilisers
Biofertilisers
1. They are traditionally used on a large scale to obtain more crop yield. 1. They are renewable nutritional resources, mostly nitrogen-fixing microbes.
2. Overuse of chemical fertilisers causes soil pollution. 2. They enrich soil with nutrients.
3. They are in the form of chemicals.
3. They are free-living or symbiotic bacteria, cyanobacteria or fungi.
4. They are not part of sustainable agriculture. 4. They are part of organic farming, promoting better and sustainable agriculture.
13. Why is excess use of fertilisers detrimental for the environment? (NCERT Exemplar)
Ans. Excess use of fertilisers can be harmful to the environment because it can:
• pollute the air, water and soil. Unused fertiliser can become a pollutant.
• release harmful greenhouse gases into the atmosphere.
• stimulate microorganism growth, which reduces the amount of dissolved oxygen in water.
• introduce pathogens and nitrates into drinking water.
• emit odours and gases into the air.
• increase the acidity of soil.
• disturb the soil ecosystem, which can reduce soil productivity.
14. List some useful traits of improved crops. (NCERT Exemplar)
Ans. Useful traits of improved crops are:
(a) Higher yield
(b) Improved nutritional quality
(c) Resistance to biotic and abiotic stresses
(d) Change in maturity
(e) Wide range of adaptability
(f) Desired agronomic characteristics
15. Discuss the role of hybridisation in crop improvement. (NCERT Exemplar)
Ans. Hybridisation refers to crossing genetically dissimilar plants. It may be intervarietal, interspecific or intergeneric. Two crops with good traits (desired characteristics) are selected and crossed to obtain a new crop with the desired characteristics of the parental crops.
Benefits of hybridisation
• Disease resistance: Hybrids are resistant to diseases, pests and environmental stressors.
• Improved yield: Hybrids have increased yield and robustness.
• Nutritional profile: Hybrids have improved nutritional profiles.
• Stress tolerance: Hybrids are more tolerant to abiotic stress.
• Adaptability: Hybrids are more adaptable and resilient.
16. What would happen if poultry birds were larger but could not adapt to summer conditions? What method would be used to get smaller poultry birds with summer adaptability? (NCERT Exemplar)
Ans. If poultry birds were larger but could not adapt to summer conditions, their egg production would decrease. This is because larger birds require more food and have a larger surface area, which makes it harder for them to maintain their temperature. To get smaller birds with summer adaptability, we can crossbreed the local birds with exotic birds.
• Egg production: Poultry birds need to maintain a certain temperature to produce eggs. Larger birds require more food and have a larger surface area, which makes it harder for them to maintain their temperature.
• Crossbreeding: Crossbreeding local birds with exotic birds can produce smaller birds with better summer adaptability.
• Maintenance: Farms with larger birds may need to spend more on heat maintenance.
17. Differentiate between the following. (NCERT Exemplar)
(a) Capture fishery and Culture fishery
(b) Mixed cropping and Inter cropping
(c) Bee keeping and Poultry farming
Ans. (a) Capture fishery is the method of obtaining fish from natural resources while culture fishery is the method of obtaining fish by fish farming.
(b) Mixed cropping is growing two or more crops simultaneously on the same piece of land; while intercropping is growing two or more crops simultaneously on the same field in a definite pattern. i.e., in different rows.
(c) Bee keeping is the practice to rear the honey bee for obtaining honey; while poultry farming is the practice to raise the domestic fowl for egg and meat production.
18. Why is organic matter important for crop production? (NCERT Exemplar)
Ans. Organic matter is important for crops because
• it helps in improving soil structure.
• it helps in increasing water holding capacity of sandy soil.
• in clayey soil large quantity of organic matter helps in drainage and in avoiding water logging.
19. How is compost different from vermicompost.
Ans. Compost: Compost formation is the process in which farm waste like livestock excreta, vegetable waste, animal refuse, domestic waste, straw and eradicated weeds are decomposed and used as manure.
Vermicompost: This compost is prepared from organic matter by using earthworms, which hastens the process of decomposition.
Long Answer Questions (80–120 words)
1. Discuss the measures to improve food resources through sustainable practices.
Ans. • Crop management involves practices like crop rotation, mixed cropping and intercropping to enhance productivity while maintaining soil fertility.
• Nutrient management focuses on using organic manures, biofertilisers and green manure to reduce dependency on chemical fertilisers and ensure sustainable soil health.
• Pest and weed control includes employing Integrated Pest Management (IPM) techniques to minimise the use of chemical pesticides. Also, mechanical and biological methods are used to control weeds effectively.
• Water management aims to conserve water by adopting efficient irrigation methods such as drip irrigation and sprinkler systems.
• Animal husbandry involves rearing improved breeds of cattle, poultry and fish to sustainably meet the growing food demands.
2. Explain the different types of farming practices in India.
Ans. • Mixed Cropping: Growing two or more crops on the same field to reduce the risk of crop failure (e.g., wheat and mustard)
• Intercropping: Growing crops in a specific row pattern (e.g., 1:1, 1:2) to enhance productivity and reduce pest attacks
• Crop Rotation: Growing different crops in succession to maintain soil fertility (e.g., legumes followed by cereals)
• Organic Farming: Using natural resources like compost, biofertilisers and biological pest control to grow crops
• Integrated Farming: Combining farming with activities like poultry, fishery or beekeeping to maximise income and resource utilisation
3. Describe the composite fish culture system. What is the major problem in fish farming? How is it overcome?
Ans. By adopting composite fish culture systems, intensive fish farming can be done. Both local and imported fish species are used in such systems. In such a system, a combination of five or six fish species is used in a single fish pond. These species are selected in such a way that they have different types of food habits and don’t compete for food among themselves. As a result, the food available in all the parts of the pond is used.
For example: Catlas are surface feeders, Rohus feed in the middle zone of the pond, Mrigals and Common Carps are bottom feeders, and Grass Carps feed on the weeds. Together, these species can use all the food in the pond without competing with each other. This naturally increases the fish yield in the pond.
One problem with such a composite fish culture is that many of these fish breed only during monsoon. Even if fish seed is collected from the wild, it can be mixed with that of other species as well. So, a major problem in fish farming is the lack of availability of good quality seed. To overcome this problem, some ways have been worked out to breed these fish in ponds using hormonal stimulation. This has ensured the supply of pure fish seed in the desired quantities.
4. Why is crop variety improvement important in cultivation? Describe the important factors for which variety improvement is done.
Ans. As we know, weather conditions, soil quality and availability of water are the main factors on which crop yield depends. As weather conditions like drought and floods are unpredictable, it is important to have varieties that can grow in adverse climatic conditions. In the same way, varieties that are tolerant to high soil salinity have also been developed. Some of the factors for which crop variety improvement is done are as follows:
• High Yield: To increase the productivity of the crop per acre
• Improve Quality: Quality considerations of crop products vary from crop to crop. For instance, baking quality is important in wheat, protein quality in pulses, oil quality in oilseeds and preserving quality in fruits and vegetables.
• Biotic and Abiotic Resistance: Crop production can fall due to biotic and abiotic stresses under different situations. Thus, varieties resistant to these stresses can improve crop production.
• Change in Maturity Duration: The shorter the duration of the crop from sowing to harvesting, the more economical the variety. It reduces the cost of crop production and allows the farmers to grow multiple crops in a year.
• Wider Adaptability: Developing varieties for wider adaptability helps stabilise crop production under different environmental conditions. Also, one variety can then be grown under different climatic conditions in different areas.
• Desirable Agronomic Characteristics: Height and profuse branching are desirable characteristics for fodder crops. Dwarfness is desired in cereals such that fewer nutrients are consumed by these crops. Thus, developing varieties of desired agronomic characters also help in higher yield.
5. Explain the advantages and disadvantages of using chemical fertilisers in crop production.
Ans. Advantages:
• Quick Supply of Nutrients: Fertilisers provide immediate nutrients to crops, boosting their growth.
• Higher Yield: They help to increase agricultural productivity, essential to meet food demands.
• Ease of Use: Chemical fertilisers are easy to transport, store and apply.
• Specific Nutrient Addition: Fertilisers like urea (nitrogen-rich) can target specific nutrient deficiencies. Disadvantages:
• Water pollution: Excess fertilisers leach into water bodies, causing eutrophication.
• Health hazards: Residues in crops may harm humans and animals.
• Costly: Frequent use increases production costs for farmers.
6. Discuss the main irrigation systems that are adopted in India.
Ans. Different kinds of irrigation systems are adopted to supply water to agricultural lands depending on the kinds of water resources available. These include wells, canals, rivers and tanks.
• Wells: These are of two types—dug wells and tube wells. In dug wells, water is collected from waterbearing strata, while in tube wells, water is tapped from the deeper strata. From these wells, water is lifted by pumps for irrigation.
• Canals: A canal system is usually an elaborate and extensive irrigation system. Canals receive water from one or more reservoirs or from rivers. The main canal is divided into branch canals with further distributaries to irrigate fields.
• River lift systems: In this system, water is directly drawn from the rivers for supplementing irrigation in areas lying close to rivers. This system is used in areas where canal flow is insufficient or irregular due to inadequate reservoir release.
• Tanks: Tanks are small storage reservoirs, which intercept and store the run-off of smaller catchment areas.
Apart from the above systems, some new initiatives have been undertaken to increase the water available for agriculture. These include rainwater harvesting systems and watershed management systems. This involves building small check-dams which lead to an increase in groundwater levels. These check-dams stop the rainwater from flowing away and also reduce soil erosion.
7. Discuss various methods for weed control.
(NCERT Exemplar) Ans. Various methods for weed control are:
(a) Manual method: Removal of weeds by hand or with the help of simple tools like a hoe
(b) Mechanical removal:
• Ploughing and tilling: Uprooting weeds using agricultural implements like a plough or cultivator before sowing seeds
• Weed cutters and mowers: Using machines to cut or chop weeds in large fields
(c) Chemical method: Chemicals like 2,4-D are sprayed to kill specific weeds without harming crops.
(d) Proper seedbed preparation to avoid weed growth
(e) Timely sowing of crops to avoid the growth of weeds
(f) Intercropping and crop rotation for weed control
(g) Combining multiple methods (manual, mechanical, chemical and biological) to achieve effective weed control while minimising environmental impact
8. Give the merits and demerits of fish culture.
(NCERT Exemplar) Ans.
Merits of Fish Culture
1. Fish are a rich source of high-quality protein, omega-3 fatty acids and essential nutrients.
2. They provide a low-fat alternative to other meat sources.
3. Fish culture allows controlled breeding, feeding and rearing, leading to higher yields than wild fishing.
4. Fish can be cultured in ponds, tanks or reservoirs, utilising land and water resources effectively. Integrated fish farming (with agriculture or poultry) improves resource utilisation.
5. Fish culture generates employment and income for farmers, especially in coastal and rural areas.
Demerits of Fish Culture
1. Setting up fish farms, including pond construction, equipment and seeds, requires significant capital.
2. Fish in dense populations are more prone to infections and diseases, which can cause mass mortality.
3. Waste and uneaten feed can pollute water, causing eutrophication and harming local ecosystems.
4. Fish farming often requires artificial feed, which may increase costs and affect sustainability.
5. Large-scale fish farming demands a consistent supply of clean water, which may lead to conflicts over water resources.
9. Answer the following questions.
(a) Name two common sources for capturing fish.
(b) Why are mussels and shellfish cultivated?
(c) How can the demand for more fish be met when marine fish stock is depleting?
(d) How are marine fish caught?
(e) Name some marine fish cultured in seawater and of high economic value.
Ans. (a) The two common sources from which fish are captured are natural resources, including the sea, oceans and rivers. The second is fish farming in fresh water and marine water.
(b) Mussels have high nutritional value. They are a good source of vitamins and have desirable fatty acids that improve brain function. They also contain zinc which boosts immunity. Shellfish is also a rich source of vitamins, minerals and fish oil, and brings good revenue as seafood.
(c) Mariculture is the practice of cultivating marine organisms for commercial purposes and meeting the ever-increasing fish demand. The advantages of mariculture are:
• It provides food and food products that contribute to the economy.
• It is important for industrial fishing.
• Marine fish, prawns and shellfish are farmed in an open ocean or artificial ponds filled with seawater.
(d) The various ways to catch fish are hand gathering, netting, spearfishing, angling and trapping. For commercial purposes, fishes are captured from high seas by long lines, gill nets, purse seines and bottom trawlers.
(e) The marine fish of high economic value are bhetki, mullet and pearl spot.
10. Write the modes by which insects affect the crop yield.
(NCERT Exemplar)
Ans. Insects affect crop yield through various modes, which can significantly impact agricultural productivity. Here are the modes by which insects affect crop yield:
Feeding of plant parts
• Leaves: Insects like caterpillars and locusts feed on leaves, reducing the plant’s ability to photosynthesise.
• Roots: Root-eating insects like root grubs damage the root system, leading to poor nutrient and water absorption.
• Fruits and seeds: Insects like fruit flies and seed borers attack fruits and seeds, reducing the quality and quantity of the produce.
Sucking of plant sap
• Insects like aphids, whiteflies and leafhoppers suck the sap from plants, weakening them and causing stunted growth.
• This can lead to wilting, reduced vigour and lower crop yields.
Transmitting diseases
• Insects act as vectors by spreading pathogens like bacteria, viruses and fungi.
• Example: Whiteflies transmit the Yellow Mosaic Virus in crops like soybean, and aphids spread viral diseases in vegetables.
Boring of plant tissues
• Borers, such as stem borers or fruit borers, tunnel into stems, fruits or grains, damaging internal tissues and affecting crop development.
• This often leads to secondary infections by pathogens.
Competency Based Questions
Multiple Choice Questions (Choose the most appropriate option/options)
1. Match column A with column B. Choose the correct answer.
9. An experiment was designed to understand the growth requirements of crops. Mustard seeds were chosen and were exposed to different temperature conditions. The table lists the locations of the seeds sown with respective temperature conditions.
Location of the Seeds
A 15°C to 18°C
B 35°C to 38°C
C –1°C to 2°C
D 45°C to 48°C
At which location would mustard grow most effectively?
(a) Location A (b) Location B (c) Location C (d) Location D
10. Crop Y is grown only in a few areas due to specific temperature requirements. To increase the productivity of crop Y, it is recommended to develop its different varieties. Which feature should be included while developing the different varieties of crop Y in order to increase its productivity? (CBSE QB)
(a) Developing varieties with strong biotic resistance
(b) Developing varieties with less dependence on water
(c) Developing varieties with extended maturity duration
(d) Developing varieties adaptable to different climatic conditions
11. Which of these would make a crop resistant to biotic stresses?
(a) Using insecticides to kill insects and other pests
(b) Developing crop varieties that are tolerant to high soil salinity
(c) Developing crop varieties that can grow in scarce water conditions
(d) Growing crops in artificial setups with fixed temperature and moisture content
12. A farmer in town X changed the cropping pattern of the farm. Earlier, only soybean was grown on the farm, but then the farm was divided into rows of different crops. Two rows had soybean and the alternate two rows had maize, and the next two had cowpea. What would be the most likely effect of the new cropping pattern?
(a) Increase in yield
(c) Increased growth of weeds
(b) Degradation of land
(d) Reduced intake of nutrients by crops
13. In order to obtain good quality and quantity of yield in dairy and poultry farming, which of the following management practices should be followed?
(i) Proper housing facilities with hygienic conditions
(ii) Preventing sunlight from entering cattle farms
(iii) Prevention and control of diseases and pests
(iv) Maintaining proper temperature
(a) (i) and (iii) (b) (i), (ii) and (iii) (c) (i), (iii) and (iv) (d) All of the above
14. Which of the following breeds of cattle is considered a dual-purpose breed?
(a) Gir (b) Sahiwal (c) Ongole (d) Red Sindhi
15. Which of the following shows the correct combination for improving crop yield?
16. Which of the following is not true for the Murrah buffalo?
(a) It is a high milk-yielding variety of buffalo. (b) It gives goods quality milk.
(c) It is a very popular exotic breed. (d) It is a breed with a massive body.
17. Four groups capture marine fish with different instruments as listed in the table.
Group Tool/Equipment Used to Capture Marine Fish
A Hand-pulled fishing nets
B Fishing nets guided by echo sounders
C Fishing nets guided by satellites
D Pulley-based fishing nets
Which set of groups will most likely get the maximum fish catch?
(a) Groups A and B
(c) Groups C and D
(b) Groups B and C
(d) Groups D and A
18. The table lists the characteristics of a few Milch cattle breeds of cattle.
Milch Breed of Cattle
Characteristics
Jersey Long lactation period
Sahiwal Resistant to diseases
Brown Swiss Long lactation period
Red Sindhi Resistant to diseases
Based on the characteristics listed in the table, what will be the characteristics of the new breed if Jersey and Red Sindhi are cross-bred? (CBSE QB)
(a) The new breed will have a long lactation period and will be resistant to diseases.
(b) The new breed will have a small lactation period but will be resistant to diseases.
(c) The new breed will have a long lactation period but will not be resistant to diseases.
(d) The new breed will have a small lactation period and will not be resistant to diseases.
19. What is the desirable trait in poultry for developing new varieties?
(a) Tolerance to cooler temperatures (b) Decrease in the number of chicks
(c) Increase in the size of the egg-laying birds (d) Control the occurrence of diseases in chicks
20. The table lists four types of food for broiler chickens.
Type of Food Nutritional Value
P Rich in fats
Q Rich in proteins
R Rich in fats with low levels of vitamin A
S Low levels of fats and vitamin K
Which type of food is desirable for broiler chickens?
(a) Food P (b) Food Q (c) Food R (d) Food S
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
1. Assertion (A): Crop rotation is an effective method to control pests and diseases.
Reason (R): Different crops attract different types of pests, and changing crops prevents pest.
2. Assertion (A): Increasing the use of chemical fertilisers improves soil fertility in the long run.
Reason (R): Continuous use of chemical fertilisers depletes essential soil microorganisms.
3. Assertion (A): Animal husbandry focuses on improving the genetic traits of livestock.
Reason (R): Crossbreeding is a common method to obtain animals with desirable traits.
4. Assertion (A): The use of biofertilisers improves soil fertility and is eco-friendly.
Reason (R): Biofertilisers are chemical substances that enhance the growth of plants.
5. Assertion (A): Organic farming does not use genetically modified seeds.
Reason (R): Organic farming practices are based on the use of natural and traditional methods of farming.
6. Assertion (A): High-yielding variety seeds are drought-resistant and require less irrigation.
Reason (R): High-yielding variety seeds are genetically modified to enhance nutrient uptake.
Case-Based/Source-Based/Passage-Based Questions
1. Study the paragraph and answer the questions.
Mohan is practicing organic farming on his land, avoiding chemical fertilisers and pesticides. Instead, he uses green manure and biofertilisers to improve soil fertility.
To enhance soil health, he also grows pulses such as beans and lentils along with cereals as part of his crop rotation. However, he has observed that the yield of cereals is slightly lower compared to when he used chemical fertilisers. Despite this, Mohan remains committed to organic farming.
(a) Why does organic farming promote the use of green manure and biofertilisers?
(b) How does growing legumes along with cereals help in improving soil fertility?
(c) Why might the yield be lower in organic farming compared to chemical-based farming?
(d) Suggest two ways the farmer can improve the yield while continuing organic farming.
2. Study the paragraph and answer the questions.
A group of scientists developed a new hybrid variety of rice. This variety has a high yield potential and is resistant to diseases. A farmer decides to use this variety in his field and observes an increase in production. However, the farmer also notices that the crop requires more irrigation and fertilisers compared to traditional varieties.
(a) What are the advantages of hybrid rice varieties?
(b) Why does the hybrid variety require more irrigation and fertilisers?
(c) What is the role of fertilisers in increasing crop yield?
(d) Suggest one sustainable practice the farmer can adopt to reduce the environmental impact of using hybrid seeds.
3. Look at the table. It shows the nutritional values of animal products. Study the data and answer the following questions. Animal
Products
(a) Compare the mineral content of the four animal products. Which product has the highest mineral content, and how much higher is it than the one with the lowest?
(i) Egg, 0.30%
(iii) Meat, 0.40%
(ii) Fish, 0.60%
(iv) Milk (Cow), 0.50%
(b) If a person consumes 100 g of each product, which product would provide the maximum combined amount of protein and minerals?
(c) Which two animal products have the closest percentage of water content, and what is the difference between their water percentages?
(d) Which animal product provides vitamins that are unique compared to the other three products? Identify the product and the unique vitamins.
4. Read the paragraph and answer the following questions.
A school organised a workshop on sustainable agriculture practices, where the speaker discussed Integrated Pest Management (IPM) techniques such as using natural predators, crop rotation and biopesticides instead of chemical pesticides.
The speaker highlighted how IPM minimises environmental damage, preserves beneficial organisms like pollinators and reduces the build-up of pesticide resistance in pests. When a student asked how IPM could help reduce crop loss and improve sustainability, the speaker explained all the benefits of IPM.
(a) What is Integrated Pest Management (IPM)?
(b) How does using natural predators in IPM benefit the environment?
(c) Why is crop rotation an important part of IPM?
(d) Suggest two biopesticides that can be used for pest control.
5. Figure shows that two crop fields—Plot A and Plot B—have been treated by manure and chemical fertilisers respectively while keeping other environmental factors the same. Observe the graph and answer the following questions.
(a) Why does plot B show a sudden increase and then a gradual decrease in yield?
(b) Why is the highest peak in the plot A graph slightly delayed?
(c) What is the reason for the different patterns of the two graphs?
(d) What is advantage an disadvantage of chemical fertiliser?
Answers
Multiple Choice Questions
1. (d) Catla are surface feeders while Rohu is a middle zone feeder.
2. (a) Milk production is related to cattle farming.
3. (a) The cross between two varieties is called intervarietal hybridisation. The cross between two species is called interspecific hybridisation.
4. (b) Manure aerates clayey soil by improving its structure, which helps loosen its small pores, and increases the water-holding capacity of sandy soil due to its larger particle size. Also, it is made from animal excretory waste. It is completely organic and replenishes the soil with its nutrient content.
5. (a) Egg production does not apply to cattle farming.
6. (d) Animal husbandry is the scientific management of animal breeding, livestock and rearing. It is a branch of agriculture that involves the domestication, care and breeding of animals for profit.
7. (b) Jersey and Brown swiss are exotic or foreign breeds.
8. (b) Sahiwal, Red Sindhi are local breed of cattle. Aseel meat is Indian breed of chicken.
9. (a) Mustard is a rabi crop, grown in winter.
10. (d) As cultivation of crop Y is restricted to few regions due to specific temperature requirements
11. (a) Biotic stresses include the harm caused to the crops by all living organisms such as insects and other pests.
12. (a) The method is intercropping. This method is useful in suppressing weeds and controlling pests.
13. (c) (i), (iii) and (iv)
Proper sunlight and ventilation are needed in cattle farms.
14. (a) Gir is considered as dual purpose because of its milk production potential and its ability to be used for draught work.
15. (c) Biofertilisers, organic manure, crop rotation are correct combinations for improvement in crop yield.
16. (c) It is a very popular exotic breed, kept for milk production.
17. (b) Echo sounders and satellites provide information regarding the location of fish.
18. (a) The cross breeds have characteristics of both varieties.
19. (d) The desirable traits in poultry for developing new varieties include disease resistance, high egg production, faster growth rate, better feed efficiency, and adaptability to diverse environmental conditions.
20. (b) A broiler is any chicken that is bred and raised for meat production.
Assertion-Reason Based Questions
1. (a) 2. (d) 3. (a) 4. (c) 5. (a) 6. (c)
Case-Based/Source-Based/Passage-Based Questions
1. (a) Organic farming promotes green manure and biofertilisers as they are eco-friendly and improve soil health by providing organic matter and nutrients like nitrogen.
(b) Legumes form a symbiotic relationship with nitrogen-fixing bacteria in their root nodules, which enrich the soil with nitrogen.
(c) Organic farming may result in lower yields initially because natural methods take time to restore soil fertility and do not provide nutrients as quickly as chemical fertilisers.
(d) The farmer can improve yield by:
• Practicing mixed cropping or intercropping with high-yielding varieties.
• Using compost and vermicompost to enhance nutrient content in the soil.
2. (a) Advantages include higher yield potential and resistance to diseases like rust, which reduces crop loss.
(b) Hybrid varieties are genetically designed for higher productivity, which increases their nutrient and water requirements.
(c) Fertilisers provide essential nutrients such as nitrogen, phosphorus and potassium that are required for plant growth and increased yield.
(d) The farmer can adopt drip irrigation and use organic fertilisers to reduce water and chemical use.
3. (a) Fish, 0.60%
Fish has the highest mineral content (1.30%) and milk (cow) has the lowest mineral content (0.70%). The difference is 1.30% - 0.70% = 0.60%.
(b) Meat
Milk (Cow): Protein 4.00% + Minerals 0.70% = 4.70%, Egg: Protein 13.00% + Minerals 1.00% = 14.00%, Meat: Protein 21.10% + Minerals 1.10% = 22.20%, Fish: Protein 19.00% + Minerals 1.30% = 20.30%
(c) Egg and Meat, 0.00%
Both egg and meat have the same water content of 74.00%, so the difference is 0.00%.
(d) Fish, Niacin, D and A
4. (a) IPM is an eco-friendly approach to controlling pests using biological, cultural and mechanical methods, reducing the reliance on chemical pesticides.
(b) Natural predators like ladybugs and spiders control pests without harming the environment or beneficial organisms.
(c) Crop rotation breaks the life cycle of pests by altering the host crop, reducing pest populations naturally.
(d) Two biopesticides are neem oil and Bacillus thuringiensis (Bt).
5. (a) With the addition of chemical fertilisers, there is a sudden increase in yield due to the release of nutrients N, P and K in high quantities. The gradual decline in the graph may be due to continuous use and high quantity of chemicals which kill microbes useful for replenishing the organic matter in the soil. This decreases the soil fertility.
(b) Manure slowly supplies small quantities of nutrients to the soil as it contains large amounts of organic matter.It enriches soil with nutrients thereby increasing soil fertility continuously.
(c) The difference in the two graphs indicates that the use of manure is beneficial for long durations in cropping as the yield tends to remain high when the quantity of manure increases. In the case of plot B, the chemical fertilisers may cause various problems when used continuously for a long time. Loss of microbial activity reduces the decomposition of organic matter and as a result, soil fertility is lost affecting the yield.
(d) Advantage: Chemical fertilisers provide essential nutrients to plants quickly, boosting crop yield and growth.
Disadvantage: Excessive use can degrade soil health, reduce microbial activity, and cause environmental pollution.
Practice Questions
Multiple Choice Questions
1. Composite fish culture involves combining different species in one pond. Which feature is crucial for selecting species in this method?
(a) All fish must eat the same type of food
(b) All fish must grow to the same size
(c) Fish must have different food habits to avoid competition
(d) Fish must breed only during monsoon
2. Which of the following is an advantage of intercropping?
(a) Increases the use of chemical fertilizers
(b) Ensures better utilization of nutrients and prevents pest spread (c) Allows only one type of crop to grow effectively
(d) Reduces crop yield
3. Cyperinus and Parthenium are types of (a) diseases (b) pesticides (c) weeds (d) pathogens
4. Poultry feed is rich in (a) vitamin A, vitamin B and Zinc (b) carbohydrates and folic acid (c) vitamin A, vitamin K and Protein (d) fat, protein and vitamin C
5. Which of the following elements do not come under the category of macronutrients? (I) Phosphorous (II) Manganese (III) Magnesium (IV) Iron (a) Both (I) and (II) (b) Only (II) (c) Both (II) and (IV) (d) Both (II) and (III)
Assertion-Reason Based Questions
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
6. Assertion (A): Organic farming practices result in higher yields compared to conventional farming using chemical fertilisers.
Reason (R): Organic farming enhances soil health and maintains ecological balance by avoiding chemical inputs.
7. Assertion (A): Poultry birds are vaccinated to increase their growth rate and egg production.
Reason (R): Vaccination protects poultry from diseases, reducing mortality and maintaining productivity.
Very Short Answer Questions (30-50 words)
8. Differentiate between inland fishery and marine fishery.
9. Give one word for the following.
(a) Farming without the use of chemicals as fertilisers, herbicides and pesticides is known as (b) Growing of wheat and groundnut on the same field is called as (c) Planting soybean and maize in alternate rows in the same field is called as (d) Growing different crops on a piece of land in pre-planned succession is known as
10. Give two advantages of mixed cropping.
11. Define the term hybridisation and photoperiod.
Short Answer Questions (50-80 words)
12. Why is organic matter important for crop production? Mention three points.
13. What mineral nutrients are supplied to the plants by air, water and soil?
14. Differentiate between the following.
(a) Capture fishery and Culture fishery
(b) Mixed cropping and Intercropping
15. How do you classify manure? How is green manure different from vermicompost?
16. Look at the picture. Identify the fish.
17. Explain genetic manipulation. How is it useful in agricultural practices?
Long Answer Questions (80-120 words)
18. Answer the following questions.
(a) Explain the process of intercropping with the help of a diagram.
(b) Give two benefits of crop rotation.
19. Cultivation practices and crop yield are related to weather, soil quality and availability of water. Justify the statement.
20. Differentiate between manure and fertilisers. Mention the advantages of manure.
Brain Charge
1. Complete the gaps in the following flow chart.
Step 1: ___________ (Improving crop variety through hybridisation)
Step 2: ___________ (Providing sufficient water through irrigation)
Step 3: ___________ (Using organic manure or fertilisers for soil enrichment)
Step 4: ___________ (Protecting crops from weeds, pests and diseases)
Step 5: ___________ (Harvesting crops at the right time for better yield)
2. Read the paragraph and complete the gaps.
To ensure food _____1_____, modern agricultural practices focus on enhancing ____2 and livestock productivity. Farmers adopt methods like organic farming, crop rotation, and mixed cropping to maintain soil fertility. The use of _____3_______ vermicomposting, and manure reduces dependency on chemical _____4_____ while sustaining soil health. In addition, ______5______using biopesticides helps protect crops. The role of ______6 _______ such as drip irrigation and sprinkler irrigation is crucial for efficient water _______7_______. Furthermore, aquaculture, _______8________, and sericulture contribute to diversified food sources.
Fig. 12.3
1. Oil yielding plant (9)
3. Crop grown in winter season (4)
5. Fixed by Rhizobium (8)
9. Common honey bee (4)
4. Solve the following riddles.
2. Animal feed (6)
4. A micronutrient (5)
6. Unwanted plant in a crop field (4)
7. An exotic breed of chicken (7)
8. Bottom feeders in fish pond (7)
(a) I live in the soil, and you’ll find me near, I make the soil rich and help crops appear.
Farmers love me, as I am green and pure, To chemical fertilisers, I’m the cure.Who am I?
(b) I am a practice that protects plants, From weeds and pests, I take a stand. I use rotation and natural foes, My goal is sustainability that grows.Who am I?
Challenge Yourself
1. What do you mean by hypophysation? What are its advantages?
Scan me
2. How does organic farming, though requiring more labour and time, contribute to long-term food security in comparison to conventional farming methods that rely heavily on chemical inputs? Consider both economic and environmental factors in your response.
3. Mixed cropping is practiced to reduce the risk of crop failure. Which of the following is the most suitable example of mixed cropping?
(a) Wheat and Mustard
(c) Maize and Cotton
(b) Rice and Sugarcane
(d) Bajra and Groundnut
4. Which of the following is NOT a component of Integrated Pest Management (IPM)?
(I) Use of chemical pesticides as the first resort
(II) Biological control using natural predators
(III) Use of pest-resistant crop varieties
(IV) Monitoring pest populations regularly
(a) Both (I) and (II) (b) Only (I)
(c) Both (I) and (III) (d) Both (II) and (IV)
5. The table below lists different methods of improving crop yield and their purposes.
Methods
Crop rotation
Use of pesticides
High-yielding varieties
Drip irrigation
Maintaining soil fertility
Purpose
Protecting crops from pests
Increasing production of grains
Efficient use of water resources
Which method is primarily used to ensure efficient water usage in crop production? (a) Crop rotation (b) Use of pesticides (c) High-yielding varieties (d) Drip irrigation
4. (a) Green Manure (b) Integrated Pest Management (IPM)
Challenge Yourself
1. Hypophysation includes the extraction of hormones from the pituitary glands of donor fishes and injecting the same in carps in captivity, that is, either in hatcheries or ponds.
Advantages:
• Carps breed in rivers and not in captivity. By hypophysation, carps can be made to breed in captivity.
• We get healthy and pure seeds for fish farming, which ensures the supply of the same in the desired quantity.
2. Organic farming contributes to long-term food security by fostering healthier soil, reducing environmental pollution and promoting biodiversity.
3. (a) 4. (b) 5. (d) Scan me for Exemplar Solutions
SELF-ASSESSMENT
Time: 1.5 Hour
Scan me for Solutions
Max. Marks: 50
Multiple Choice Questions (10 × 1 = 10 Marks)
1. Look at the picture. The Italian bee, named Apis mellifera, is the most common breed domesticated in India.
Fig. 12.4
Which of the following is not a reason for the preferential use of this breed?
(a) It is gentle in nature.
(b) It has the ability to protect itself from enemies.
(c) It has good honey collection capacity.
(d) It has a prolific queen with more swarming.
2. Which of the following statements is incorrect regarding animal husbandry practices?
(a) Crossbreeding improves milk yield and disease resistance in cattle.
(b) Poultry birds require no vaccination since they have natural immunity.
(c) Proper housing and sanitation reduce the chances of diseases in livestock.
(d) Fodder quality directly affects the productivity of farm animals.
3. Drones in the honeybee colony are stingless and are unable to gather food. Their main role is to mate with the queen bee and help in breeding.
Drones are born from:
(a) Fertilised eggs and well-nourished larvae
(b) Unfertilised eggs
(c) Fertilised eggs giving heat treatment
(d) Same way as the worker bee is born
4. In order to obtain good quality and quantity of yield in dairy and poultry farming, which of the following management practices should be followed?
(I) Proper housing facilities with hygienic conditions
(II) Preventing sunlight from entering cattle farms
(III) Prevention and control of diseases and pests
(IV) Maintenance of proper temperature
(a) (I) and (III)
(c) (I), (III) and (IV)
(b) (I), (II) and (III)
(d) All of the above
5. Among the following statements, which might not be the correct statement about the Murrah breed buffalo?
(a) It is a high milk-yielding variety of buffalo.
(b) It gives goods quality milk.
(c) It is a very popular exotic breed.
(d) It is a breed with a massive body.
6. Identify which of the following are not infectious diseases of dairy animals?
(I) Ringworm (II) Foot and mouth disease (III) Salmonellosis (IV) Anaemia (a) (I) and (II) (b) (II) and (III) (c) (I) and (IV) (d) (I) and (III)
7. Which of the following are exotic breeds?
(I) Brawn (II) Jersey (III) Brown Swiss (IV) Jersey Swiss (a) (I) and (III) (b) (II) and (III) (c) (I) and (IV) (d) (II) and (IV)
8. Which among the following is a kharif crop?
(a) Soybean (b) Wheat (c) Mustard (d) Linseed
9. To enhance fish seed quality, fish are bred in ponds using: (a) Hormonal modification (b) Gene transfer (c) Hormonal stimulation (d) Gene abbreviation
10. Weeds are unwanted plants that grow alongside cultivated plants. How do they affect the crop plants?
(a) By killing plants in the field before they grow
(b) By dominating the plants to grow
(c) Competing for various plant resources causing low availability of nutrients
(d) All of the above
Assertion–Reason Based Questions (4 × 1 = 4 Marks)
In each of the following questions, a statement of Assertion (A) is given by the corresponding statement of Reason (R). Of the statements, mark the correct answer as
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
11. Assertion (A): Overuse of pesticides can lead to the development of resistant pest strains.
Reason (R): The continuous use of a single pesticide increases the survival rate of pests that are resistant to that chemical.
12. Assertion (A): The use of synthetic fertilisers has led to an increase in food production worldwide.
Reason (R): Synthetic fertilisers provide an immediate supply of essential nutrients to plants, enhancing their growth in a short period.
13. Assertion (A): Organic farming is less productive than conventional farming in the long run.
Reason (R): Organic farming often involves more labour-intensive methods and lower yields due to the absence of synthetic chemicals.
14. Assertion (A): The Green Revolution led to long-term environmental sustainability in agriculture.
Reason (R): The Green Revolution introduced high-yielding varieties of crops and increased the use of synthetic fertilisers and pesticides.
15. Read the following paragraph and answer the following questions.
The Green Revolution, which began in the 1960s, introduced high-yielding varieties (HYVs) of crops, particularly wheat and rice, along with the use of chemical fertilisers, pesticides and advanced irrigation techniques. These innovations significantly increased food production in many parts of the world, particularly in Asia. However, while the Green Revolution helped achieve food security in the short term, it also led to environmental challenges such as soil degradation, water scarcity and loss of biodiversity. Additionally, the heavy reliance on chemical inputs created long-term sustainability issues, as it led to the development of pest resistance and soil fertility depletion.
(a) What was the main outcome of the Green Revolution in the 1960s?
(i) Decreased food production in Asia
(ii) Significant increase in food production, particularly in Asia
(iii) Decline in the use of chemical fertilisers
(iv) Introduction of organic farming techniques
(b) Which of the following environmental challenges was a result of the Green Revolution?
(i) Soil improvement and biodiversity increase
(ii) Soil degradation, water scarcity and loss of biodiversity
(iii) Increased crop yields with no environmental impact
(iv) Sustainable farming practices
(c) What long-term issue was caused by the heavy reliance on chemical inputs during the Green Revolution?
(i) Increased pest resistance and soil fertility depletion
(ii) Improved crop diversity
(iii) Decrease in water usage for irrigation
(iv) Development of sustainable farming methods
(d) Which of the following was a key feature of the Green Revolution's approach to agriculture?
(i) Use of high-yielding varieties of crops (HYVs)
(ii) Emphasis on organic fertilisers and pesticides
(iii) Focus on reducing water usage in agriculture
(iv) Shift to traditional farming practices
16. Read the following paragraph. Answer the questions.
Dev practices mixed farming on his land. He has recently ventured into animal husbandry, apiculture and pisciculture to diversify his income. He has a small herd of cows for dairy production, a few beehives for honey production and a small fish pond for fish farming.
He follows best practices, ensuring that his cows are well-fed with a balanced diet and clean water. He uses modern techniques in apiculture, maintaining healthy beehives and monitoring the bees' activity. In pisciculture, he ensures proper water quality management, feeds the fish nutritious food and provides an environment free of pollutants. However, he faces challenges such as managing pests in the fish pond, maintaining the health of the bees and dealing with fluctuating feed costs for his cattle.
(a) What is one of the key practices Dev follows for his cows?
(i) Adding artificial fertilisers to cow food
(ii) Ensuring a balanced diet and clean water for the cows
(iii) Using pesticides for better milk production
(iv) Keeping cows in confined spaces for high milk yield
(b) Which of the following is a challenge Dev faces in pisciculture?
(i) Maintaining high soil fertility
(ii) Managing pests in the fish pond
(iii) Controlling the humidity in beehives
(iv) Managing fluctuating feed costs for cows
(c) In apiculture, Dev is focused on:
(i) Increasing the bee population through artificial means
(ii) Maintaining healthy beehives and monitoring bee activity
(iii) Using chemical pesticides to protect the hives
(iv) Storing honey for long-term commercial use
(d) Which practice would help Dev improve the sustainability of his farming activities in the long term?
(i) Reducing water quality management in pisciculture
(ii) Overfeeding cows for higher milk yield
(iii) Practicing integrated farming techniques by balancing animal husbandry, apiculture and pisciculture
(iv) Using synthetic fertilisers and pesticides in all farming activities
Very Short Answer Questions (30-50 words)
17. What are the different types of fisheries? Explain.
18. What is apiculture? Name a few products obtained from apiculture.
19. Cultivation practices and crop yield are related to environmental conditions. Explain.
Short Answer Questions (50-80 words)
(3 × 2 = 6 Marks)
(4 × 3 = 12 Marks)
20. Why has improving crop production become more important these days? List the major group of activities for improving crop yields. Which one of the activities is the most important and why?
21. What are the three advantages of beekeeping? Explain.
22. Enlist some desirable traits for crossbreeding poultry birds.
23. Differentiate between compost and vermicompost.
Long Answer Questions (80-120 words)
(2 × 5 = 10 Marks)
24. List two desirable agronomic traits that are considered for crop improvement. How can farmers introduce these traits into new crop varieties? Mention two conditions that must be met for the acceptance of these newly developed varieties.
25. Answer the following questions.
(a) Look at the diagram. It shows the culture of bees.
(i) List any two factors to consider when selecting a site for beekeeping.
(ii) How does beekeeping benefit agriculture?
(b) Explain the factors to be considered before deciding the nature of feed for cattle.
(c) S represents nuclear pores that regulate the transport of molecules between the nucleus and the cytoplasm of eukaryotic cells.
(d) T refers to the nucleolus which is responsible for ribosome production.
16. (a) Cytosol is the fluid present in the cell and is a constituent of the cytoplasm.
(b) In prokaryotic cells, the true nucleus is absent. Also, membrane-bound organelles are absent.
(c) No, the nucleoplasm is a part of the nucleus.
(d) Protoplasm includes the cytoplasm and the nucleus of the cell.
Chapter 6
1. (a) 2. (c) 3. (a) 4. (d) 5. (c)
11. (a) 12. (a) 13. (a) 14. (d)
15. (a) (ii) Vessels
(b) P refers to tracheids. Tracheids are elongated, dead cells in the xylem that help in:
(c)
(a)
• Water conduction – Transport water and minerals from the roots to different parts of the plant.
• Mechanical support – The thick, lignified walls provide strength and rigidity to the plant.
• Preventing water loss – The narrow lumen reduces water loss by slowing down water movement.
(c) Both tracheids (P) and vessels (Q) are components of xylem and help in water conduction and mechanical support in plants.
(d) R refers to xylem parenchyma. Xylem parenchyma is the only living component of xylem, and it plays an essential role in:
• Storage – Stores starch, fats and other nutrients.
• Lateral transport – Helps in the sideways movement of water and minerals.
• Wound healing – Assists in repair and regeneration of xylem tissues.
16. (a) Skeletal muscle
(b) During anaerobic respiration, there is an accumulation of lactic acid.
(c) During rest, oxygen neutralises the lactic acid and relives muscle fatigue.
(d) Skeletal muscles: Striated, voluntary, multinucleated, attached to bones and responsible for body movements.
Smooth muscles: Non-striated, involuntary, uninucleated, found in the walls of internal organs and responsible for peristalsis and other involuntary movements.
Chapter 7
(b)
16. (a) aX = 2 m/s2 and aY = 1.5 m/s2 (b) 500 m (c) v avg = 15m/s
17. 10 s 19. (b) 20. 0.5 m/s2 22. 19.6 m, 19.6 m/s
24. (a) 2 m/s2 (b) 36 m (c) -3 m/s2 (d) 180 m
25. (a) 3.2 m/s-2 (b) 16 m/s (c) 10 s
Chapter 8
1. (b) 2. (c) 3. (a) 4. (d) 5. (b)
11. (d) 12. (a) 13. (a) 14. (b)
15. (a) (iv) (b) (iii) (c) (i) (d) (ii)
16. (a) (ii) (b) (iv) (c) (i) (d) (ii)
18. 1 g·cm/s2 22. 2 m/s2 23. a = 2 m/s2; v = 10 m/s; S = 25 m
25. a = 5 m/s2 , t = 2 s; a n = 3 m/s2 , t = 3.33 s
Chapter 9
1. (a) 2. (c) 3. (b) 4. (b) 5. (d) 6. (d)
11. (a) 12. (a) 13. (d) 14. (d)
15. (a) (ii) (b) (ii) (c) (iv) (d) (iv) Both (b) and
16. (a) 2.45 ms–2 (c) 2R
19. u = 50 m/s and t = 5 sec
20. (a) 7.8 N (b) 1 N (c) 6.8 N
22. 2 sec; 60 m from the ground
23. (b) mass – 1 : 1 and weight – 1 : 6 (c) 50 kgm/s
Read the following instructions very carefully and strictly follow them
I. All questions would be compulsory.
Max Marks: 80
II. Section A would have 16 simple/complex MCQs and 4 Assertion-Reasoning type questions carrying 1 mark each.
III. Section B would have 6 Short Answer (SA) type questions carrying 2 marks each.
IV. Section C would have 7 Short Answer (SA) type questions carrying 3 marks each.
V. Section D would have 3 Long Answer (LA) type questions carrying 5 marks each.
VI. Section E would have 3 source based/case based/passage based/integrated units of assessment (4 marks each).
SECTION-A
Question 1 to 16 are multiple choice questions. Only one of the choices is correct. Select and write the correct choice as well as the answer to these questions. They are one mark each.
1. Which of these is common for all chemical changes?
(a) Change in shape (b) Absorption of heat (c) Increase in volume (d) Formation of a new substance
2. What would be the valency of an element that is chemically inactive? (a) 0 (b) 1 (c) 2 (d) 5
3. To prepare iron sulphide, by heating a mixture of iron filings and sulphur powder, we should use a: (a) Copper dish (b) China dish (c) Watch dish (d) Watch glass
4. Choose the correct statement of the following.
(a) Conversion of solid into vapours without passing through the liquid state is called vaporisation. (b) Conversion of solid into vapour without passing through the liquid state is called sublimation.
(c) Conversion of vapours into solid without passing through the liquid state is called freezing. (d) Conversion of solid into liquid is called sublimation.
5. The proteins and lipids, essential for building the cell membrane, are manufactured by: (a) rough endoplasmic reticulum (b) golgi apparatus (c) plasma membrane (d) mitochondria
6. When we change feeble sound to loud sound we increase its: (a) frequency (b) amplitude (c) velocity
7. Amoeba acquires its food through a process, termed:
(a) Exocytosis
(c) Plasmolysis
(d) wavelength
(b) Endocytosis
(d) Both exocytosis and endocytosis
8. Which of these involves the conversion of kinetic energy to potential energy?
(a) A person diving into a pool of water from a board.
(b) A person gliding in the air with the help of a parachute.
(c) A person sliding down from the top of a water slide.
(d) A person riding a motorbike to the top of an overbridge.
Question No. 17 to 20 consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, and R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.
17. Assertion (A): Solids do not diffuse in air.
Reason (R): The particles are closely packed in solids.
18. Assertion (A): Lysosomes are known as suicidal bag of cells.
Reason (R): Lysosomes contain powerful enzymes capable of breaking down all organic material.
19. Assertion (A): Motion of satellites around their planets is considered an accelerated motion. Reason (R): During their motion, the speed remains constant, while the direction of motion changes continuously.
20. Assertion (A): Atom is electrically neutral.
Reason (R): A neutral particle, neutron, is present in the nucleus of an atom.
SECTION-B
Question No. 21 to 26 are very short answer type questions. They are 2 marks each.
21. (a) Is ice water a homogeneous or heterogeneous substance? Is it pure or impure? (b) Is fresh air free of dust particles and impurities of all other kinds, a pure substance?
22. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer. OR
A boy of mass 40 kg runs up a flight of 50 steps each 10 cm high in 5 second. Find (a) the work done by the boy (b) the power developed. (g = 9.8 ms2)
23. Why do substances undergo change in physical state? Give one example.
24. When a vertical jerk is given to a string, transverse waves are formed. Give three features of these waves.
25. A person takes a concentrated solution of salt, after sometime, he starts vomiting. What is the phenomenon responsible for such a situation? Explain.
26. Differentiate between voluntary and involuntary muscles. Give one example of each type.
SECTION-C
Question No. 27 to 33 are short answer questions. They are 3 marks each.
27. State the law of inertia. Why do we fall in forward direction if a moving bus stops suddenly and fall in the backward direction if it suddenly accelerates from rest?
28. (a) Define power. Derive its SI unit.
(b) An electric bulb is rated 15 watts. What does it mean?
(c) What is the energy consumed in joules if it is used for 10 minutes?
29. (a) How can we say that sugar is a pure substance whereas milk is not?
(b) Which of the following materials fall in the category of a pure substance? (i) Ice (ii) Iron (iii) Wood (iv) Brick
30. Draw a plant cell and label the parts which (a) determines the function and development of the cell.
(b) packages materials coming from the endoplasmic reticulum.
(c) provides resistance to microbes to withstand hypotonic external media without bursting.
31. Give reasons for
(a) Meristematic cells have a prominent nucleus and dense cytoplasm but they lack vacuole.
(b) Intercellular spaces are absent in sclerenchymatous tissues.
(c) We get a crunchy and granular feeling, when we chew pear fruit.
32. Verify by calculating that
(a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.
(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3 : 5.
33. Write three differences between prokaryotic and eukaryotic cells.
SECTION-D
Question No. 34 to 36 are long answer questions. They are 5 marks each.
34. How does the water change into vapours at a temperature below its boiling point? List the factors affecting evaporation. Mention two examples from daily life where evaporation causes cooling.
OR
(a) Explain the term diffusion. Illustrate with an activity that rate of diffusion increases with temperature.
(b) Name two compressed gases.
(i) used in our homes for cooking.
35. Analyse the reason behind the following statements:
(ii) supplied to the hospital in cylinders.
(a) Epidermis is thicker in desert plants though it is usually single-layered.
(b) Presence of a waxy layer (secreted by epidermis) on the outer surface of plants.
(c) Discuss the cell arrangement which supports the fact that epidermis is a protective tissue.
Name the following tissues:
(a) That forms the inner lining of our mouth.
(c) Found in the iris of the eye.
(e) Epithelial tissue present on the tongue.
OR
(b) Present in the brain.
(d) That connects two bones.
36. (a) How does the force of attraction between the two bodies depend upon their masses and distance between them? A student thought that two bricks tied together would fall faster than a single one under the action of gravity. Do you agree with his hypothesis or not? Comment.
(b) Suppose gravity of earth suddenly becomes zero, then in which direction will the moon begin to move if no other celestial body affects it?
SECTION–E
Question No. 37 to 39 are Science-based/case-based questions. They are 4 marks each
37. Study the following graph and choose the correct options to answer the following questions given below:
The velocity-time graph of an object is shown in the following figure.
(a) State the Kind of motion that objects have, from A to B and from B to C.
(b) What does the area enclosed by the velocity-time graph represent?
(c) Identify the part of the graph where the object has zero acceleration. Give reasons for your answer.
Identify the part of the graph where the object has negative acceleration. Give reasons for your answer.
38. Read the following text carefully and answer the questions that follow:
India has the maximum cattle population in the world. However, their productivity is less than half the productivity of many exotic breeds of cattle. The exotic breeds live in cold countries. They cannot live comfortably in hotter India. The only way to improve the productivity of Indian cattle is to produce hybrids which are acclimatised to Indian conditions and are resistant to most local diseases. For hybridisation exotic bulls are kept in colder climates. Their semen is collected and cryopreserved. The same is sent to various parts of the country for artificial insemination.
(a) What is artificial insemination?
(b) State one advantage of artificial insemination.
(c) Mention some factors governing the yield of milk.
OR
When can cattle be inseminated?
39. In the given activity below, on heating the solution, water evaporates, and we get back the ink dye in the watch glass. The different substance has a different boiling point. We use this property to separate the components of the mixture. Here, the boiling point of ink is much higher than that of water. On heating the ink solution, water evaporates while ink dye remains in the china dish.
(a) Name the process shown in the diagram.
(b) Which type of substance can be separated by this method?
(c) Is ink a pure substance? Justify your answer.
(d) Name the component which gets evaporated.
Sample Question Paper – 1 (Solved)
Answer Key
SECTION-A
SECTION-B
21. (a) Ice water is heterogeneous because it has both a solid and liquid phase. It is a pure substance since ice is made up of water only.
(b) No, air is a mixture of gases and so, not a pure substance.
22. When a satellite moves around the Earth in a circular path, then the force of gravity acts on it directed towards the centre. The motion of the satellite is in the horizontal plane. Therefore, the force of gravity of Earth on the satellite and the direction of motion of the satellite are perpendicular to each other. Therefore, net work done = Fs cos 90 = 0. That is, the work done by the force of gravity on a satellite moving around the Earth is zero.
OR
The boy has to overcome the force of gravity. Hence force of gravity on the boy
F = mg = 40 × 9.8 = 392 N
Total distance covered s = 50 × 10 = 500 cm = 5 m
(a) Work done by the body in climbing (W) = force (F) × distance (s) = W = 392 × 5 = 1960 J
(b) Power developed = 1960 392 5 W W t ==
23. Substances undergo change in physical state because both inter-particle spaces and inter-particle forces can be changed by changing the conditions of temperature and pressure. Ex: Melting of ice.
24. Three features of transverse waves are:
(a) The particles of the medium vibrate at right angles to the direction of propagation of the wave.
(b) Transverse waves travel in the form of crests and troughs.
(c) They cannot travel through a vacuum.
25. When a person consumes a highly concentrated salt solution, the concentration of salt in the stomach becomes higher than that inside the cells of the stomach lining. This creates an osmotic imbalance, leading to exosmosis, where water moves out of the cells (which have a lower concentration of salt) to the surrounding area (which has a higher concentration of salt). As a result, the cells in the stomach lining lose water and become dehydrated. This dehydration triggers the body's natural response of vomiting to expel the excess salt and restore fluid balance.
26. Voluntary muscles can be moved by conscious will when we want them to move. For example, muscles of limbs or skeletal muscles. Involuntary muscles function on their own. We cannot start or stop them from working by our desire. For example, cardiac muscles and smooth muscles.
SECTION-C
27. Law of inertia: An object remains in its state of rest or of uniform motion in a straight line until an external unbalanced force acts on it. When a moving bus stops suddenly, the bus slows down but our body tends to remain in the state of forward motion due to inertia of motion, hence we experience a forward jerk.
Sudden start of bus brings motion to the bus as well as our feet but rest of the body still has inertia of rest due to which we fall backward.
28. (a) Power is the rate of doing work.
Power = Work Time
1 water or 1 W = 1 Joule 1 Second
(b) If the power of an electric bulb is 15 W it consumes 15 Joules of energy per second.
(c) Energy consumed by the bulb in 10 minutes = P × t = 15 W × 600s = 9000 Joules
29. (a) Sugar is a pure substance because it cannot be separated and is formed of only a single type of molecule. In the case of milk, it can be separated by physical process into its components. It has components like water, fat and proteins, etc.
(b) Ice and iron are pure substances as they contain particles of only one kind of matter while wood and brick contain more than one kind of matter.
30. Diagram of plant cell with the labels: (a) Nucleus (b) Golgi apparatus (c) Cell wall
31. (a) Meristematic cells are actively dividing cells, so they require a prominent nucleus to control cell division and dense cytoplasm to provide necessary materials and enzymes for metabolism. They lack vacuoles because vacuoles mainly serve storage functions and are not needed during division process.
(b) Sclerenchyma cells have very thick, lignified cell walls and are tightly packed to provide mechanical strength to the plant. Intercellular spaces are absent because any space between cells would weaken the tissue and reduce its ability to provide support.
(c) This is due to the presence of sclereids (stone cells), a type of sclerenchyma cells with thick and lignified walls found in pear fruit. These hard cells give the crunchy and gritty texture when chewing.
32. (a) CO2 has molar mass = 44 g/mol
5 moles of CO2 has molar mass = 44 × 5 = 220 g
H2O has molar mass = 18 g/mol
5 moles of H2O have mass = 18 × 5 g = 90 g
Hence, their molar mass differs.
(b) Number of moles in 240g Ca metal = 240 40 = 6
Number of moles in 240g of Mg metal = 240 40 = 10
Ratio = 6 10 which is equal to 3 : 5.
33. Difference between Prokaryotic and Eukaryotic cells.
Feature
Size
Nuclear Region
Generally small (1–10 μm)
Contains only nucleic acid; undefined due to absence of nuclear membrane; called nucleoid
34. The phenomenon of change of a liquid into vapour at a temperature below its boiling point is called evaporation. Fractions of particles at the surface having higher kinetic energy, are able to break away from the forces of attraction of other particles and get converted into vapour.
Factors which affect the rate of evaporation are:
• Surface area
• Temperature
• Humidity
• Wind speed
Two examples from daily life where evaporation causes cooling:
Sprinkling of water on the roof, cooling of water kept in earthen pots, etc.
OR
(a) Diffusion is the movement of particles from a region of higher concentration to a region of lower concentration until they are evenly spread out.
Activity to show diffusion rate increases with temperature:
(i) Take two identical jars filled with water, one with cold water and the other with hot water.
(ii) Add a few crystals of potassium permanganate (or any colored substance) to each jar simultaneously.
(iii) Observe how the purple color spreads in both jars. The color spreads faster in hot water than in cold water.
This shows that the rate of diffusion increases with temperature because particles move faster at higher temperatures.
(b) (i) Used in our homes for cooking: Liquefied Petroleum Gas (LPG), Propane
(ii) Supplied to hospitals in cylinders: Oxygen (O₂), Nitrous Oxide (N₂O)
35. (a) In desert habitats, water loss is a major challenge due to extreme heat and dryness. To reduce water loss, desert plants have a thicker epidermis compared to plants in other habitats. This thicker layer provides enhanced protection and helps conserve water by minimizing evaporation.
(b) The waxy covering aids in protecting the plant against loss of water, mechanical injury and invasion by parasitic fungi.
(c) Epidermis is the outermost covering of cells in plants. It is usually made up of a single layer of cells. On aerial parts of a plant, epidermal cells often create a waxy, water-resistant layer on their outer surface to prevent loss of water from the plant.
The cells of epidermis are present in a continuous layer without intercellular spaces. Small pores are present on the epidermis of the leaf. These pores are called stomata, which help in gaseous exchange and transpiration. As the plant grows older, a strip of secondary meristem replaces the epidermis in the stem and forms a thick cork further emphasizing the protective role.
(a) Epithelial tissue-squamous epithelium
(b) Nervous tissue
(c) Involuntary muscular tissue.
(d) Ligament
(e) Stratified squamous epithelium
OR
36. From Newton’s law of gravitation, force of attraction between two bodies is directly proportional to the product of their masses, and inversely proportional to the square of distance between their centres. This relationship is given by Newton’s law of gravitation:
F = G(m1 × m2)
r2
where F = gravitational force, m1 and m2 = masses of the two bodies, r = distance between centres, and G = gravitational constant.
(a) A student's hypothesis is not correct. The two bricks, like a single body, fall with the same speed to reach the ground at the same time in case of free-fall. This is because acceleration due to gravity is independent of the mass of the falling body.
(b) If Earthʼs gravity suddenly vanished, the Moon would no longer be pulled towards Earth. It would move tangentially (in a straight line) to its orbit at the point where gravity ceased, following Newton’s first law of motion. This means the Moon would continue moving forward in the direction it was moving at that instant, flying off into space in a straight line.
SECTION-E
37. (a) Uniform motion from A to B and non-uniform motion from B to C.
(b) Displacement
(c) AB because velocity remains constant from A to B. In this interval, velocity is constant (horizontal line), so acceleration = 0. No change in velocity over time means no acceleration. OR
BC because velocity decreases from B to C. The velocity decreases with time in this section (downward sloping curve), which indicates negative acceleration (retardation).
38. (a) Artificial insemination is a reproductive technique in which there is introduction of semen into the vagina or cervix of a female by any method other than sexual intercourse.
(b) It increases the potential for genetic selection as it enables selective breeding by using semen from genetically superior males, thereby improving desirable traits in the offspring and increasing overall productivity.
(c) Some factors governing the yield of milk are:
(i) Number of milkings per day
(ii) Amount of milk at each milking
(iii) Length of the lactation period, i.e., the time span over which the animal produces milk. OR
Cattle should be inseminated only during the estrus (heat) period, when the female is fertile and capable of conceiving.
39. (a) Evaporation
(b) The substances which have differences in their boiling points can be separated by this method.
(c) No, ink is not a pure substance. It is a mixture of various coloured dyes.
(d) It is the water which gets evaporated.
Sample Question Paper – 2 (Solved)
Time Allowed: 3 hours
General Instructions:
Read the following instructions very carefully and strictly follow them
I. All questions would be compulsory.
Max Marks: 80
II. Section A would have 16 simple/complex MCQs and 4 Assertion-Reasoning type questions carrying 1 mark each.
III. Section B would have 6 Short Answer (SA) type questions carrying 2 marks each.
IV. Section C would have 7 Short Answer (SA) type questions carrying 3 marks each.
V. Section D would have 3 Long Answer (LA) type questions carrying 5 marks each.
VI. Section E would have 3 source based/case based/passage based/integrated units of assessment (4 marks each).
SECTION-A
Question 1 to 16 are multiple choice questions. Only one of the choices is correct. Select and write the correct choice as well as the answer to these questions. They are one mark each.
1. Latent heat of fusion is the amount of heat energy required to change:
(a) 1 kg of solid into its gaseous state at its boiling point
(b) 1 kg of liquid into its gaseous state at its boiling point
(c) 1 kg of solid into its liquid state at its melting point
(d) 1 kg of liquid into its solid state at its freezing point
2. Argon has atomic number 18 and mass number 40. Calcium has atomic number 20 and mass number 40. Calcium and argon are examples of:
3. Distance between two alternate crests in a wave is equal to. (a) λ 2 (b) λ (c) 3λ 2 (d) 2λ
4. A permanent slide shows thin-walled isodiametric cells with a large vacuole. The slide contains (a) Parenchyma cells (b) Nerve cells (c) Sclerenchyma cells (d) Collenchyma cells
5. The number of hydrogen atoms present in one gram of hydrogen molecule is approximately: (given weight of one hydrogen atom = 1.67 × 10−27 kg) (a) 3 × 1023 (b) 6 × 1023 (c) 12 × 1023 (d) 1023
6. In which one of the following cases is work scientifically said to be done?
(a) Mira is studying for her exam.
(b) Manish pulls the trolley up to a certain distance.
(c) Manoj pushes the wall with no change in the position of the wall.
(d) Aruna standing at a bus stop holding a bag in her hand.
7. Which of the following statements is true for colloids?
(a) A colloid is a homogeneous mixture.
(b) Particles of a colloid can be seen by the naked eye.
(c) Particles of a colloid scatter a beam of light passing through it.
(d) All the above.
8. Valency of hydrogen is 1 and that of sulphur is 2 . What should be the formula of hydrogen sulphide?
(a) HS (b) HS2
(c) H2S (d) H2S2
9. The given figure shows an important cell organelle. This organelle is regarded as the of the cell.
(a) Suicidal bag
(b) Powerhouse
(c) Food factory
(d) Storage bag
10. When a ball is thrown vertically upwards its velocity keeps on decreasing. What happens to its kinetic energy when it reaches half of the maximum height?
(a) Kinetic energy is maximum.
(c) Kinetic energy is zero.
(b) Kinetic energy is half the initial value.
(d) Kinetic energy is double.
11. Which of the following is a likely advantage of composite fish culture?
(a) Fish grow better when different species live together.
(b) All areas of the pond are utilised for better fish production.
(c) Fish eat less food when grown in composite culture ponds.
(d) Dissolved oxygen level is higher in composite culture ponds
12. If the velocity of a body changes uniformly from u to v in time t, then what will be the value of vavg t + a ?
(a) 3v – u t (b) v – 3u t (c) 3v – u
13. According to the law of conservation of mass, mass of reactants will be equal to the mass of:
(a) Products
(c) Gases evolved
(b) Catalysts
(d) Apparatus used for reaction
14. It is due to which one of the following reasons that a body floats in a liquid?
(a) When the weight of the body is exactly half of the weight of the liquid displaced.
(b) When the weight of the body is greater than the weight of the liquid displaced.
(c) When the weight of the body is equal to the weight of the liquid displaced.
(d) When the weight of the body is less than the weight of the liquid displaced.
15. A and B are plant tissues. Cells of tissue A have small inter-cellular spaces and walls are thickened at the corners. Cells of tissue B do not have inter-cellular spaces and the cell walls are thickened. Identify tissues A and B.
16. The sound wave is produced by a vibrating tuning fork shown in the diagram below. Half the wavelength is represented by:
Question No. 17 to 20 consist of two statements – Assertion (A) and Reason (R). Answer these questions by selecting the appropriate option given below:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true
17. Assertion (A): It is difficult to balance our body when we accidentally slip on a peel of banana.
Reason (R): According to Newton's third law of motion "Every action have an equal and opposite reaction."
18. Assertion (A): Epithelial tissues are covering or protective tissues.
Reason (R): Materials are exchanged at the surface across the epithelial tissues.
19. Assertion (A): An atom is the smallest particle in the element that has the properties of element.
Reason (R): Molecules are formed by the combination of two or more atoms.
20. Assertion (A): Chromosomes are constituted by RNA and proteins.
Reason (R): Chromosomes are thread-like structures present in the nucleus of a cell.
SECTION–B
Question No. 21 to 26 are very short answer questions. They are 2 marks each.
21. Differentiate between speed and velocity.
22. Define:
(a) Law of constant proportion
(b) Law of conservation of mass
23. The given figure represents a tissue found in the human body.
(a) Identify the tissue and state its function.
(b) Mention any two locations where this tissue is found in the body.
24. An object of volume V is fully immersed in a liquid of density ρ. Calculate the magnitude of buoyant force acting on the object due to the liquid. Mention the direction of buoyant force.
25. How does crop rotation help maintain soil fertility?
26. Nishant is a lab technician working in a pharmaceutical company, and he has to study different types of colloids used in drug formulations. He needs to take different samples based on the combinations as given below. Suggest an example for each.
(a) Liquid and gas
(b) Liquid and solid Give one example of each.
SECTION-C
Question No. 27 to 33 are short answer questions. They are 3 marks each.
27. We eat food composed of nutrients like carbohydrates, proteins, fats, vitamins, minerals, and water. After digestion, these are absorbed in the form of glucose, amino acids, fatty acids, and glycerol. What mechanisms are involved in the absorption of digested food and water?
28. (a) Why is large part of the atom neglected when calculating the mass of the atom?
(b) Give reason why atoms combine with each other.
(c) In the notation of an atom, how are the atomic number, mass number and symbol of the element written?
OR
(a) Calculate formula unit mass of K2SO4
(b) Write the formula of oxide for an element with valency 3?
29. The given figure represents a tissue found in the human body.
(a) Identify the tissue. Give reason for your identification.
(b) State its location in the body.
(c) Is it voluntary or involuntary in nature?
30. Deduce the formulae for below compounds.
(a) Ammonium sulphate
(b) Sodium phosphate
(c) Carbon dioxide
31. A 10 kg ball is thrown upward with a velocity of 5m/s.
(a) Find the kinetic energy of ball at its initial position.
(b) Find its potential energy when it reaches the highest point.
(c) Calculate the maximum height the ball reaches. (g = 10 m/s2).
32. What are the different ways of improving crop production?
33. (a) List the parameters which characterise a sound wave.
(b) Given below are graphical wave shapes of two sound waves. Identify the sound which has:
Wave disturbance
Wave disturbance
Give reason for your answer.
(c) A sound wave has a frequency of 2 kHz and wavelength 35 cm. Find its velocity.
(A)
(B)
SECTION-D
Question No. 34 to 36 are long answer questions. They are 5 marks each.
34. Sejal is a graduate student presenting at a science conference, and her topic is the historical development of atomic models. During her presentation, she focuses on Niels Bohr and his contributions to atomic theory. The audience is particularly interested in understanding how Bohr's model improved upon earlier models of the atom.
Answer the following questions:
(a) What were Bohr's key contributions to the understanding of atomic structure?
(b) Describe Bohr's model of the atom in detail, including a neat and labelled diagram.
OR
(a) Describe Rutherford's model of an atom.
(b) Give reasons:
(i) After a hot sunny day, people sprinkle water on the roof or open ground.
(ii) A wooden table should be called a solid.
35. (a) The given figure shows a type of animal tissue.
(i) Identify the tissue.
(ii) Label parts 1 and 2.
(iii) State the characteristic features of this tissue.
(b) Give reasons:
(i) Intercellular spaces are absent in sclerenchymatous tissues.
(ii) Meristematic cells have a prominent nucleus and dense cytoplasm, but they lack vacuole.
OR
(a) Imagine you are a scientist studying the lining of the small intestine. You notice that it is composed of cells that are tall and narrow, with microvilli on their surface. What type of epithelial tissue is this, and what is the likely function of the microvilli?
(b) Suresh was studying a plant species that grows very rapidly. He noticed that it has a thick layer of actively dividing cells near the tips of its roots and shoots. What type of meristematic tissue is this, and what is its role in plant growth?
36. (a) Why does the rider fall in the forward direction when a running horse stops suddenly?
(b) Why is it easier to stop tennis ball in comparison to a cricket ball moving with the same speed?
(c) An athlete always runs some distance before taking a jump. Why?
(d) Which is having a higher value of momentum - A bullet of mass 10 g moving with a velocity 400 m/s or or a cricket ball of mass 400 g thrown with the speed of 90 km/h?
SECTION–E
Question No. 37 to 39 are Source-based/Case-based questions. They are 4 marks each.
37. The following diagram shows a simple pendulum consisting of a bob of mass
100 g. Initially the bob of the pendulum is at rest at 0. It is then displaced to one side at A. The height of A above 0 is 5 cm (Take g = 10 m/s2)
(a) What is the value of potential energy of bob at A and where does it come from?
(b) What is the value of total energy of the bob at position A?
(c) What is the value of kinetic energy of the bob at mean position O ?
(d) What is the value of kinetic energy and potential energy of the bob at position P whose height above O is 2 cm?
38. We know that the gases are highly compressible as compared to solids and liquids. The liquefied petroleum gas (LPG) cylinder that we get in our home for cooking, or the oxygen supplied to hospitals in cylinders is compressed gas. Now-a-days compressed natural gas (CNG) is used as fuel in vehicles. The liquid takes up the shape of the container in which they are kept. Liquids flow and change shape, so they are not rigid but can be called fluids. Solids and liquids can diffuse into liquids. The aquatic animals can breathe underwater. The rate of diffusion of liquids is greater than solid.
Solids
Liquids
(a) (i) How aquatic plants and animals can breathe underwater?
Gases
(ii) Among solids, liquids and gases which can be termed as fluids and why?
(b) Why compressed natural gas is used as fuel in vehicles these days?
39. Suppose a plant cell or a group of cells (beetroot cells) like the one in figure (A) is placed in a very concentrated solution of sucrose. After few minutes, it will appear as in figure (B). This condition is said to be plasmolysed. Based on this, answer the following questions:
Cellulose Cell wall
Cell Membrame
Cytoplasm
(a) Which of the two, the cell wall or the cell membrane, acts as a semipermeable membrane?
(b) In the given experiment, what fills the space between the cell wall and the cell membrane in the plasmolysed cell?
(c) Suppose you are given both, fresh potato cells and fresh beetroot cells. Which cells would you prefer for demonstrating plasmolysis? Give reason. OR
Define plasmolysis.
Vacuole
(A)
(B)
Sample Question Paper – 2 (Solved)
Answer Key
SECTION-A
SECTION-B
21. Speed Velocity
1. is the rate of change of distance with respect to time.
2. It has only magnitude.
3. Speed does not depend on direction of motion.
4. It is always positive or zero.
22. (a) Law of constant proportion:
1. Velocity is the rate of change of displacement with respect to time.
2. It has both magnitude and direction.
3. Velocity depends on the direction of motion.
4. It can be positive, negative, or zero depending on direction.
According to the law of constant proportion, in a chemical substance the elements are always present in a definite proportion by mass.
(b) Law of conservation of mass:
According to the law of conservation of mass, the mass can neither be created nor destroyed in a chemical reaction.
Total Mass of the Reactant = Total Mass of the Product
23. (a) The figure shows simple columnar epithelial tissue. Its main functions include absorption and secretion.
(b) Simple columnar epithelial tissue forms the lining of the stomach, small intestine, colon, gall bladder and oviducts.
24. Volume of the immersed body = Volume of the liquid displaced = V
Mass of the liquid displaced = vp
Thus, weight of the liquid displaced = buoyant force = Vpg
The buoyant force acts in the upward direction.
25. Crop rotation helps maintain soil fertility by alternating crops with different nutrient requirements, such as cereals followed by legumes.
Benefits of crop rotation
• Improves soil health
• Reduces the need for chemical fertilisers and pesticides
• Improves crop emergence, growth and health
• Reduces the risk of pests and disease
• Increases biodiversity on the farm
26. (a) Aerosol, e.g., clouds
(b) Gel, e.g., jelly
SECTION-C
27. The mechanisms involved in the absorption of digested food and water are diffusion and osmosis, respectively.
• Absorption of digested food occurs through diffusion. Diffusion is a process in which molecules move from their region of higher concentration to their region of lower concentration till they are uniformly distributed throughout the available space.
• Absorption of water occurs through osmosis. Osmosis involves the passage of water from a region of higher water concentration through a semi-permeable membrane to a region of lower water concentration. In this case, from small intestine (high concentration) to blood (low concentration).
28. (a) The large or extra nuclear part of the atoms contains electrons which have negligible mass. Heavier particles like protons and neutrons are contained in the nucleus. So, only nucleus is considered when calculating the mass of an atom.
(b) The atoms combine with other atoms to achieve the electronic configuration of nearest noble gas and thus to become more stable.
(c) In the notation of an atom, the atomic number (Z), mass number (a), and symbol of the element (X) are to be written as: X A Z
(a) Formula unit mass of K2SO4 = (2 × Atomic mass of potassium) + Atomic mass of Sulphur + (4 × Atomic mass of oxygen)
= (2 × 39u) + 32u + (4 × 16u)
= 78u + 32u + 64u
= 174u
(b) An element with valency 3 is aluminium.
The formula of aluminum oxide = AI2O3
29. (a) The given figure shows cardiac muscle tissue because the cells are striated, uninucleated and branched.
(b) Cardiac muscles are found in the walls of the human heart.
(c) Cardiac muscles contract and relax continuously. We do not have control over their working. Hence, they are involuntary in nature.
30. (a) Ammonium sulphate
• Ammonium ion = NH₄⁺
• Sulphate ion = SO₄²⁻
To balance the charges, 2 ammonium ions are needed for 1 sulphate ion:
Ammonium sulphate formula = (NH₄)₂SO₄
(b) Sodium phosphate
• Sodium ion = Na⁺
• Phosphate ion = PO₄³⁻
To balance the charges, 3 sodium ions are needed for 1 phosphate ion:
Sodium phosphate formula = Na₃PO₄
(c) Carbon dioxide
• Carbon has valency 4
• Oxygen has valency 2
To attain noble gas configuration, 1 carbon combines with 2 oxygen atoms:
Carbon dioxide formula = CO₂
31. (a) Given that,
m = 10kg, u = 5 m/s
KE = 1 2 mu2 = 1 2 × 10 × 5 × 5 = 125 J
(b) At the highest point, all the kinetic energy gets converted to potential energy.
∴ PE = 125 J
(c) Using 3rd equation of motion, v2 – u2 = 2as
h = v2 – u2 2 g = 02 – 52 2 × (–10) = 25 20
∴ h = 1.25 m
32. The different ways of improving crop production are:
• Crop production management: Efficient use of resources like irrigation, fertilisers and proper farming practices
• Crop protection management: Protecting crops from pests, diseases and weeds using chemical, biological or cultural methods
33. (a) Pitch, Loudness, quality and Speed characterizes sound waves.
(b) (i) B has higher pitch because it has higher frequency. (ii) A has more loudness because it has larger amplitude.
(c) v = vλ = 2000 × 0.35 = 700 m/s
SECTION-D
34. (a) Bohr developed the Bohr model of the atom, in which he proposed that energy levels of electrons are discrete and that the electrons revolve in stable orbits around the atomic nucleus but can jump from one energy level (or orbit) to another.
Bohrʹs Model of an Atom
• Niels Bohr revised Rutherford's atomic model and put forth the following suggestions:
• Niels Bohr proposed that the electrons possess a specific amount of energy which allows them to revolve around the nucleus.
• An atom contains discrete orbits which correspond to specific amount of energy. Hence, these orbits are also known as energy levels.
• The energy levels of an atom are represented as K, L, M, N and so on or the numbers n = 1, 2, 3, 4 and so on. Electrons
(b) Niels Bohr's Atomic Model
• Electrons are confined to these energy levels. While revolving in these discrete orbits, the electrons do not radiate energy. Hence, these orbits are also known as stationary orbits or stationary shells. Smaller the size of the orbit, smaller is its energy.
• As we move away from nucleus, the energy of the orbit increases progressively.
• The transfer of an electron from one orbit to another is always accompanied with absorption or emission of energy.
• When an electron jumps from a lower energy level to a higher energy level, it absorbs energy.
• When an electron returns from a higher energy level to a lower energy level, it emits energy.
OR
(a) Rutherford's Atomic Model
• Based on the results of the α-particle scattering experiments, Rutherford put forth his atomic model.
• An atom contains a positively charged centre called the nucleus of the atom. Almost all the mass of the atom is concentrated in the nucleus.
• The electrons of the atom revolve around the nucleus in fixed, circular orbits.
• The size of the nucleus is many times smaller than the size of the atom. The nucleus of an atom is 10,000 times smaller than the atom.
(b) (i) After a hot sunny day, people sprinkle water on the roof or open ground because the large latent heat of vaporisation of water helps to cool the hot surface.
(ii) Particles of a wooden table are rigid and have a fixed location. They also possess a definite shape and volume. Because of these properties, we should call a wooden table a solid substance.
35. (a) (i) Cartilage (Connective tissue)
(ii) 1 - Matrix; 2 - Chondroblast/Chondrocyte
(iii) Cartilage supports the external ear, nose, etc. It is an elastic tissue.
(b) (i) Sclerenchyma tissues are dead simple permanent tissues of the plant. The cells of sclerenchyma are closely packed without intercellular spaces, so that they can provide strength, rigidity, flexibility, and elasticity to the plant to withstand various strains. The presence of lignified cell walls makes the cells compact and leaves no intercellular spaces.
(ii) Meristematic cells are actively dividing cells. Therefore, they have a prominent nucleus, dense cytoplasm, and a thin cell wall. Vacuoles possess cell sap and provide rigidity and turgidity to the cell. The presence of vacuoles might pose a problem for cell division in meristematic cells. Hence, they lack vacuole.
OR
(a) The epithelial tissue in the small intestine is likely to be columnar epithelium. This type of epithelium is characterised by its tall, column-shaped cells, which are wellsuited for absorption and secretion. The microvilli on the surface of these cells significantly increase the surface area for absorption. This increased surface area allows for more efficient uptake of nutrients from the digested food. Therefore, the primary function of the microvilli is to enhance nutrient absorption in the small intestine.
(b) The meristematic tissue found in the tips of roots and shoots is called apical meristem. This type of meristem is responsible for primary growth in plants.
Apical meristems contain actively dividing cells that produce new cells, leading to the elongation of roots and shoots. This growth is essential for plants to access water, nutrients, and sunlight. Additionally, apical meristems contribute to the formation of new organs, such as leaves and flowers.
36. (a) The rider falls in the forward direction when a running horse stops suddenly because of inertia of motion which is the tendency to be in state of motion and continue even when the horse stops suddenly.
(b) Mass is the measure of inertia. A cricket ball has larger mass than a tennis ball. Thus, it has larger inertia than the tennis ball. Therefore, the cricket ball needs a larger force to be stopped.
(c) If the athlete runs some distance before taking the jump, then inertia of motion helps the athlete to continue the state of motion which aids him to take a longer jump.
(b) Given that,
Mass of Bullet = 10 g = 0.01 kg
Velocity = 400 m/s
Momentum of bullet = m × v = 0.01 × 400 = 4 kg m/s
Mass of Cricket ball = 400 g = 0.4 kg
Velocity = 90 km/h = 5 18 × 90 = 25 m/s
Momentum of ball = 0.4 × 25 = 10 kg m/s
Thus, the cricket ball has higher momentum than the bullet.
SECTION-E
37. (a) PE = mgh = 0.1 × 10 × 5 × 10–2 = 0.05 J
The work done in raising the bob through a height of 5 cm (against the gravitational attraction) gets stored in the bob in the form of its potential energy.
(b) At position A, PE = 0.05 J, KE = 0
So, Total energy = 0.05 J
(c) At mean position, potential energy is zero, hence KE at O = PE at A = 0.05 J
(d) PE at P = mgh = 0.1 × 10 × 2 × 10–2 = 0.02 J
K.E Total energy – PE = 0.05 – 0.02 = 0.03 J
38. (a) (i) Aquatic plants and animals can breathe underwater because atmospheric oxygen gas dissolves in water due to diffusion.
(ii) Only liquids and gases can be termed fluids because they both have the unique property of being fluids, which is that they can flow.
(b) Compressed natural gas (CNG) is used as fuel in vehicles these days because,
(i) CNG has high compressibility,
(ii) Large volumes of a gas can be compressed into a small cylinder,
(iii) It can be transported easily.
39. (a) Cell membrane acts as a semipermeable membrane. Cell wall is freely permeable.
(b) In the given experiment, the space between the cell wall and the cell membrane in the plasmolysed cell is filled by sucrose solution.
(c) Beetroot cells contain soluble sugar, whereas potato cells contain insoluble starch. Osmotic processes readily occur in beetroot cells than in potato cells. Hence, fresh beetroot cells are preferred over fresh potato cells as a material for demonstrating plasmolysis.
OR
(c) Plasmolysis occurs due to the presence of sucrose solution. The cell membrane withdraws itself from the cell wall and the cell gets plasmolysed.
Plasmolysis is the process of contraction or shrinkage of the protoplasm of a plant cell caused due to the loss of water in the cell.
Sample Question Paper – 3 (Unsolved)
Time Allowed: 3 hours
General Instructions:
Read the following instructions very carefully and strictly follow them
I. All questions would be compulsory.
II. Section A would have 16 simple/complex MCQs and 4 Assertion-Reasoning type questions carrying 1 mark each.
III. Section B would have 6 Short Answer (SA) type questions carrying 2 marks each.
IV. Section C would have 7 Short Answer (SA) type questions carrying 3 marks each.
V. Section D would have 3 Long Answer (LA) type questions carrying 5 marks each.
VI. Section E would have 3 source based/case based/passage based/integrated units of assessment (4 marks each).
SECTION-A
Question 1 to 16 are multiple choice questions. Only one of the choices is correct. Select and write the correct choice as well as the answer to these questions. They are one mark each.
1. If the tip of the sugarcane plant is removed from the field, even then it keeps on growing in length. It is due to the presence of:
(a) Auxins
(c) Lateral meristem
2. The slope of a velocity-time graph gives:
(a) Distance
(c) Acceleration
(b) Apical meristem
(d) Intercalary meristem
(b) Displacement
(d) Speed
3. The gravitational force between two objects is F. If masses of both objects are halved without changing the distance between them, then the gravitational force would become: (a) F 4 (b) F² (c) F (d) 2F
4. In which of the following conditions would the distance between the molecules of hydrogen gas increase?
(i) Increasing pressure on hydrogen contained in a closed container.
(ii) Some hydrogen gas leaking out of the container
(iii) Increasing the volume of the container of hydrogen gas
(iv) Adding more hydrogen gas to the container without increasing the volume of the container (a) (i) and (ii) (b) (i) and (iv) (c) (ii) and (iii) (d) (ii) and (iv)
5. Which fish species is widely used in composite fish culture due to its ability to feed at the bottom of the pond?
(a) Catla (b) Rohu (c) Mrigal (d) Hilsa
6. The pictures show the arrangement of electrons in the shells of different atoms.
Which atom has the highest valency?
(a) Atom 1 (b) Atom 2 (c) Atom 3 (d) Atom 4
7. Which of the following does not lose its nucleus at maturity?
(a) Companion cells
(b) Red blood cells
(c) Vessel (d) Sieve tube cells
8. In case of negative work, the angle between the force and displacement is: (a) 0° (b) 45° (c) 90° (d) 180°
9. A sample of pure water, irrespective of its source, contains 11.1% hydrogen and 88.9% oxygen. The data supports:
11. Dry ice is often used for refrigeration and creating fog effects. Chemically, it is:
(a) solid SO2 (b) solid NO2
(c) solid CO2 (d) solid H2
12. Which structure in a plant cell is responsible for providing the energy required to drive cellular processes?
(a) Chloroplast
(c) Nucleus
(b) Mitochondrion
(d) Golgi apparatus
13. A person met with an accident in which two long bones of the hand were dislocated. Which among the following may be the possible reason?
(a) Tendon break
(c) Ligament break
(b) Cartilage
(d) Areolar tissue break
14. A plant cell placed in a hypotonic solution will not burst because of the presence of: (a) Plasma membrane (b) Cell wall (c) Chloroplast (d) Cytoplasm
15. Which of the following statements is/are correct?
1. Mass of an object is the measure of its inertia.
2. Heavier the object, smaller is the inertia.
3. The mass of an object is variable.
(a) Only 1 (b) 1 and 3
(c) 2 and 3
16. The sound waves having a frequency more than 20,000 Hz are called:
(a) Infrasonic waves
(c) Ultrasonic waves
(d) 1 and 2
(b) Intersonic waves
(d) Hypersonic waves
Question No. 17 to 20 consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true
17. Assertion (A): A gas exerts pressure on the walls of the container.
Reason (R): Rate of diffusion of gases is more than that of liquids.
18. Assertion (A): Permanent tissue is composed of mature cells.
Reason (R): Meristematic tissue is a group of actively dividing cells.
19. Assertion (A): On the Moon, humans feel lighter than on Earth.
Reason (R): It is due to more gravitational force exerted by the Moon on man.
20. Assertion (A): Guard cells are specialised epidermal cells.
Reason (R): Stomata are found in the epidermis of leaves.
SECTION-B
Question No. 21 to 26 are very short answer type questions. They are 2 marks each.
21. State why Newton’s first law of motion is called law of inertia.
22. What is endoplasmic reticulum? Name the two types of endoplasmic reticulum, and write their main functions.
OR
Name the tissue which helps in transportation of oxygen that we inhale to various parts of our body. Write the composition of this tissue.
23. A man pushes four boxes of different masses. The table shows the acceleration produced for each box during the push.
What amount of force does the man exert on each box? Is the force acting on each box unbalanced? Explain your answer.
24. How can you test the purity of a given substance?
25. “The atomic mass of an element is in fraction.” What does it mean?
26. What are the three advantages of shorter duration of the crop in between sowing and harvesting?
SECTION-C
Question No. 27 to 33 are short answer questions. They are 3 marks each.
27. After winters, people pack off their woollens by keeping naphthalene balls in them. With passage of time these balls become smaller in size.
(a) Why does this happen?
(b) What type of change is involved during this process?
(c) How can you convert a saturated solution into an unsaturated solution?
28. A body can have zero average velocity but not zero average speed. justify giving an example.
OR
A car is moving with an initial speed of 10 m/s on a straight road. The driver accelerates, causing the car to move with a uniform acceleration of 2 m/s². The driver stops accelerating once the car reaches a speed of 108 km/h. Calculate the duration for which the car accelerated and the distance it covered during this time.
29. How is a bacterial cell different from an onion peel cell?
OR
Why do plant cells possess large sized vacuoles?
30. (a) Name the different components of xylem.
(b) Which structure protects the plant body against the invasion of parasites?
31. Differentiate between transverse and longitudinal waves. Also, provide examples of each wave.
32. (a) Helium atom has 2 electrons in its valence shell but its valency is not 2, Explain.
(b) Write any two observations which support the fact that atoms are divisible.
33. (a) Why do we hear the sound produced by the humming bees while the sound of vibrations of pendulum is not heard?
(b) If any explosion takes place at the bottom of a lake, what type of shock waves in water will take place?
SECTION-D
Question No. 34 to 36 are long answer questions. They are 5 marks each.
34. (a) What are the consequences of the following conditions?.
(i) A cell having higher water concentration than the surrounding medium.
(ii) A cell having lower water concentration than the surrounding medium.
(iii) A cell having equal water concentration to its surrounding medium.
(b) Name the materials of which the cell membrane and cell wall are composed of.
35. Define isotopes. Why do isotopes have the same atomic number but different mass number? Explain with the help of an example.
36. A vehicle of 1 tonne travelling with a speed of 60 m/s notices a cow on the road 9 m ahead and applies brakes. It stops just in front of the cow.
(i) Find out the kinetic energy (KE) of the vehicle before applying brakes.
(ii) Calculate the retarding force provided by the brakes.
(iii) How much time did it take to stop after the brakes were applied?
(iv) What is the work done by the braking force?
SECTION–E
Question No. 37 to 39 are Science-based/case-based questions. They are 4 marks each
37. The following data represents the distribution of electrons, protons, and neutrons in atoms of three elements A, B, and C. Understand the data carefully and answer the following questions.
(a) What is the atomic number of element A? Also, Identify the element.
(b) What will be the atomic mass number of element B?
(c) Atomic number of element C is 12. Identify the element and write its electronic configuration. OR
What will be the valency of element A? Justify your answer.
38. Observe the diagram and answer the following below:
F1 = 20 N
F2 = 30 N
Two forces F1 = 20 N and F2 = 30 N are acting on an object as shown in figure:
(a) Find the net force acting on the object.
(b) State the direction of the net force acting on the object.
(c) If the object still does not move under the application of these forces, what can be the possible reason for this?
OR
Why is no force required to move an object with a constant velocity?
39. It is necessary for both plants and animals to perform the necessary life processes for their survival. Division of labour with their body has helped to reduce the workload from a single cell. It increases the cell efficiency and ensures greater organization within the body of an organism. Hence organism can easily survive in the given or changing surrounding conditions.
However, the structure of plant and animal tissues is quite different. This is largely because of the functions they perform and type of external environment they are subjected to. For example, plants are not as mobile as animals.
They are fixed at a place. So they require less energy than animals. But they do require supportive tissues which provide mechanical strength to the body of the plant and require less or no energy (i.e., they are made of dead cells). Animals, on the other hand, have more complex structural organization as compared to plants.
Plants grow indefinitely for their whole life while animals grow upto a given period of time. Growth of plants occurs due to the presence of meristematic tissues in root and shoot apex. However, in animals, the growth of the body is quite uniform and proportionate to all body parts. Hence animals don’t possess any growing and non-growing regions in their body.
(i) Which meristem helps in increasing the girth?
(ii) Which tissue provide flexibility and mechanical support to the plant organs?
(iii) What are parenchyma containing air cavities called?
(iv) What is the wall of sclerenchyma cells made up of?
Which types of cells are most likely to divide?
Sample Question Paper – 4 (Unsolved)
Time Allowed: 3 hours
General Instructions:
Read the following instructions very carefully and strictly follow them
I. All questions would be compulsory.
Max Marks: 80
II. Section A would have 16 simple/complex MCQs and 4 Assertion-Reasoning type questions carrying 1 mark each.
III. Section B would have 6 Short Answer (SA) type questions carrying 2 marks each.
IV. Section C would have 7 Short Answer (SA) type questions carrying 3 marks each.
V. Section D would have 3 Long Answer (LA) type questions carrying 5 marks each.
VI. Section E would have 3 source based/case based/passage based/integrated units of assessment (4 marks each).
SECTION–A
Question 1 to 16 are multiple choice questions. Only one of the choices is correct. Select and write the correct choice as well as the answer to these questions. They are one mark each.
1. Look at the following figure and suggest in which of glass container the rate of evaporation will be highest?
(a) II (b) I (c) IV (d) III
2. The proteins and lipids, essential for building the cell membrane, are manufactured by, (a) Mitochondria
(b) Golgi apparatus
(c) Rough endoplasmic reticulum
(d) Plasma membrane
3. Bones are connected to muscles at the joints by (a) tendon (b) adipose tissue (c) areolar tissue (d) Ligament
4. A permanent slide shows thin-walled isodiametric cells with a large vacuole. The slide contains (a) Parenchyma cells (b) Nerve cells (c) Sclerenchyma cells (d) Collenchyma cells
5. Area under a velocity-time graph gives (a) the displacement (b) the acceleration (c) the distance (d) the speed
6. An archer shoots an arrow. Consider the action force to be the bowstring against the arrow. What will be the reaction force?
(a) Arrow pushing against the bowstring.
(b) Weight of the arrow
(c) Air resistance against the bow
(d) Grip of the archer’s hand on the bow
Moving Fan
(I) (II)
(III)
(IV)
7. The given figure shows parenchyma tissue. Identify parts 1-4.
8. The statement ‘cells arise only from pre-existing cells’ was given by:
(a) Louis Pasteur (b) Schwann (c) Schleiden (d) Rudolf Virchow
9. When a ball is thrown vertically upwards its velocity keeps on decreasing. What happens to its kinetic energy when it reaches the maximum height?
(a) Kinetic energy is maximum at maximum height.
(b) Kinetic energy is half the initial value.
(c) Kinetic energy is zero.
(d) Kinetic energy is double.
10. Which of the following solutions has the highest mass by mass percentage?
(a) 20 g of sodium carbonate in 90 g of water
(b) 15 g of sugar in 160 g of water
(c) 10 g of sodium chloride in 200 g of water
(d) 60 g of potassium permanganate in 200 g of water
11. When a bird flies off a branch, the branch starts swinging. Which of the following laws best describes the above statement?
(a) Newton’s first law (b) Newton’s second law
(c) Newton’s third law (d) Law of Inertia
12. It is due to which one of the following reasons that a body floats in a liquid?
(a) When the weight of the body is exactly half of the weight of the liquid displaced.
(b) When the weight of the body is greater than the weight of the liquid displaced.
(c) When the weight of the body is equal to the weight of the liquid displaced.
(d) When the weight of the body is less than the weight of the liquid displaced.
13. The atoms usually exist in nature as:
(a) In the form of molecules (b) In the form of ions
(c) In a free state (d) In the form of molecules and ions
14. A car travels 10 m in 5 seconds, 20 m in the next 10 seconds, and 30 m in the last 10 seconds. The average speed of the motion is:
(a) 30 ms–1
15. The cells of the ciliated epithelium have cilia at their free ends which constantly keep lashing and move the materials which enter these regions. The ciliated epithelium is most likely to be found in the (a) Walls of the heart (b) Lining of the trachea (c) Sweat glands (d) Around the muscles, blood vessels and nerves
16. Which of the following pair of elements represents a mole ratio of 1:1?
(a) 7 g of nitrogen and 12 g of sodium
(b) 20 g of sodium and 20 g of calcium
(c) 14 g of nitrogen and 24 g of magnesium
(d) 10 g of calcium and 6 g of carbon
Question No. 17 to 20 consist of two statements - Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true, and R is the correct explanation of A (b) Both A and R are true, and R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true
17. Assertion (A): It is difficult to balance our body when we accidentally slip on a peel of banana.
Reason (R): According to Newton’s third law of motion “Every action have an equal and opposite reaction.”
18. Assertion (A): Watt hour is unit of energy.
Reason (R): Kilo-watt hour is unit of electric power.
19. Assertion (A): Golgi bodies store, modify and pack products in vesicles.
Reason (R): They are involved in the formation of lysosomes.
20. Assertion (A): The sensation of sound persists in our brain for about 0.1 s.
Reason (R): The sound returning towards the source after getting reflected from a distant obstacle is called an echo.
SECTION-B
Question No. 21 to 26 are very short answer questions. They are 2 marks each.
21. Explain how sound is produced by your school bell.
22. How much water should be mixed with 12 mL of alcohol so as to obtain a 12% alcohol solution?
23. Figures A, B and C represent cross-section of three different types of plant tissues.
(a) Which tissue – A, B or C, provides mechanical strength and flexibility to the plant body?
(b) Which tissue – A, B or C, can be modified to form air cavities in aquatic plants?
24. Explain law of conservation of energy with an example.
25. An egg sinks in fresh water but floats in highly salty water. Give reason.
OR
(a) What is the force between the Earth and a body called?
(b) What will happen to the force of gravitation between two objects if the distance between them is reduced to half?
26. Raghu observed that the growth of weeds in his field reduced the crop yield. What can the farmer do to prevent the growth of weeds without investing money on weedicides?
SECTION-C
Question No. 27 to 33 are short answer questions. They are 3 marks each.
27. Kalpesh is a chemistry student working on a project to understand the concept of formula unit mass in ionic compounds. His teacher has assigned him the following task to demonstrate his understanding. Calculate the formula unit masses of the following compounds:
• Zinc oxide (ZnO)
• Sodium oxide (Na2O)
(Atomic mass of Zn = 65u, Na = 23u, 0 = 16u)
28. (a) What remains unchanged in the oscillation of a pendulum if it is released from point A?
(b) Take the lowest position of pendulum (B) during the motion as reference point for potential energy, the height of the pendulum from reference point at any time t as h and its speed at that time as v. Derive an expression relating v and h.
(c) If the mass of pendulum is 5 kg and the difference in height of point A and point B is 0.5 m. What will be the energy of the pendulum at the point A? (g = 10 m/s²)
29. What is a permanent tissue? Classify permanent tissues and describe them.
30. Explain the advantages and disadvantages of using chemical fertilisers in crop production.
31. Distinguish between Plant cell and Animal cells. OR
State reason for the following.
(a) Mitochondria is known as the powerhouse of the cell.
(b) Plastids can make their own proteins.
(c) Plant cell shrinks when kept in hypertonic solution.
32. Write the names of compounds:
(a) Al2(SO4)3 (b) CaCl2 (c) K2SO4
33. (a) How will you differentiate a high pitch sound from a low pitch sound with the help of a graph?
(b) Give reasons for the following:
(i) The reverberation time of a hall used for speeches should be very short.
(ii) Sounds of same loudness and pitch but produced by different musical instruments like a violin and flute are distinguishable.
SECTION-D
Question No. 34 to 36 are long answer questions. They are 5 marks each.
34. (a) Animals of colder region and fishes of cold water have thicker layer of subcutaneous fat. Give reason. (b) What will happen if all the blood platelets are removed from the blood? OR
State one point of difference between:
(a) Blood and lymph (b) Bone and cartilage (c) Tendon and ligament
(d) Areolar and adipose tissues (e) Xylem and phloem
35. (a) Swimmers are provided with an inflated rubber jacket/tube. Explain why?
(b) A body whose volume is 100 cm3 weighs 10 N in air. Find its weight in water. (Take g = 10 ms–2), density of water = 1000 kg m–3)
(c) A body is weighed first in air, then in liquid A and then in liquid B. The observations are 100 N, 50 N and 60 N, respectively.
(i) Which liquid is denser? (ii) What is the density ratio of liquid A to liquid B?
OR
(a) On what factors does the gravitational force between the two bodies depend? How does the gravitational force between the two bodies change if the distance between them is tripled?
(b) What is meant by free fall? A man weighs 600 N on earth. What is his mass? (g = 10 ms–2). On moon, his weight would be 100 N. What is the acceleration due to gravity on the moon?
(c) Distinguish between mass and weight of an object.
36. (a) What factors affect the solubility of solvent and solute?
(b) State the electronic configuration of (i) aluminum atom and (ii) sulphur atom.
(c) State the differences between compounds and mixtures.
SECTION-E
Question No. 37 to 39 are case-based/data-based questions with 2 to 3 short subparts. Internal choice is provided in one of these sub-parts. They are 4 marks each.
37. Cell is the basic structural and functional unit of life. The activities of the cell are carried out by membrane-bound cell organelles like mitochondria, Golgi bodies, ribosomes, plastids, etc. Each cell organelle performs a designated function. A particular cell organelle is responsible for the synthesis of substances like lipid, protein, starch, etc. while the other generates energy. One secretes hormones, enzymes, etc. while the other participates in the digestion of substances taken up by the cell.
(a) Identify W, X, Y and Z in the given figure.
(b) What is the role of Z?
(c) Explain the functions of X.
OR
What will happen to the cell in the absence of W?
38. Homogeneous mixtures are regarded as solutions or true solutions. Heterogeneous mixtures are of two types. These are suspensions and colloidal solutions. These differ in the size of the particles responsible for the difference in their properties. In a suspension, the particle size is more than 10–5 cm whereas in a colloidal solution, it ranges between 10–5 cm to 10–7 cm. The two phases which constitute colloidal solutions, are dispersed phase and dispersion medium. Based upon their nature, the colloidal solutions are classified into eight types. The mixture of the non-reacting gases is always homogeneous irrespective of their nature. Therefore, it is not a colloidal solution.
(a) Scattering of light occurs when a beam of light is passed through Blood. Why?
(b) What is Tyndall effect?
(c) What is called colloidal solution?
OR
Differentiate between Physical change and Chemical change.
39. Ramesh conducts an experiment where he simultaneously drops a sheet of paper and a stone from the first floor of his school building. He observes that the paper reaches the ground later than the stone due to air resistance. Now, he imagines conducting this experiment on the moon, where there is no air.
(a) Predict what would happen if Ramesh repeated his experiment on the moon, where there is no air. Justify your answer.
(b) What would gravity affect the paper and stone on Earth versus on the moon?
(c) If Ramesh conducts the same experiment in the vacuum chamber. Will there be any difference in observation?
Sample Question Paper – 5 (Unsolved)
Time Allowed: 3 hours
General Instructions:
Read the following instructions very carefully and strictly follow them
I. All questions would be compulsory.
II. Section A would have 16 simple/complex MCQs and 4 Assertion-Reasoning type questions carrying 1 mark each.
III. Section B would have 6 Short Answer (SA) type questions carrying 2 marks each.
IV. Section C would have 7 Short Answer (SA) type questions carrying 3 marks each.
V. Section D would have 3 Long Answer (LA) type questions carrying 5 marks each.
VI. Section E would have 3 source based/case based/passage based/integrated units of assessment (4 marks each).
SECTION–A
Question 1 to 16 are multiple choice questions. Only one of the choices is correct. Select and write the correct choice as well as the answer to these questions. They are one mark each.
1. What is the purpose of growing cover crops in agricultural fields?
(a) To increase the cost of farming
(b) To provide shade to the main crop
(c) To protect the soil from erosion and improve fertility
(d) To increase the water retention capacity of the plants
2. Argon has atomic number 18 and mass number 40. Calcium has atomic number 20 and mass number 40. Calcium and argon are examples of:
3. On the distance-time graph, the Y-axis should be labelled as (a) Speed (b) Displacement (c) Distance (d) Time
4. It is due to which one of the following reasons that a body floats in a liquid?
(a) When the weight of the body is exactly half of the weight of the liquid displaced.
(b) When the weight of the body is greater than the weight of the liquid displaced.
(c) When the weight of the body is equal to the weight of the liquid displaced. (d) When the weight of the body is less than the weight of the liquid displaced.
5. Hydrogen and oxygen combine in the ratio 1:8 by mass to form water. How much mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?
(a) 8 g (b) 16 g (c) 24 g (d) 32 g
6. Which of the following statements is correct?
(a) A solution is a homogenous mixture.
(b) A suspension is a homogenous mixture.
(c) In a solution, the substance present in lesser quantity is called solvent. (d) In a solution, the substance present in greater quantity is called solute.
7. The correct labelling for the given structure is
8. The sea water is denser than fresh water due to (a) Evaporation (b) Mixing of sand (c) Mixing of salts (d) Stagnation
9. Following figure shows three states of matter and its interconversion. Which process display in A and B?
(a) (A) Sublimation (B) condensation
(b) (A) Fusion (B) Solidification
(c) (A) Vapourisation (B) Condensation
(d) (A) Fusion and (B) Condensation
10. A body moving with uniform acceleration has velocities 20 ms–1 and 30 ms–1, when passing points A and B, respectively. Then the velocity midway between A and B is:
(a) 25.5 ms–1 (b) 24
11. Small intestine absorbs the digested food materials. Which type of epithelial cells present in the small intestine carry out this function?
(a) Stratified squamous epithelium
(c) Glandular epithelium
(b) Columnar epithelium
(d) Cuboidal epithelium
12. Latent heat of vapourisation is the amount of heat energy required to change:
(a) 1 kg of solid into its gaseous state at its boiling point
(b) 1 kg of liquid into its gaseous state at its boiling point
(c) 1 kg of solid into its liquid state at its melting point
(d) 1 kg of liquid into its solid state at its freezing point
13. Most of the alpha particles passed through the gold foil without deflection because
(a) Atoms are stable.
(b) Most of the space inside the atom is empty.
(c) Atoms are unstable.
(d) All the space inside the atom is covered by nucleus.
14. What is the nature of the velocity-time graph of a uniformly accelerated object?
(a) Straight line sloping upwards
(c) Straight line sloping downwards
(b) Curved line
(d) Straight line parallel to the y-axis
15. Which of the following cell functions will stop, if its ribosomes are destroyed?
(a) Formation of complex sugars (b) Lipid metabolism
(c) Protein synthesis (d) ATP synthesis
16. When a ball is thrown vertically upwards its velocity keeps on decreasing. What happens to its kinetic energy when it reaches the maximum height?
(a) Kinetic energy is maximum at maximum height.
(b) Kinetic energy is half the initial value.
(c) Kinetic energy is zero.
(d) Kinetic energy is double
Question No. 17 to 20 consist of two statements - Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:
(a) Both A and R are true, and R is the correct explanation of A (b) Both A and R are true, and R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true
17. Assertion (A): During evaporation of liquid, the temperature of liquid remains unaffected.
Reason (R): Kinetic energy of the molecules is directly proportional to absolute temperature.
18. Assertion (A): Work is done in lifting a book through a height of 10 cm.
Reason (R): Work done is equal to ratio of force and displacement.
19. Assertion (A): It is difficult to balance our body when we accidentally slip on a peel of banana.
Reason (R): According to Newtonʼs third law of motion “Every action have an equal and opposite reaction.”
20. Assertion (A): Sclerenchyma cells are dead and provide mechanical support to plants.
Reason (R): The walls of sclerenchyma cells are uniformly thickened with lignin.
SECTION-B
Question No. 21 to 26 are very short answer questions. They are 2 marks each.
21. Write the parameters required to describe a waveform.
22. In chemistry class, Radhika learns that CO2 is a gas. Can you write two properties of CO2 to justify this? Also, how can Radhika liquefy CO2 gas? Finally, why is solid CO2 called dry ice?
23. If there is low rainfall in a village throughout the year, what measures will you suggest to the farmers for better cropping? List any two such measures.
24. Explain how defects in a metal block can be detected using ultrasound.
25. What will happen if the apical meristem is damaged or cut?
26. Find the percentage composition of sucrose C12H22O11.
SECTION-C
Question No. 27 to 33 are short answer questions. They are 3 marks each.
27. (a) What is the difference between echo and reverberation?
(b) How can we reduce reverberation in an auditorium or a big hall?
(c) Why we can hear more clearly in a room having curtains?
28. Deduce the formulae for below compounds.
(a) Ammonium sulphate
(b) Sodium phosphate
(c) Carbon dioxide
29. How do biotic and abiotic factors affect crop production?
30. What is the difference between plasma membrane and cell wall? Give the functions of each one.
31. (a) What is the basic difference between two isotopes of an element?
(b) What is the basic difference between two isobars?
32. What will happen to the red blood cells if they are placed in
(a) Hypotonic solution
(b) Isotonic solution
(c) Hypertonic solution
33. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40,000 N and the track offers a frictional force of 5,000 N, then calculate:
(a) Net accelerating force.
(b) Acceleration of the train.
(c) Force of the 1st wagon on the 2nd wagon s11, 32
SECTION-D
Question No. 34 to 36 are long answer questions. They are 5 marks each.
34. (a) State the Universal Law of Gravitation.
(b) Calculate the force of gravitation between the earth and the sun.
[Given that the mass of the earth = 6 × 1024 kg, mass of the sun = 2 × 1030 kg, average distance between the two is 1.5 × 10–11 m and G = 6.67 × 10–11 Nm2 kg–2 ].
(c) A planet's weight is twice that of earth and the radius is 3 times that of earth. Find the acceleration due to gravity of that planet.
OR
(a) Explain why swimmers are provided with an inflated rubber jacket/tube?
(b) It is easier to swim in seawater than in river water. Why?
(c) State the conditions under which the object will float? Ships are made of iron and steel, yet they float on water. Why?
(d) Identify and write the names of substances that sink in turpentine at the same temperature. [Given: The density of turpentine at 293 K is 870 kg/m3]
35. Describe Rutherford’s model of an atom. Write down its limitations.
(a) What is an octet? How do elements attain an octet?
(b) Na+ has completely filled K and L-shells. Explain.
(c) If, Z = 13, what would be the valency of the element? Also, name the element?
36. (a) Animals of colder region and fishes of cold water have thicker layer of subcutaneous fat. Give reason.
(b) What will happen if all the blood platelets are removed from the blood?
State one point of difference between:
(a) Blood and lymph
(b) Bone and cartilage
(c) Tendon and ligament (d) Areolar and adipose tissues
(e) Xylem and phloem,
SECTION-E
Question No. 37 to 39 are case-based/data -based questions with 2 to 3 short subparts. Internal choice is provided in one of these sub-parts. They are 4 marks each.
37. Epithelial tissues are widespread throughout the body. They form the covering of all body surfaces, line body cavities and hollow organs. Depending upon the shape and function of the constituent cells, epithelial tissues can be squamous, columnar, cuboidal, glandular, ciliated, and stratified.
(a) What is the role of ciliated epithelium?
(b) Where is stratified epithelium located in the body?
(c) Give two points of differences between cuboidal and columnar epithelium. OR
Describe the structure, location and one function of squamous epithelium.
38. In 1911, Earnest Rutherford, a scientist from New Zealand, overturned Thomson's atomic model by his gold foil experiment. Rutherford selected a gold foil as he wanted a very thin layer. The gold foil used by Rutherford was 0.004 millimetres in thickness. That is, the foil was about 1000 atoms thick. In his experiment, fast moving α-particles (alpha particles) were made to fall on a thin gold foil. The α-particles are helium ions with a+2 charge. Their atomic mass is 4u. Hence, a high velocity beam of α-particles has a lot of energy. These particles were studied by means of flashes of light they produced on striking a zinc sulphide screen. The α-particles are much heavier than the sub-atomic particles present in gold atoms. Hence, he expected the α-particles to pass through the gold foil with little deflection and strike the fluorescent screen. However, the observations he made were quite unexpected.
(i) What are the alpha particles in Rutherford scattering experiment?
(ii) What happened to the particles that Rutherford used in his experiment?
(b) What information does it give about the structure of the atom? OR
(b) How did Rutherford detect where the particles were strike at the gold foil?
39. (a) Define potential and kinetic energy.
(b) Give an example where potential energy is acquired by a body due to change in its shape.
(c) A skier of mass 50 kg stands at A, at the top of a ski jump. He takes off from A for his jump to B. Calculate the change in his gravitational potential energy between A and B.
Ground
About the Book
LUMO is a comprehensive and student-friendly one-stop resource for Science (CBSE subject code 086) for Grades IX–X. It offers a brief overview of the theory and a wide variety of practice questions, designed to promote conceptual clarity, core skill development and exam readiness. The book presents the entire NCERT curriculum in a clear, accessible, and exam-oriented format to support students at every stage of their learning journey.
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