Grade 9 Physics

IL FOUNDATION SERIES

PHYSICS

A Reliable Companion for JEE | NEET | Olympiads

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ISBN 978-81-985385-8-1 Second Edition

Contents 1.

Units and Measurement

01

2. Motion

18

3. Force and Laws of Motion

43

4. Gravitation

68

5. Work and Energy

117

6. Sound

140

1

1.1

UNITS AND MEASUREMENT

THE INTERNATIONAL SYSTEM OF UNITS

1.1.1 Quantity Quantity means size, amount, vastness, and magnitude or simply stated as the answer for 'how much?' or 'how many?’.

Physical quantity: A quantity which can be 'measured’ is called physical quantity.

Examples: Length, Area, Volume, Speed, Force, Energy, etc.

Physical quantities are organised in various ways in various systems like FPS, CGS, MKS and SI. Now, the scientists of all the countries follow SI for all their research work.

In SI, physical quantities are organised into three groups. They are: • Fundamental physical quantities • Supplementary physical quantities • Derived physical quantities Set of fundamental physical quantities: Consider a set of physical quantities. If no quantity can be derived from the remaining physical quantities, then the set is called a set of fundamental physical quantities.

Examples: a) {Mass, Length, Time} In the above set, mass cannot be derived from the remaining physical quantities. It's the same for the other two quantities too. So, it is a set of fundamental quantities. b) {Force, Length, Time} In the above set, the force cannot be derived from the remaining physical quantities, nor can the other two quantities be derived from the remaining ones. So, it is a set of fundamental quantities.

In SI, there are seven fundamental physical quantities. They are length, mass, time, thermodynamic temperature, strength of electric current, amount of substance and luminous intensity.

Supplementary physical quantities: In SI, there are two supplementary physical quantities. They are plane angle and solid angle.

Derived physical quantities: Physical quantities derived from fundamental or supplementary physical quantities are called derived physical quantities.

1

UNITS AND MEASUREMENT

Examples: Force, Area, Volume, Speed, etc.

Note: Fundamental physical quantities are like the letters of the English alphabet and are derived physical quantities are like words.

1.1.2 Unit Unit is a 'standard measure' of any physical quantity.

Measurement of a physical quantity involves two steps.

Step - 1: Choose a standard value as a unit for measurement.

Step - 2: Find how many times that unit is contained in the given physical quantity.

Unit is also a value. We should not think of it as a name. Hence, all algebraic operations are possible with units.

Examples: 1) cm ◊ cm = cm2 [Note: cm2 is equal to (cm)2 ] m2 =m m Note: A standard unit should be consistent, reproducible, invariable, and easily available for usage.

2)

Fundamental units: The units of fundamental physical quantities are called fundamental units.

In SI, the fundamental units are as follows: Physical Quantity

Unit

Symbol

Length

metre

m

Mass

kilogram

kg

Time

second

s

Strength of electric current

ampere

A

Thermodynamic temperature

kelvin

K

Amount of substance

mole

mol

Luminous Intensity

candela

cd

Table 1.1 Fundamental units

Note: The fundamental units in some of the old systems are as follows: Fundamental Physical Quantities

System

Length

Mass

FPS

foot (ft)

pound (lb)

second (s)

CGS

centimetre (cm)

gram (g)

second (s)

MKS

metre (m)

kilogram (kg)

second (s)

Table 1.2 Fundamental units in the old system 2

Time

IL Foundation Series Class 9

Supplementary units: The units of supplementary physical quantities are called supplementary units. The SI units of supplementary physical quantities are as follows: Supplementary Quantity

Unit

Symbol

plane angle

Radian

rad

solid angle

Steradian

sr

Table 1.3 Supplementary units

Derived units: The units of derived physical quantities are called derived units. The SI units of some derived physical quantities are as follows: Derived Physical Quantities

Derived Units

Area

m2

Volume

m3

Velocity

ms-1

Force

kgms-2 or newton (N)

Energy

kgm2 s-2 or joule (J) Table 1.4 Derived units

Rules for writing units in SI • All the symbols of the fundamental units should be written as they appear in the table of SI units.

Examples: metre ⇒ m or second ⇒ s 5M

(✕)

5m

(✓)

• In compound units, all the symbols of the fundamental units should be written as they appear in the table of SI units.

Example: kg sec −1

(✕)

kgs −1

(✓)

• Some of the units are represented by the names of the scientists as an honour for their research in the respective fields. The full name of such units should be written with a lower initial letter.

Example: newton, joule, watt, ampere, kelvin, etc.

7 Newton

(✕)

7 newton

(✓)

The unit of force is kgms −2, but it is named as newton (N).

• The symbol of the unit named after a scientist should always be written with the capital letter.

3

UNITS AND MEASUREMENT

Example: 5n

(✕)

5N

(✓)

• Punctuations like full stop (.), comma (,), etc., should not be used after the symbol of the unit.

Example: 5 s.

(✕)

5s

(✓)

• A unit should always be represented in a singular form.

Example: 5kgs

(✕)

5 kg

(✓)

Need of multiples and submultiples Multiples and submultiples are introduced to change the size of the units to fulfil the needs of various branches of physics.

• In SI, multiples and submultiples are as follows: S. No

Multiplication factor

Prefix

Symbol

1.

* 10

deca

da

2.

* 10

hecto

h

3.

103

kilo

k

4.

10

6

mega

M

5.

109

giga

G

6.

10

12

tera

T

7.

1015

peta

P

8.

10

18

exa

E

9.

* 10-1

deci

d

10.

* 10-2

centi

c

11.

10

-3

milli

12.

10-6

micro

m µ

13.

10

-9

nano

n

14.

10-12

pico

p

15.

10

-15

femto

f

16.

10-18

atto

a

2

Table 1.5 Multiple and submultiple units

Note: The multiplication factors with '*’ are not SI prefixes, but are commonly in use.

4

IL Foundation Series Class 9

• Some practical units of length in various branches of physics: (These are not SI units.) −6 ( µ m) 10 = m 10−4 cm Micron =

 −10 = = m 10−8 cm  A  10 Angstrom   −15 Fermi = 10 = m 10−13 cm Light year = 9.5 × 1015 m = 9.5 × 1012 km Astronomical unit ( A.U= ) 1.5 × 1011 m 1 parsec = 3.26 light years.

(Light year is the distance travelled by light in a vacuum in one year.)

• Some practical units of mass in various branches of physics: (These are not SI units.)

Commercial Units: 1 metric ton = 1000 kg 1 quintal = 100 kg 1 slug = 14.59 kg 1 pound = 0.453 kg Atomic mass unit (1 amu = ) 1.67 ×10−27 kg Astronomical Unit of mass (Solar mass) = 2 ×1030 kg • Some practical units of time in various branches of physics: (These are not SI units.)

1 Shake 10 8 s 1 Minute 60 s 1 Hour 60 minute 3600 s 1 mean solar day 1 24 hour 1 24 60 minute 1 24 60 60 s 86400 s Month 30 days (April, June, September and November) 31 days (January, March, May, July, August, October and December) 28 days (February in other than leap year)

t

29 days (February in a leap year ) Lunar month 4 weeks 27.3 days (approximately) Year 365.25 days Decade 10 years Centtury 100 years

5

UNITS AND MEASUREMENT

• Special units of pressure: 1 atm = 760 mm of Hg = 76 × 13.6 × 980 dynes / cm 2 = 1.013 × 106 dynes / cm 2 = 1.013 × 105 Pa

1 bar = 750 mm of Hg (approximately) = 106 dynes / cm 2 = 105 Pa = dynes / cm 2 133.3 Pa = 1 mm of Hg 1333 1 torr=

This unit is named after the scientist, Torricelli.

1.2 DIMENSIONS OF PHYSICAL QUANTITIES The nature of a physical quantity is described by its dimensions.

Dimensions: The dimensions of a physical quantity are the powers to which the base quantities are raised to represent that quantity.

The physical quantity that is expressed in terms of the base quantities is enclosed in square brackets '[ ]’. mass mass = volume ( length )3 (or ) density = (mass) ( length ) −3

Example: = density

Thus, the dimensions of density are ’1' in mass and '-3' in length. In this case, the dimensions of all other fundamental quantities are zero.

1.3 DIMENSIONAL FORMULAE AND DIMENSIONAL EQUATIONS 1.3.1 Dimensional formulae The expression which shows how and which of the base quantities represent the dimensions of a physical quantity is called the dimensional formula of the given physical quantity.

Dimensional formulae of fundamental quantities in SI are as follows.

Quantity

Dimensional formula

Mass

[M]

Length

[L]

Time

[T]

Strength of electric current Thermodynamic temperature

[I] [θ]

Amount of substance

[mol]

Luminous Intensity

[cd]

Table 1.6 Dimensional formulae of fundamental quantities 6

IL Foundation Series Class 9

Note: 1.

The supplementary physical quantities have no dimensions.

2. Some authors use [K] for thermodynamic temperature and [A] for the strength of electric current instead of [θ] and [I], respectively.

Examples: 1.

Dimensional formula of the area is M 0 L2 T 0  .

[M 0 L2 T 0 ]= [M 0 ] × [L] × [L] × [T 0 ]= [M 0 ] × [L]2 × [T 0 ]= [M 0 ] × [L2 ] × [T 0 ] Here, the dimension of area is 0 in mass.

The dimension of the area is 2 in length.

The dimension of the area is 0 in time.

2.

Dimensional formula of volume is M 0 L3 T 0  .

[M 0 L3 T 0 ]= [M 0 ] × [L] × [L] × [L] × [T 0 ]= [M 0 ] × [L]3 × [T 0 ]= [M 0 ] × [L3 ] × [T 0 ] Here, the dimension of volume is 0 in mass.

The dimension of volume is 3 in length.

The dimension of volume is 0 in time.

3.

Dimensional formula of speed is M 0 L1 T −1  . M 0 L1 T −1  = M 0  ×  L1  / T1  = M 0  ×  L1  × T −1  Here, the dimension of speed is 0 in mass.

The dimension of speed is 1 in length.

The dimension of speed is -1 in time.

1.3.2 Principle of homogeneity of dimensions If x = y + z is dimensionally correct and if x represents the physical quantity, the force, then y and z also must represent the same physical quantity, i.e., force. It means the terms on both sides of a dimensional equation should have the same dimensions. This is called the principle of homogeneity of dimensions.

Example: The velocity 'v' of a particle depends upon the time according to the equation, c v =a + bt + . Write the dimensions of a, b, c and d. d+t Solution: From the principle of homogeneity,

[a] = [ v ] or 7

UNITS AND MEASUREMENT

[a] =  LT −1  [ bt ] = [ v ] or [= b]

−1 v ]  LT  [= [t ] [T]

−2 or [ b ] =  LT 

] Similarly, [d=

[t=] [T ]

[c]

= [v] [d + t ] [c] [ v ][d + t ] or= or [ c] =  LT −1  [ T ] or [ c] = [ L ] Further,

1.3.3 Dimensional equation An equation obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of the physical quantity.

Examples: 1.

The dimensional equation of area is expressed as M 0 L2 T 0 

2.

0 3 0 The dimensional equation of volume is expressed as M L T 

3.

The dimensional equation of speed is expressed [V ]as= M 0 L1 T-1 

1.4 DIMENSIONAL ANALYSIS AND ITS APPLICATION 1.

Dimensional formulae can be used to convert one system of units into another system.

2.

Dimensional formulae can be used to check the correctness of an equation.

3.

Dimensional formulae can be used to derive relationships among different physical quantities.

4.

Dimensional formulae can be used to find a unit of a given physical quantity.

5.

Dimensional formulae can be used to design our new system of units.

S.No

8

Physical quantity

Formula

Dimensional

S.I unit

1.

Volume

length × breadth × height

M ° L3 T°

m3

2.

Density

mass/volume

ML−3 T°

kgm −3

3.

Linear density

mass/length

ML−1 T°

kgm −1

IL Foundation Series Class 9

4.

Relative density

density of substance density of water

M ° L°T°

no units

5.

a) Velocity

displacement/time

M ° LT −1

ms −1

b) Speed

distance/time

M ° LT −1

ms −1

c) Acceleration

force/mass

M ° LT −2

ms −2

a) Momentum (linear) mass × velocity

ML1 T −1

kgms −1

b) Force

mass × acceleration

MLT −2

kgms −2

c) Impulse

Force × time

ML1 T −1

kgms −1

a) Work

Force × displacement

ML2 T −2

joule

ML2 T −2

joule

ML2 T −3

watt

6.

7.

b) All energies c) Power

Force × velocity

Table 1.7 Dimensional formulae of physical quantities

Note: Understanding the concepts of dimensions, which play a crucial role in describing physical behaviour, holds fundamental significance. This importance arises from the fact that only physical quantities with identical dimensions can be added or subtracted. A comprehensive grasp of dimensional analysis aids in deriving relationships among diverse physical quantities and helps us to verify the accuracy, derivation, and dimensional consistency or homogeneity of various mathematical expressions.

When multiplying the magnitudes of two or more physical quantities, their units should be treated akin to ordinary algebraic symbols. The cancellation of identical units in both the numerator and denominator is permissible. This principle extends to the dimensions of a physical quantity as well.

Additionally, in a mathematical equation where symbols represent physical quantities on both sides, it is imperative that these quantities share identical dimensions.

1.4.1 Checking the dimensional consistency of equations • We know only similar physical quantities can be added or subtracted, adhering to the principle of homogeneity of dimensions within an equation. This principle proves highly valuable for validating the accuracy of an equation. If the dimensions of the terms within an equation are not uniform, the equation is deemed incorrect.

• For instance, when deriving an expression for the length or distance of an object, irrespective of the symbols utilised in the original mathematical relation, the resulting simplified dimensions must exclusively represent length.

• Similarly, in the derivation of an equation for speed, the dimensions on both sides of the equation, upon simplification, must be consistent with length/time or [L T-1].

9

UNITS AND MEASUREMENT

Dimensions are a preliminary test for the equation’s consistency when doubts arise regarding its correctness. However, dimensional consistency does not guarantee the accuracy of equations, especially when dealing with dimensionless quantities or functions.

Special functions, such as trigonometric, logarithmic, and exponential functions, necessitate dimensionless arguments. Pure numbers or ratios of similar physical quantities, such as angles as (length/length) or refractive index as (speed of light in vacuum/speed of light in medium), are examples of entities without dimensions.

Now, let’s examine the dimensional consistency or homogeneity of the equation, x =x o + v o t + (1/ 2 ) at 2 , representing the distance x traveled by a particle or body in time t.

The dimensions of each term are as follows:

[x] = [L] [ x0 ] = [ L] [ v 0t ] =  LT −1  [T ] = [L] (1/ 2 ) at 2  =  LT −2  T 2  = [L] Each term on the right-hand side shares the same dimension (length), which matches the dimension of the left-hand side, this equation is considered dimensionally correct.

Note: It is crucial to note that a dimensional consistency test is equivalent to a test of unit consistency but offers the advantage of not being tied to a specific choice of units or concerns about unit conversions. While a dimensionally correct equation may not necessarily be an exact or correct equation, a dimensionally incorrect or inconsistent equation is conclusively incorrect.

1 mv 2 = mgh , 2 where m is the mass of the body, a its velocity, g is the acceleration due to gravity and h is the height. Verify if this equation adheres to dimensional correctness.

Example: Let us consider an equation,

Solution: The dimensions of LHS are 2

= = [M ]  LT −1  [M ]  L2T −2  ML2T −2  The dimensions of RHS are, [ M ]  LT −2  [ L ] == [ M ]  L2 T −2  = ML2 T −2  The dimensions of the left-hand side (LHS) and right-hand side (RHS) are identical, confirming the dimensional correctness of the equation.

10

IL Foundation Series Class 9

1.4.2 Deducing relations among the physical quantities The technique of dimensional analysis is occasionally employed to infer relationships among physical quantities. To achieve this, it is necessary to understand the dependency of a given physical quantity on other quantities, typically up to three physical quantities or linearly independent variables. The approach involves treating this dependence as a product-type relationship. To illustrate, let’s consider an example.

Example: Let’s consider a simple pendulum consisting of a suspended bob connected to a string, undergoing oscillations influenced by the force of gravity. Assume that the period of the pendulum’s oscillation relies on its length (l), the mass of the bob (m), and the acceleration due to gravity (g).

Establish the expression for its time period utilizing the method of dimensions.

Solution: The relationship between the time period T and the variables l, g, and m as a product can be expressed as: T = kl x g y m z , where k is a dimensionless constant, and x, y and z are the exponents.

By considering dimensions on both sides, we have: x

y

z

 L0 M 0 T1  =  L1   L1 T -2  M1  =Lx+y T -2y M z On equating the dimensions on both sides, we get: x + y = 0; −2y = 1; and z = 0

1 1 ,y = − ,z = 0 So that, x = 2 2 Then, T = kl1/2 g −1/2 or,

T=k

l g

It is important to highlight that the constant k cannot be determined through the method of dimensions. In this context, it is irrelevant if the right side of the formula is multiplied by a numerical factor, as it does not impact its dimensions.

Actually, k = 2π so that, T = 2π

l g

Dimensional analysis proves valuable for deducing relationships among interdependent physical quantities. Nevertheless, it falls short of obtaining dimensionless constants through this method.

The technique of dimensions is confined to testing dimensional validity and cannot precisely establish the relationship between physical quantities within an equation. Furthermore, it cannot differentiate between physical quantities that share identical dimensions.

11

UNITS AND MEASUREMENT

QUICK REVIEW • A quantity which can be 'measured’ is called physical quantity.

• In SI, there are seven fundamental physical quantities. They are length, mass, time, thermodynamic temperature, strength of electric current, amount of substance and luminous intensity.

• In SI, there are two supplementary physical quantities. They are plane angle and solid angle.

• Physical quantities derived from fundamental or supplementary physical quantities are called derived physical quantities.

• Unit is a standard measure of any physical quantity.

• Units of fundamental physical quantities are called fundamental units.

• Units of supplementary physical quantities are called supplementary units.

• Units of derived physical quantities are called derived units.

• The dimensions of a physical quantity are the powers to which the base quantities are raised to represent that quantity.

• The expression which shows how and which of the base quantities represent the dimensions of a physical quantity is called the dimensional formula of the given physical quantity.

• If x = y + z is dimensionally correct and x represents the physical quantity, i.e., force, then y and z also must represent the same physical quantity, i.e., force. It means the terms on both sides of a dimensional equation should have the same dimensions. This is called the principle of homogeneity of dimensions.

• An equation obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of the physical quantity.

WORKSHEET - 1 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER I.

The international system of units 1. A measurable quantity is taken as a:

a. Physical quantity

b. Fundamental quantity

c. Derived quantity

d. Supplementary quantity

2. In a set of physical quantities, if no quantity can be derived from the remaining physical quantities, then the set is called a:

12

a. Set of physical quantities

b. Set of derived quantities

c. Set of fundamental quantities

d. Set of supplementary quantities

IL Foundation Series Class 9

3. A physical quantity that can be derived from other quantities is called: a. Fundamental quantity

b. Derived quantity

c. Supplementary quantity

d. Complementary quantity

4. Among the following, the supplementary quantity is: a. Temperature

b. Current

c. Plane angle

d. Time

5. Among the following, the set of fundamental physical quantities is/are: a. {Mass, length, time}

b. {Force, length, time}

c. {Energy, length, time}

d. All of the above

6. Among the following, the derived unit is: a. ampere

b. candela

c. newton

d. kelvin

c. ampere

d. kelvin

b. steradian

c. degree

d. theta

b. newton

c. dyne

d. joule

7. The SI unit of luminous intensity is: a. kilogram

b. candela

8. The SI unit of plane angle is: a. radian 9. The SI unit of Force is: a. erg

10. The distance travelled by light in a vacuum in one year is: a. light year

b. sound year

c. energy year

d. leap year

b. 1015 m

c. 106 m

d. 10−6 m

11. The value of 1 fermi is: a. 10−15 m

12. The thickness of a layer is 10-6. Its value in micron is: a. 106

b. 109

c. 10−6

d. 1

13. Among the following, the arrangement of multiples in increasing order is: a. milli, kilo, mega, giga

b. giga, mega, kilo, milli

c. kilo, giga, mega, milli

d. milli, giga, mega, kilo

14. 10−2 is called: a. centi

b. deca

c. hecto

d. deci

b. 10−4

c. 10−3

d. 103

15. 1 ton = _____ kg. a. 104

16. The most suitable unit to measure the mass of an atom is: a. kilo

b. amu

c. ton

d. quintal

13

UNITS AND MEASUREMENT

II. Dimensions of physical quantities, formulae, equations and analysis 1. The dimensional formula for mass in SI is: a.  F1 

b.  L1 

c. T1 

d. M1 L0 T 0 

c.  L1 T -1 

d.  L1 T1 

2. The dimensional formula for speed in SI is: b. T1 

a.  L1 

3. Among the following, the unit of luminous intensity is: a. kelvin

b. mole

c. candela

d. ampere

c. May have a unit

d. Does not exist

4. A dimensionless quantity: a. Never has a unit

b. Always has a unit

5. A unit less quantity: a. Never has a non-zero dimension

b. Always has a non-zero dimension

c. May have a non-zero dimension

d. Does not exist

6. If the units of length, mass and time are doubled, the unit of force will be: a. Doubled

b. Halved

c. Quadrupled

d. Unchanged

7. Among the following, the set that cannot enter the list of fundamental quantities in SI is: a. {length, mass, speed}

b. {length, time, speed}

c. {mass, time, speed}

d. {length, time, mass}

8. The dimensional formula for volume is: a.  L1 

b.  L2 

c.  L3 

d.  L-3 

9. A force is given by F = at + bt 2 , where 't' is time. The dimensions of a and b are: a. MLT −4  and MLT l 

b. MLT −1  and MLT 0 

c. MLT −3  and MLT −4 

d. MLT −3  and MLT 0 

2

a-t a 10. The equation P = bx , where P is pressure, x is distance and t is time. The dimensions of is: b −2 a. M 2 LT -3  b. MT  c.  LT −3  d. ML3 T -1 

b 11. The velocity v of a particle at time t is given by v = at + , where a,b and c are constants. t+c The dimensions of a,b and c, respectively, are: a.  LT −2  , [ L ] and [ T ]

c.  LT 2  , [ LT ] and [ L ]

b.  L2  ,[T] and  LT 2 

d. [ L ] , [ LT ] and T2 

12. The physical quantity which has dimensional formula as that of a. Force 14

b. Power

c. Pressure

Energy is: (mass × length) d. Acceleration

IL Foundation Series Class 9

13. A pair of physical quantities having the same dimensional formula is: a. Force and work

b. Work and energy

c. Force and torque

d. Work and power

WORKSHEET - 2 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. In SI, the number of supplementary quantities is/are: a. 1

b. 2

c. 3

d. 4

2. Among the following, the fundamental quantity in SI is: a. Plane angle

b. Solid angle

c. Mass

d. Density

3. Among the following, the derived quantity in SI is: a. Mass

b. Time

c. Force

d. Temperature

4. Among the following, the derived quantity in SI is/are: a. Force

b. Work

c. Power

d. All of the above

5. At present the system adopted by all the scientists in the world is: a. MKS

b. CGS

c. FPS

d. SI

6. The number of fundamental physical quantities in SI is: a. 2

b. 5

c. 3

d. 7

7. The derived quantities in SI are: 1. Strength of electric current

2. Volume

3. Speed

4. Amount of substance

a. 1 and 2 are correct.

b. 2 and 3 are correct.

c. 3 and 4 are correct.

d. 1 and 4 are correct.

8. Among the following, the supplementary physical quantities in SI are: i. Luminous Intensity ii. Plane angle

iii. Area

iv. Solid angle

a. i, ii and iii only

c. ii, iii and iv only

d. ii and iv only

b. ii and iii only

9. Among the following, the derived quantities in SI are: i. Force

ii. Strength of Electric current

iii. Density

iv. Plane angle

a. i, ii and iii only

b. ii and iii only

c. i and iii only

d. iv only

c. m

d. Mole

10. The SI unit of the amount of substance is: a. Mol

b. mole

15

UNITS AND MEASUREMENT

11. Among the following, the odd one is: a. kilogram

b. newton

c. candela

d. mole

c. 6a

d. 6Am

12. 6 ampere can be represented as: a. 6Amp

b. 6 A

13. Among the following, the physical quantity having steradian as SI unit is: a. Strength of electric current

b. Plane angle

c. Solid angle

d. Amount of substance

14. The unit ms −2 is: a. Fundamental

b. Derived

c. Supplementary

d. Complementary

15. In SI, ampere is the unit of: a. Strength of electric current

b. Thermodynamic temperature

c. Luminous intensity

d. Amount of substance

16. The unit of derived quantity is called: a. Fundamental unit

b. Supplementary unit c. Derived unit

d. Physical unit

17. The unit of fundamental quantity is called: a. Fundamental unit

b. Supplementary unit c. Derived unit

d. Physical unit

18. Statement (A): Ampere is a fundamental unit in SI. Statement (B): Kelvin is a derived unit in SI. a. A is true, but B is false.

b. Both A and B are true.

c. A is false, but B is true.

d. Both A and B are false.

19. Statement (A): The SI unit of force is newton.

Statement (B): The first letter of a scientist’s name should not be written in capital letters when it is represented as a unit. a. A is wrong, but B is correct.

b. A is correct, but B is wrong.

c. Both A and B are correct.

d. Both A and B are wrong.

20. Statement (A): Mass is the measure of inertia.

Statement (B): Mole is the amount of substance. a. A is true, but B is false.

b. A is false, but B is true.

c. Both A and B are true.

d. Both A and B are false.

21. Arrange the following units of length in the increasing order of their magnitude:

16

i. Tm

ii. pm

iii. µm

iv. Pm

a. ii, i, iii, iv

b. iii, ii, i, iv

c. iv, iii, ii, I

d. ii, iii, i, iv

IL Foundation Series Class 9

22. 400 nm can be written as: a. 400×106 m

b. 400×10 -6 m

c. 400×10 -9 m

d. 400×109 m

b. tera

c. femto

d. atto

23. 10−15 is called : a. peta

24. Among the following, the appropriate unit used to measure the size of a plant cell is: a. nm

b. fm

d. µm

c. mm

25. Statement (A): Dimensional formula for area is M 0 L2 T 0  .

Statement (B): Dimensional formula for speed is M 0 L1 T -1  . a. Both A and B are true.

b. Both A and B are false.

c. A is true, but B is false

d. B is true, but A is false

PASSAGE: Q26-Q30: My name is Mr. Gold, and I am 200 years old. The name of my country is Strange. In my country, we have a system of units called the strange system. In the strange system, the set of fundamental quantities are {force, length, time}, and the remaining quantities are derived quantities.

Quantity

Unit

Dimensional formula

Force

strange newton

[ F0 ]

Length

strange metre

[L0 ]

Time

strange second

[T0 ]

26. In a strange system, mass is a: a. Fundamental quantity

b. Supplementary quantity

c. Derived quantity

d. Complementary quantity

27. In India, mass is a: a. Fundamental quantity

b. Supplementary quantity

c. Derived quantity

d. Trigonometrical quantity

28. The unit of force in India in the MKS system is: a. kgms −2

b. gcms

−2

c. strange newton

d. Newton

29. The dimensional formula for force in a strange system is: a. [ F0 ]

b. [ L 0 ]

c. [ T0 ]

d. [ N 0 ]

c.  F1 L1 T -2 

d.  F1 L-1 T2 

30. The dimensional formula for force is: a. M1 

b. M1 L1 T -2 

17

MOTION

2

MOTION

2.1 DESCRIBING MOTION To determine the location of an object, we designate it as a reference point. For instance, if we say that a school is 2 km north of a railway station, the railway station serves as the reference point.

This reference point, known as the origin, helps us describe the position of the object. We can choose various reference points based on our convenience.

Rest: If the position of an object does not change with respect to a reference point as time passes, it is said to be at rest.

Motion: If the position of an object changes with respect to a reference point as time passes, it is said to be in motion.

Rest and motion are not absolute; they are relative. Objects may move in a straight line or in a circular path. An object may also exhibit rotatory and vibratory motions.

2.1.1 Motion in a straight line Distance

The length of the total path covered by an object is called distance. Distance has only magnitude; it is a scalar physical quantity. Its SI unit is metre (m).

Displacement

The shortest distance measured from the initial to the final position of an object is known as displacement. Displacement is independent of the path followed by the object and is a vector physical quantity. Its SI unit is metre (m).

The magnitude of the displacement of an object is less than or equal to the distance covered by it.

The displacement for a course of motion may be zero, while the corresponding distance covered is not zero.

Example: Consider someone walking from their home to a nearby market, which is 2 km away, and then returning home. While their displacement, the change in position from the initial point to the final point, is zero as they end up back at their starting point, the distance covered will be 4 km.

2.1.2 Uniform motion and non-uniform motion Motion can be uniform or non-uniform in nature.

Uniform motion

When an object covers equal distances in equal intervals of time, it is known as uniform motion. It means a constant motion.

18

IL Foundation Series Class 9

0m

2 sec

O

4m

2 sec

8m

2 sec

B

A

12m

C

Fig. 2.1 Uniform motion

Example: A car travelling at a constant speed.

Imagine a car moving on a straight highway at a steady speed of 60 kilometres per hour. This means that the car travels a fixed distance of 60 km each hour. Thus, it is said to be in a state of uniform motion.

Non-uniform motion

When an object covers unequal distances in equal intervals of time, it is known as non-uniform motion.

2 AM

3 AM

60 Km

4 AM

40 Km

5 AM

50 Km

Fig. 2.2 Non-uniform motion

Example: A car travelling at a non-constant speed.

Imagine a car moving in traffic; its speed will change continuously, every instant. Thus, it is said to be in a state of non-uniform motion.

2.2 MEASURING THE RATE OF MOTION Different objects may require different durations to traverse a specified distance. Some exhibit a high pace, while others move at a slower pace.

2.2.1 Rate of motion Speed: Different objects exhibit varying rates of motion with different speeds and time intervals.

Speed tells us how fast something is moving. It's a measure of how much distance an object covers in a certain amount of time.

The SI unit for speed is metres per second (m/s), often represented as ms-1.

19

MOTION

Additional units include centimetres per second (cms-1) and kilometres per hour (kmh-1).

Speed, described by its magnitude, doesn’t have to be constant, especially in cases of non-uniform motion. In non-uniform motion scenarios, the average speed becomes a relevant metric.

The formula for average speed is expressed as: Average speed =

Total distance travelled Total time taken

For an object covering a distance, s in time, t, its speed, v is given by: s v= t To illustrate, consider a car covering 150 km in 3 hours, resulting in an average speed of 50 km/h.

It’s important to note that this average speed doesn’t imply constant motion at 50 km/h throughout; the car could have varied its speed during the journey.

2.2.2 Speed with direction Velocity, a vector quantity, provides a more comprehensive description of an object’s motion by specifying its speed and direction. It can be uniform, variable, and subject to a change in speed, direction, or both.

Average velocity, akin to average speed, captures the overall motion characteristics, considering both magnitude and direction.

In the case of an object moving along a straight line with variable speed, average velocity is calculated similarly to average speed: Average velocity = Total displacement Total time taken When velocity changes uniformly, the average velocity ( v av ) is determined by the arithmetic mean of the initial velocity (u) and final velocity (v) during a specific time period: v av =

u+v 2

The units for speed and velocity are the same, expressed as metres per second (m/s) or ms-1.

2.3 RATE OF CHANGE OF VELOCITY The velocity of a particle changes when: a) Its magnitude is changed (or) b) Its direction is changed (or) c) Both magnitude and direction are changed.

20

IL Foundation Series Class 9

     The change in velocity ∆v = v f − vi , where v f is the final velocity, and vi is the initial velocity.

2.3.1 Acceleration The change in velocity per unit time is called average acceleration.   • If the velocity of a particle changes from u to v in an interval of time, t, then its average     (v - u)  acceleration is a = . Here, u is called the initial velocity, and v is called the final velocity. t • Its SI unit is ms–2.

• Its dimensional formula is M 0 L1T −2  .

 • It is a vector quantity. Its direction is equal to the direction of ∆v, that is the direction of the change in velocity.

Instantaneous acceleration

The acceleration at an instant is called instantaneous acceleration.

• Its SI unit is ms-2.

• Its dimensional formula is M 0 L1 T −2  .

• It is a vector quantity. Its direction depends on the given situation.  dv • Mathematically, it is equal to the direction of . dt Uniform acceleration

If the acceleration of a particle is the same at every instant in a given time interval, then its acceleration is called uniform acceleration in that interval of time.

• Its SI unit is ms-2.

• Its dimensional formula is M 0 L1 T −2 .



• It is a vector quantity. Its direction is equal to the direction of ∆v (in this case, the direction of   dv and the direction of coincide with each other). ∆v dt 2.3.2 Jerk In physics, jerk refers to the rate at which the acceleration changes with respect to time. da Mathematically, jerk (j) is the derivative of acceleration (a) with respect to time (t), j = . dt The unit of jerk is m/s3 (metres per second cubed).

For example, if an object is undergoing a linear motion and its acceleration changes over time, the jerk can be calculated by finding the rate of change of acceleration with respect to time.

A real-life example of a jerk-like motion can be experienced in an elevator. When an elevator starts or stops abruptly, passengers inside may feel a sudden jolt or discomfort. This sensation is a result

21

MOTION

of the change in acceleration over a short period, and it is reflected in the jerk experienced by the occupants.

SOLVED EXAMPLES Example 1: A bus decreases its speed from 80 kmph to 60 kmph in 5 s. Find the acceleration of the bus.

Solution: Initial speed, u = 80 kmph 5 200 = 80 × ms −1 =ms −1 18 9

Final speed, v = 60 kmph 5 50 −1 = 60 × ms −1 = ms 18 3

The time taken to change its speed, t = 5 s v−u t 50 200 150 − 200 − −50 3 9 9 = = = 5 5 9×5 10 = − = −1.11 ms −2 9 ∴ acceleration, a =

Here, the negative sign indicates that the bus is in retardation.

Example 2: A train starts from a railway station. Moving with a uniform acceleration, it attains a speed of 40 kmph in 10 minutes. Find its acceleration.

Solution: Initial speed, u = 0 ms −1 Final speed, v = 40 kmph = 40 × =

5 ms −1 18

100 ms −1 9

The time taken to change its speed, t = 10 minutes = 600 s

22

= 600 s 100 −0 100 v−u 9 ∴ acceleration (a) == = t 600 9 × 600 1 ms −2 0.0185 ms −2 = = 54

IL Foundation Series Class 9

2.4 GRAPHICAL REPRESENTATION OF MOTION Graphs: A graph is a very powerful method of presenting information.

• Graphs provide a convenient method to present basic information. They are a more accessible means to grasp an overall understanding of the data.

• A graph is plotted to display the relation between two quantities.

Plotting a graph includes: a) Choosing the axes b) Choosing the scale c) Marking the points d) Joining the points.

Generally, a horizontal line from left to right is drawn to represent the independent quantity. A perpendicular line is drawn to represent the dependent quantity. These perpendicular lines meet at a point called the origin.

2.4.1 Distance-time graphs The distance-time graph of an object moving with a uniform speed is a straight line.

• Even though the distance-time graph is a straight line, the motion need not be along a straight path.

• If the distance-time graph of an object is a straight line, then the slope of the line gives the speed of an object.

• If an object moves with a non-uniform speed, its distance-time graph is not a straight line.

• The displacement-time graph for uniform accelerated motion is a parabola.

23

time

Rest

distance

distance

distance

MOTION

time

Uniform motion

time

Non-uniform motion

Fig. 2.3 Distance-time graphs

2.4.2 Velocity-time graphs If a particle moves with a constant velocity, the velocity-time graph will be a straight line parallel to the time axis.

• The area under the velocity-time graph of an object gives its displacement.

• The slope of the velocity-time graph gives the acceleration for an object moving along the straight line.

Time

Uniform velocity

Time

Uniform acceleration

Velocity

Velocity

Velocity

Velocity

• If the acceleration of an object moving along a straight line is not constant, the velocity-time graph is not a straight line.

Time

Uniform retardation

Time

Non-Uniform retardation

Fig. 2.4 Velocity-time graphs

2.5 EQUATIONS OF MOTION When an object moves along a straight line with uniform acceleration, the equations relating velocity (u, v) covered in a certain time interval (t) are: 2.5.1 Equation for velocity-time relation  Consider a particle moving along a straight line with a uniform acceleration, a. If its velocity   changes from u to v in an interval of time, t, then its acceleration is given by:    ( v − u)    = a ⇒ v= − u at t

24

IL Foundation Series Class 9

   v= u + at Important: The sign convention must be followed while using the scalar formula.

2.5.2 Equation for position-time relation

 Consider a particle moving along a straight line with a uniform acceleration, a . Let its velocity    change from u to v in an interval of time, t. Let its displacement in this duration be s.

From the definition of acceleration, we know that v = u + at. It is a linear equation. Hence, its average velocity over a time interval, t is given by:   u+v  v avg = 2 u+v or simply v avg = (∴ all the vectors are parallel to each other).

2 From the definition of average velocity, we know that displacement is given by: S = v ( avg ) t

From the above two equations, u+v Displacement, S =   t ...(1)  2  If we substitute v = u + at in the equation (1), then we get:  u + u + at   2u + at  2ut at 2  s =  t t = + =      2  2 2 2 1 ⇒ s = ut + at 2 2

1 Displacement, s= ut + at 2 2

2.5.3 Equation for position–velocity relation Similarly, if we substitute u = v – at, in equation (1), we get: 2vt at 2  v − at + v  S  ∴ = = t −    2 2 2 1 s vt − at 2 = 2

1 s vt − at 2 Displacement, = 2 In this manner, if we substitute t =

( v − u) in equation (1), we get, a

25

MOTION

v 2 − u2 2a 2 v − u2 = 2 as

s=

Displacement in the n th second: Consider a particle moving along a straight line with a uniform   acceleration, a . Let its total time of journey be n second, and its initial velocity be u.

The displacement in the last second is: S n = displacement in the n th second.

S n = displacement in the n th second. [ displacement in the first n second ] − [displacement in the first (n-1)second] [ displacement in the first n second ] − [displacement in the first (n-1)second] = S(n)-S(n-1) = S(n)-S(n-1) 1 1   = un + 1 an 22 − u(n − 1) + 1 a(n − 1) 22  = un + 2 an − u(n − 1) + 2 a(n − 1)  2 2   1 2  an 22 a 2an  = un + 1 an − un − u + an + a − 2an  = un + 2 an 2 − un − u + 2 + 2 − 2  2 2 2 2  a  =u + an − a =u + an − 2 a2 =u + an − a =u + an − 2 2 1   s n =+ u a  n − 1  s n =+ u a  n − 2  2  if we substitute v = u + an ⇒ u = v − an ,

s n =v − an + an − s n= v −

a 2

a 2

Important: • sn - sn-1 = a, i.e., the change in the displacement in two successive seconds is numerically equal to its acceleration.

• The sign convention must be followed while using all of these equations.

Note: In all of the above formulae, s represents total displacement in the time interval, t, from the initial instant.

26

IL Foundation Series Class 9

SOLVED EXAMPLES Example 1: The driver of train A, travelling at a speed of 60 km/h applies brakes and retards the train uniformly; the train stops in 5s. Another train, B, is travelling on a parallel track with a speed of 36 km/h. This driver also applies the brakes, and the train retards uniformly. Train B stops in 10s. Plot a speed-time graph for both trains on the same paper. Also, calculate the distance travelled by each train after the brakes were applied.

Solution:

Speed (kmph)

60 50 40 30 20 10

B

A

5 Time (S)

10

The distance travelled by train A = Area =

1 5 × 5 × 60 × = 41.6 m 2 18

1 5 The distance travelled by train B = × 10 × 36 × =50 m 2 18 Example 2: An object starts linear motion with a velocity, u, and under uniform acceleration, a, it acquires a velocity, v, in time, t. Draw a velocity-time graph. From this graph, obtain the following equation: v = u + at.

Solution:

From the graph, we know that slope = acceleration,

27

MOTION

a=

v−u , t

i.e., a =

v−u , t

⇒ v = u + at

Example 3: Draw a graph of velocity versus time for a body that starts to move with velocity, u, under constant acceleration, a, for time, t. Using this graph, derive an expression for distance, s, covered in time, t.

Solution: B

Velocity

v

(v - u) t

A

C u

u

O

Time

D

From the graph, we know that: displacement, s = Area under curve = Area of Trapezium OABD

Area of OACD + Area of ABC 1 =ut + (v − u)t 2 1 = ut + (at )t ( v − u = at ) 2 1 s ut + at 2 = 2 Example 4: Derive the equation: v 2 − u 2 = 2as graphically.

Solution: Consider an object with an initial velocity, u. It covers a distance, s, and attains a velocity, v, to achieve a uniform acceleration, a. Then, the graph between velocity and time will be as shown in the figure below.

28

IL Foundation Series Class 9

From the graph, we know that: displacement, s = Area under the line = Area of Trapezium OABD = Area of OACD + Area of ABC 1 =ut + (v − u)t 2 vt ut = + ut − 2 2  v + u = t  2 

⇒ v2

v − u  v + u  v − u  = a  2   a   t  v 2 − u2 = 2a 2 = u + 2aS

Example 5: A car accelerates uniformly from 18 Kmh-1 to 36 Kmh-1 in 5s. Calculate: i) the acceleration and ii) the distance covered by the car.

Solution: We are given that: = u 18 = kmh −1 5 ms −1 = = kmh −1 10 ms −1 v 36 t = 5 s.

i) We have: a=

v-u 10 ms -1 -5 ms -1 = =1 ms -2 t 5s

ii) We have: 1 s = ut + at 2 2 1 × 1 ms -2 × (5 s)2 2 = 25 m + 12.5 m = 37.5 m = 5 ms -1 × 5 s +

The acceleration of the car is 1 ms-2, and the distance covered is 37.5 m.

29

MOTION

Example 6: A stone is thrown in a vertically upward direction with a velocity of 5 ms −1 . If the acceleration of the stone during this 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?

Solution: i) Initial velocity, u = 5 ms −1 (upward direction) Acceleration, a = −10 ms −2 (direction) at maximum height, v = 0 ms −1 ∴ Maximum height attained by the stone, s = ?  v 2 − u2 = 2aS ⇒ s=

v 2 − u 2 0 − (5) 2 25 = = = 1.25 m 2a 2(−10) 20

ii) Time taken to reach maximum height, = ?  v= u + at v − u −5 ⇒= t = = 0.5 s. a −10

2.6 CIRCULAR MOTION If a particle is moving along a circular path, then the particle is said to be in a circular motion.

2.6.1 Uniform circular motion If the speed of the particle is constant and moving along a circular path, then it is called a uniform circular motion.

In the case of circular motion, radial and tangential directions are important. In the case of a circle, normal direction coincides with radially inward direction.

The direction of velocity at any point is given by the tangent drawn at that point to the circular path.

V V

V

V

V

Fig. 2.5 Circular motion 30

IL Foundation Series Class 9

From the figure, it is very clear that the direction of the velocity of the particle changes continuously, i.e., the velocity of the particle is changing continuously. Hence, the particle must possess some acceleration. This acceleration lies along the radius towards the centre at any instant.

Hence, it is called centripetal acceleration.

From the definition of acceleration, the centripetal acceleration for a particle moving with a speed v along a circular path of radius r is an =

v2 r

Where an = instantaneous acceleration of the particle at the given point v = Speed of the particle at a given point r = radius of the circle (In general, the radius of curvature of the curve at a given point) Note: 1. For a curve, the direction of the centripetal acceleration is along the normal towards the centre of curvature.

2. In the figure, the direction of acceleration changes continuously, i.e., the acceleration is nonuniform even though the speed is constant.

3. The acceleration of a particle having uniform circular motion is not constant.

Centripetal force

A particle moving in a circle is accelerated, and the acceleration can be produced only if a resultant force acts on it.

V

V

V V V Fig. 2.6 Centripetal force

The resultant force along the normal direction, directed towards the centre, is called centripetal force. It is given by: FN =

mv 2 r 31

MOTION

Centrifugal reaction

It is the radial force acting outwards on the agency which makes the body move in a circular path.

When an electron moves around the nucleus, it experiences the centripetal force, while the nucleus experiences a centrifugal reaction. When a planet moves around the sun, the planet experiences the centripetal force, and the sun experiences the centrifugal reaction.

Centripetal force and centrifugal reaction form an action-reaction pair. Centripetal force and centrifugal reaction are real forces. In magnitude, centripetal force = centrifugal force.

Centrifugal force

The pseudo force, which acts radially outwards on the body moving along a circle, is called centrifugal force.

Even though centripetal and centrifugal forces are equal in magnitude and opposite in direction, they do not form an action-reaction pair. They act on the same body in two different frames.

The magnitude of a centrifugal force is equal to that of a centripetal force.

An observer can observe the centrifugal force when he is also moving along with the body moving along a circle.

Cream can be separated from milk using a centrifuge. Wet clothes can be dried by a centrifuge.

QUICK REVIEW • The length of the path covered by an object is called distance.

• The shortest distance measured from the initial to the final position of an object is known as displacement.

• An equal distance covered in equal intervals of time is known as uniform motion.

• An unequal distance covered in equal intervals of time is known as non-uniform motion.

• Speed is defined as the distance covered in a unit time.

• The SI unit for speed is metres per second (m/s), often represented as ms-1 .

• Average speed =

Total distance travelled Total time taken

• Average Velocity =

Total displacement Total time taken

   ( v − u) • Change in velocity per unit time is called average acceleration a = t • If the acceleration of a particle is the same at every instant in a given time interval, then its acceleration is called uniform acceleration in that interval of time.

• The distance-time graph of an object moving with a uniform speed is a straight line.

32

IL Foundation Series Class 9

• If the particle moves with a constant velocity v, the velocity-time graph will be a straight line parallel to the time-axis.

• The area under the velocity-time graph of an object gives its displacement.

• Equation for velocity-time relation: v= u + at 1 • Equation for position-time relation: = s ut + at 2 2 2 as • Equation for position-velocity relation: v 2 − u 2 = • If a particle is moving along a circular path, then the particle is said to be in a circular motion.

• The centripetal acceleration for a particle moving with a speed, v, along a circular path of radius, v2 r, is a n = . r • The resultant force along the normal direction, directed towards the centre is called centripetal mv 2 force. FN = r

WORKSHEET - 1 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER I.

Motion and rate of change of velocity

1. Convert 15 m/s into km/h: a. 24 km/h

b. 30 km/h

c. 50 km/h

d. 54 km/h

2. If s is displacement and d is the distance between two points on a straight line, then the correct relation is: a. s = d

b. s > d

c. d > s

d. None

3. 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: a. Moving with no acceleration

b. At rest

c. In accelerated motion

d. Moving with uniform velocity

4. A bus decreases its speed from 80 kmph to 60 kmph, in 5 s. The acceleration of the bus is: −2

a. a = −2.11 ms −2

b. a = +2.11 ms

c. a = −1.11 ms −2

d. a = +1.11 ms −2

5. A motorcyclist drives from point A to B with a uniform speed of 30 kmph and returns back with a speed of 20 kmph. Find his average speed: a. 20 kmph

b. 24 kmph

c. 42 kmph

d. 52 kmph

33

MOTION

6. A body, thrown in the vertically upward direction, rises up to a height, h, and comes back to its starting position. What is: i) The total distance travelled by the body and ii) The displacement of the body? a. 4 h, zero

b. 2 h , zero

c. zero, 4 h

d. zero, 2 h

7. If a particle covers equal distances in equal time intervals, it is said to: a. Be at rest

b. Move with uniform speed

c. Move with a uniform velocity

d. The particle moves along a straight line

8. A quantity has a value of -6.0 m/s. It may be the: a. Speed of a particle

b. Velocity of a particle

c. Acceleration of particle

d. Position of a particle

9. A bus decreases its speed from 80 m/s to 40 m/s in 5 s. Then, the acceleration of the bus is: a. 8 ms-2

b.- 8 ms-2

c. 10 ms-2

d. -10 ms-2

10. A racing car has a uniform acceleration of 4 ms-2. The distance it will cover in 10 s after the start is: a. 100 m

b. 200 m

c. 300 m

d. 400 m

11. Assertion (A): A body may be moving with a uniform speed and a non-uniform acceleration.

Reason (R): A body may have a uniform velocity and a non-zero acceleration. a. Both A and R are correct, and R is the correct explanation of A. b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, R is incorrect. d. A is incorrect, R is correct.

12. Assertion (A): A body may be accelerated even when it is moving at uniform speed.

Reason (R): When the direction of motion of a body is changing, then the body may have an acceleration. a. Both A and R are correct, and R is the correct explanation of A. b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, R is incorrect. d. A is incorrect, R is correct.

34

IL Foundation Series Class 9

II. Graphical representation of motion 1. If the displacement of an object is proportional to the square of time, then the object moves with: a. Uniform acceleration

b. Decreasing acceleration

c. Increasing acceleration

d. Uniform velocity

time(s)

d.

time(s)

distance(m)

c.

distance(m)

b.

distance(m)

a.

distance(m)

2. Which of the following figures represents the uniform motion of a moving object correctly?

time(s)

time(s)

3. The slope of a velocity-time graph gives: a. The distance

b. The displacement

c. The acceleration

d. The speed

4. Four cars, A, B, and C, are moving on a levelled road. Their distance versus time graphs are shown in Figure. Choose the correct statement.

distance(m)

C D A B time(s)

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.

5. When a graph of one quantity versus another results in a straight line, the quantities are a. Both constant

b. Equal

c. directly proportional

d. Inversely proportional

6. The velocity-time graph of a particle is not a straight line. Its acceleration is: a. Zero

b. Constant

c. Negative

d. Variable

35

MOTION

7. Whenever an object moves with a constant speed, its speed-time graph is a: a. Parabola, parallel to the time axis b. Straight line, perpendicular to the time axis c. Straight line, parallel to the time axis d. Parabola, perpendicular to the time axis 8. The area under a speed-time graph is represented by the unit: a. m

b. m2

c. m3

d. m–1

III. Equations of motion 1. Which of the following is correct?

1 a. = s ut + t 2 2

c. v = u + at

b. v 2 − u 2 = 2a

1  d. s n =u +  n −  2 

2. A car, starting from rest, travelled a distance s and gained a velocity v0. What is the distance travelled by the car when it continues with the same acceleration and velocity doubles? a. 4s

b. 2s

c. 3s

d. s

3. A police van moving on a highway with a speed of 30 km/h, fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km/h. If the muzzle speed of the bullet is 150 m/s, at what speed does the bullet hit the thief’s car? (Note: Find the speed that is relevant for damaging the thief’s car). a. 105 m/s

b. 100 m/s

c. 95 m/s

d. 110 m/s

4. A car accelerates from rest at a constant rate, α, for some time, after which it decelerates at a constant rate, β, to come to rest. If the total time elapsed is t second, then the maximum velocity reached is:  αβ  a. Vmax =   t α + β 

 αβ  1 b. Vmax =   α + β  t

α +β c. Vmax = (α + β )t d. Vmax = αβ t αβ 5. Person x moves from point A to point B with a uniform speed of 30 kmhp, and from B to A with 60 kmhp, then the average speed is _______×101 m/s. a. 10/9 b. 20/9 c. 50/9 d. 40/9 6. A particle moving with a constant acceleration of 2 m/s2 due west has an initial velocity of 9 m/s due east. The distance covered in the fifth second of its motion will be: 1 1 3 a. m b. 1 m c. m d. m 2 4 4 36

IL Foundation Series Class 9

7. Average velocity of a particle moving in a straight line with constant acceleration 'a' and initial velocity 'u' in 't' seconds is at u + at at a. u + b. u + at c. d.

2 2 2 IV. Circular motion

1. When a particle moves in a circle with a uniform speed: a. Its velocity and acceleration are both constant b. Its velocity is constant, but the acceleration changes c. Its acceleration is constant but velocity changes d. Its velocity and acceleration both change 2. In a circular motion, the: a. Direction of motion is fixed

b. Direction of motion changes continuously

c. Acceleration is zero

d. Velocity is constant

3. When a body moves with a constant speed along a circle: a. Its velocity remains constant

b. No force acts on it

c. No work is done on it

d. No acceleration is produced by it

4. An object follows a curved path. The quantity from the following that may remain constant during the motion is the: a. Direction of velocity

b. Velocity

c. Acceleration

d. Magnitude of acceleration

5 Which of the following is incorrect about uniform circular motion? a. Speed remains the same, but direction changes. b. Acceleration is directed toward the center. c. Velocity always points toward the center. d. The object is constantly changing its velocity.

6. A 500 kg car takes a round turn of radius 50 m with a velocity of 36 kmph. The centripetal force required is: a. 10 N

b. 100 N

c. 1000 N

d. 900 N

7. If the radii of circular paths of two particles of the same masses are in the ratio of 1:2, then in order to have a constant centripetal force, their velocities should be in the ratio of: a. 1:4

b. 4:1

c. 1: 2

d.

2 :1

37

MOTION

8. If a body moves on the circumference of a circle in a horizontal plane with speed equal to that it would acquire by falling freely through half the radius of the circle. Its centripetal acceleration is: g g g b. c. g d.

2 3 4 9. A particle moves over three-quarters of a circle of radius, r. What is its displacement?

a.

a.

2r

b.

2r

c. 3 r

d.

3r

WORKSHEET - 2 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Choose the correct statement. a. Rest and motion are absolute. b. Rest is absolute, but motion is relative. c. Rest is relative, but the motion is absolute. d. Rest and motion are relative.

2. The change in velocity is: a. Absolute in an inertial frame b. Relative in an inertial frame c. Sometimes relative in an inertial frame d. Sometimes absolute in an inertial frame 3. Identify the relative quantities: 1. Rest

2. Mass

3. Motion

4. Time

a. 1 and 2 are correct

b. 2 and 4 are correct

c. 1 and 3 are correct

d. 1,3 and 4 are correct

4. A man walks 12 due North, 4 m due East and finally climbs up a vertical pole up to a height of 3 m. Then, the distance travelled by him is: a. 12 m

b. 13 m

c. 19 m

d. 17 m

5. Statement (A): It is not possible to have a body moving with constant velocity but varying speed.

Statement (B): It is possible to have a body moving with constant speed but has no displacement.

38

a. Both A and B are true.

b. Both A and B are false.

c. A is true, but B is false.

d. A is false, but B is true.

IL Foundation Series Class 9

6. Assertion (A): The displacement-time graph of a body moving with a uniform acceleration is not a straight line.

Reason (R): The displacement is proportional to the square of time. a. Both A and R are correct, and R is the correct explanation of A. b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, but R is incorrect. d. A is incorrect, but R is correct.

7. Assertion (A): The position-time graph of a body moving uniformly in a straight line parallel to the position axis.

Reason (R): The position-time graph in a uniform motion gives the velocity of an object. a. Both A and R are correct, and R is the correct explanation of A. b. A Both A and R are correct, but R is not the correct explanation of A. c. A is correct, but R is incorrect. d. A is incorrect, but R is correct.

8. A racing car has a uniform acceleration of 2 m/s2. What distance will it cover in 5 s after the start? a. 25 m

b. 35 m

c. 20 m

d. 15 m

distance(m)

9. The displacement-time graph for a body is given in Figure. The ratio of the velocities of A and B is:

B A 450

300

time(s)

a.

3 :1

b. 1:1

c. 1:2

d. 1: 3

10. Two children start at one end of a street (the origin), run to the other end, and then head back.

On the way back, Sita is ahead of Gita. Which statement is correct about the displacements from the origin at that moment? a. Gita has run a greater distance, but her displacement is less than Sita’s. b. Gita has run a greater distance, and her displacement is also greater than Sita’s. c. Sita has run a greater distance, but her displacement is less than Gita’s. d. Sita has run a greater distance, and her displacement is also greater than Gita’s.

39

MOTION

11. Ball A was dropped from the top of a tall building. At the same instant and from the same height, ball B was thrown straight downward. Neglecting the effects of air friction, compare their accelerations while they were falling. a. Their accelerations are equal. b. Ball A has the greater acceleration. c. Ball B has the greater acceleration. d. Initially A has a greater acceleration, but after some time B has the greater acceleration.

12. Which of the following speed-time graphs is not possible? v

v

a.

b.

c. t

t

v

v

d. t

t

13. A bullet is fired onto a wall with a velocity of 50 m/s. If the bullet is at a depth of 10 cm inside the wall, find the retardation produced by the wall. a. −1250 ms −2

b. 12500 ms −2

c. −2050 ms −2

d. 3250 ms-2

14. A particle experiences a constant acceleration for 20 s after starting from rest. If it travels a distance s in the first 10 s and distance s2 in the next 10 s, then: a. s2 = s1

b. s2 = 2 s1

c. s2 = 3 s1

d. s2 = 4 s1

15. A train is moving at a speed of 40 km/h at 10:00 am and at 50 km/h at 10:15 am. Assuming that the train moves along a straight track, and the acceleration is constant, find the value of the acceleration. a. 30 km/h2

b. 40 km/h2

c. 35 km/h2

d. 25 km/h2

16. A car covers 30 km in 30 minutes and the next 30 km in 40 minutes. Calculate the average speed for the entire journey. a. 51.4 km/h

b. 52.4 km/h

c. 25.4 km/h

d. 21.4 km/h

17. A particle with a velocity of 2 m/s at t = 0 moves along a straight line with a constant acceleration of 0.2 m/s2. Find the displacement of the particle in 10 s. a. 40 m

b. 30 m

c. 50 m

d. 35 m

18. A bus starting from rest moves with a uniform acceleration by 0.1 ms −2 for 2 minutes, then the speed acquired and distance travelled are_________, respectively.

40

a. 12 m, 720 ms −1

b. 72 ms −1 ,120 m

c. 12 ms −1 , 720 m

d. 720 ms −1 ,12 m

IL Foundation Series Class 9

19. The area under a v-t graph represents: a. Acceleration

b. Displacement

c. Velocity

d. Jerk

20. The direction of a centripetal force in a uniform circular motion is: a. Towards the centre

b. Tangential

c. Away from the centre

d. None

21. Which of the following can be zero for an object moving on a circular path? a. Distance b. Displacement c. Both distance and displacement d. None 22. A block slides down on the track shown below. Comment on its speed and acceleration in the direction of motion. (Assume that the friction is absent everywhere.)

a. Speed decreases, Acceleration decreases b. Speed decreases, Acceleration increases c. Speed increases, Acceleration decreases d. Speed increases, Acceleration increases 23. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first? a. The faster one

b. The slower one

c. Both will reach simultaneously

d. Depends on the masses

24. A body moves at a speed of 10 ms-1 due East. After 10 minutes, it moves at a speed of 5 ms-1 due West. The change in the velocity of the body during this time interval is: a. 15 ms-1

b. 5 ms-1

c. -5 ms-1

d. -15 ms-1

25. A body moves with a speed of 10 ms-1 due East. After 10 minutes, it moves at a speed of 5 ms-1 due West. The change in the speed of the body is: a. 5 ms-1

b. 0

c. -5 ms-1

d. 15 ms-1 41

MOTION

26. The following figure shows a particle moving with a uniform speed of 5 ms-1 along a circular path. In its journey from A to B:

Statement (A): The change in speed is zero.

Statement (B): The change in velocity is zero. a. Both A and B are true.

b. Both A and B are false.

c. A is true, but B is false.

d. A is false, but B is true.

27. A car is moving towards the East with a velocity of 20 ms-1 . Its velocity after 10 s is 20 ms-1 due West. The change in its velocity in this duration of 10 s. a. 20 ms-1 due East

b. 40 ms-1 due East

c. -20 ms-1 due East

d. -40 ms-1 due East

28. Two particles of equal masses are revolving in circular paths of radii r1 and r2, respectively, with the same period. The ratio of their centripetal force is 2

2

r  r  b. c.  1  d.  2   r1   r2  29. Statement (A): In a uniform circular motion, the magnitude of the centripetal acceleration is constant r a. 1 r2

r2 r1

Statement (B): When the magnitude of the centripetal acceleration is constant, the body must be in a uniform circular motion. a. Both A and B are true.

b. Both A and B are false.

c. A is true, but B is false.

d. A is false, but B is true.

30. Assertion (A): A body with a constant magnitude of acceleration may not speed up.

Reason (R): In the case of uniform circular motion, though the magnitude of acceleration is constant, the body moves with uniform speed. 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 the incorrect explanation of A. c. A is true, R is false. d. A is false, R is true.

42

3

FORCE AND LAWS OF MOTION

3.1 INTRODUCTION 3.1.1 Force Force is an external agent of push or pull that changes or tends to change the state of rest or the state of motion of a body.

Let us consider a few examples, Example 1: Imagine you are going to a school picnic. The bus has some trouble with its battery.

The driver asks the boys to push the bus. What would the boys do? They apply force during pushing so that the bus starts.

Example 2: Imagine that the door of your house is jammed, as often happens in the rainy season.

Your mother tries to pull it so that it opens. However, the door does not open. She calls you for help.

Both of you apply force to pull the door.

From the above examples, it is clear that the word force is associated with either pull or push, which causes some kind of motion.

3.1.2 Effects of force •

A force can cause a motion in a stationary object.

•

A force can stop the moving object or slow it down.

•

A force can make a moving object move faster.

•

Force can change the direction of a moving object.

•

Force can change the shape of an object.

3.2 TYPES OF FORCES 3.2.1 Balanced force If a set of forces acting on a body produces no acceleration in it, the forces are said to be balanced.

Fig. 3.1 Balanced force 43

FORCE AND LAWS OF MOTION

3.2.2 Unbalanced force If a set of forces produces a non-zero acceleration, the forces are said to be unbalanced.

Fig. 3.2 Unbalanced force

3.3 NEWTON'S LAWS OF MOTION 3.3.1 First law of motion Newton's First Law of Motion, also known as the Law of Inertia, states that an object remains at rest or in uniform motion in a straight line unless acted upon by an external force.

Explanation: 1st law explains two important things. They are: a) It answers the question, 'What is force?', and explains the force qualitatively. From this law we can understand force as a physical quantity that changes the state of rest or uniform motion and causes acceleration. b) Nobody can change its state of rest or uniform motion. It is the property of a body called 'inertia'.

3.3.2 The law of inertia Inertia: Inertia is the inherent property of a body to oppose change of natural state. It can be expressed as the tendency of a body to oppose any change in its state of rest or of uniform motion.

The mass of an object serves as a quantitative measure of its inertia.

To illustrate, consider a heavy truck and a lightweight bicycle. The truck, possessing greater mass, exhibits greater inertia, making it more resistant to changes in motion or rest compared to the bicycle. This concept is encapsulated by Newton's first law of motion.

I nertial mass can be measured by Newton's second law of motion (which we are going to discuss) Mass that can be measured by a common balance or spring balance is called gravitational mass.

Einstein proved in his principle of equivalence that the gravitational mass of a body is equal to the inertial mass.

44

IL Foundation Series Class 9

Types of Inertia: Inertia is of three types: 1. Inertia of rest: The inertia (resistance) offered by a body to change its state of rest is called inertia of rest. Examples are: • When a branch of a tree is shaken vigorously, ripened fruits get detached and fall. This is because the branch comes into motion, but the fruits tend to remain at rest due to inertia of rest and get detached. After the fruits get detached, gravity plays its role in making the fruits fall.

•

To remove dust from a carpet, we hang the carpet and then beat it with a stick. Due to the beating action, the carpet moves along the stick, and the dust particles, with a tendency to remain at rest, get detached from the carpet.

•

When you hit a striker on a pile of carrom counters, you will observe that only the lowest counter moves away. The rest of the pile remains in its original position.

Fig. 3.3 Inertia of rest

2. Inertia of motion: The inertia (resistance) offered by a body to change its state of uniform motion is called inertia of motion. Examples are: •

hat happens when a bus stops suddenly? The passengers standing casually experience a jerk W forward. This happens because the foot is in contact with the floor. Due to frictional force, the foot stops along with the bus, but the upper part of the body, which is in motion, tends to remain in motion and experiences a jerk forward.

•

When we switch off a fan, it continues to rotate for a while due to inertia of motion.

•

In a long jump, people utilise inertia of motion to perform the jump forward.

Fig. 3.4 Inertia of motion

45

FORCE AND LAWS OF MOTION

3. Inertia of direction: The resistance offered by a body to change its direction of motion is called inertia of direction. Examples are: •

When a bus suddenly takes a turn, the passengers sitting casually experience a jerk in the outward direction. This happens because the passenger tends to remain in its original direction of motion due to inertia of direction.

•

S uppose a stone tied to a string is whirled around in a circle. What happens when the string is cut? The stone flies off tangentially due to the inertia of direction.

Fig. 3.5 Inertia of direction

Example: There are three solid balls made up of aluminium, steel, and wood of the same shape and volume. Which of them would have the highest inertia? Why? Solution: The natural tendency of objects to resist a change in their state of rest or uniform motion is called inertia. m We know that d = , and all three balls have the same volume (V). As we can see, density (d) is V directly proportional to mass. Since steel has the highest density among the three materials, it will have the highest inertia.

3.4 SECOND LAW OF MOTION 3.4.1 Momentum  Momentum or p is the property of a body which depends on its mass and velocity and is given by the product of mass and velocity.

= mass × velocity momentum   ⇒p = mv

46

IL Foundation Series Class 9

• •

Its SI unit is kgms −1 .

The dimensional formula is [ MLT−1 ] .

•

The magnitude of linear momentum (p) = mv.

Relation between momentum (p) and kinetic energy (K) 1 Kinetic energy K = mv2 2

1 m m2 v 2 p2 mv2 × = = 2 m 2m 2m 2 p = 2mK =

3.4.2 Second law of motion Newton's second law of motion states that the rate of change of momentum is directly proportional to the force applied and acts along the direction of the applied force.

Mathematical formulation of the second law of motion

 Consider a body of mass 'm' is moving with a velocity u initially, after a time 't', its velocity       becomes v , then the change in momentum is (p) = pf − pi = mv − mu   mv − mu rate of change of momentum = t according to the above statement,  mv − mu ⇒F ∝ t  m(v − u ) ⇒F ∝ t   ⇒ F ∝ ma   kma ⇒F= Here, k is a proportionality constant, and its value depends on the definition of unit force. Here, k = 1. ⇒F= ma •

The SI unit of force is newton (N) or kgms−2 .

•

The dimensional formula is [MLT ].

•

It is a vector quantity.

-2

1 newton: One newton force can be defined as the force required to produce an acceleration of 1 ms-2 in a body of 1 kg mass.

Note: 1 kgwt = 9.8 N

47

FORCE AND LAWS OF MOTION

3.4.3 Impulse The product of force and time, or a large force handled in a short period of time is called impulse. ∴I = F × t    = I m(v − u)    = I Pf − Pi •

SI unit: Ns

•

The dimensional formula is [MLT-1].

•

It is the direction of change in momentum produced by the force.

Example: Force acting between two billiard balls during a collision.

Fig. 3.6 Billiard balls during collision

SOLVED EXAMPLES Example 1: A ball of mass 10 g is moving with a velocity of 50 ms-1. On applying a constant force on the ball for 2 s, it acquires a velocity of 70 ms-1. Calculate the magnitude of force.

Solution: Given, m= = 10 g

10 = kg 0.01 kg, 1000

−1 = u 50 ms = , t 2.0 = s, v 70 ms −1.

Initial momentum of the = ball mass × initial velocity = mu = 0.01 × 50 = 0.5 kgms −1 Final momentum of the = ball mass × final velocity = mv = 0.01 × 70 =

48

0.7 kgms -1

IL Foundation Series Class 9

Rate of change of momentum =

=

Final momentum - Initial momentum Time interval

(0.7 − 0.5) = 0.1kgms −2 ( or 0.1N) 2.0

v − u (70 − 50) = = 10 ms −2 t 2 Force = mass × acceleration = 0.01 kg × 10 ms-2 = 0.1 N

Acceleration = a

Example 2: For how long should a force of 100 N act on a body of mass 20 kg so that it acquires a velocity of 100 m/s?

Solution: Given, = F 100 = N, m 20 kg F 100 a = = = 5 m / s2 m 20

Now u = 0, v = 100 m/s, a = 5 m/s2 = t

v − u 100 − 0 = = 20 s a 5

Example 3: An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and the road if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2?

Solution: Mass of the automobile, m = 1500 kg Acceleration of the automobile, a = −1.7 m / s2 Force between vehicle and road F= ?

F = ma = 1500 × (−1.7) F = −2,550 kgm / s2 The '-ve' sign indicates that the force acts in a direction opposite to the motion of the vehicle.

3.5 THIRD LAW OF MOTION 3.5.1 Newton's third law of motion Newton's third law states that for every action, there exists an equal and opposite reaction. With this law, it is clear that forces always exist in action reaction pairs. Single force can never exist.

Example 1: Let us consider a block suspended by a string. The Earth exerts force on the block (action), and an equal amount of force is exerted by the block on the Earth (reaction). Similarly, 49

FORCE AND LAWS OF MOTION

the block exerts force on the support through the string (action), and an equal amount of reaction is exerted by support on the block (reaction).

In this case, the action and reaction are equal to the mg.

Support T Action - reaction pair T

Action - reaction pair Earth

Fig. 3.7 Block suspended by a spring

Example 2: While walking, we press the ground (action) with our feet slanted in the backward direction. The ground exerts an equal and opposite force on us. The vertical component of the reaction force balances our weight, and the horizontal component provides us with forward motion.

FV R

FH

FH R

W

Ground

FV

Fig. 3.8 Pressing the ground while walking

Example 3: Let us consider a horse that starts pulling a cart from rest, gradually gains speed.

The ground exerts force on the horse, whose horizontal component is FH. The force exerted by the horse on the cart in forward direction is T. The cart exerts a force T on the horse in the backward direction. In addition to this, there is a frictional force on the wheels of the cart in the backward direction.

Normal Force (N) Reaction (FR) Friction (Ff)

Action (FA)

Weight (W) Fig. 3.9 The horse and the cart

50

IL Foundation Series Class 9

  Note: If FAB is the force exerted on body A by body B (i.e., action), and FBA is the force exerted on   body B by body A(i.e. reaction), then according to the third law FAB = −FBA 3.5.2 Conservation of momentum Law of conservation of momentum

When two (or more) bodies act upon one another, their total momentum remains constant (or conserved), provided no external force acts on them. m1 A

u1

m2 B

Before collision

u2

m1 m2

m1

B

A

A

FAB FBA During collision

v1

m2 B

v2

After collision

Fig. 3.10 Law of conservation of momentum

•

Suppose two balls, A and B, of masses m1 and m2, are travelling in the same direction with different velocities u1 and u2 ( u1 > u2 ) before the collision, v1 and v2 ( v2 > v1 ) after the collision.

Then according to the law of conservation of momentum: m1u1 + m2 u2 = m1 v1 + m2v2

•

Total momentum of system before collision = Total momentum of the system after collision.

•

The law of conservation of linear momentum is based on Newton's third law of motion. This is the fundamental law of nature, and there is no exception to it.

•

This law is valid only for linear motion.

•

Internal forces cannot change the total momentum of the system. However, they may change the momentum of each particle of the system.

•

The motion of a rocket, the firing of a bullet from a gun, and the explosion of a shell fired from a cannon are some examples where we can apply the law of conservation of linear momentum.

Recoil of gun

• •

•

• •

If a stationary gun fires a bullet horizontally, the total momentum of the gun and the bullet is zero before and after firing.   According to the conservation of linear momentum, m1v1 + m2 v2 = 0  −m v  ∴ v2 = 1 1 m2 The negative sign shows that as the bullet moves in the forward direction, the gun moves in the backward direction, i.e., the gun recoils.  If ' v ' is the muzzle velocity of the bullet (velocity of bullet w.r.t. gun), the velocity of the gun   −m1v V= . m1 + m2 The magnitude of the gun's momentum is equal to that of the bullet's.

P2 KE = .

2m 51

FORCE AND LAWS OF MOTION

• •

As p = constant, the bullet has greater kinetic energy than the gun. m K.E.gun = 1 K.E.bullet m2

Tension in a string

A block of mass m hangs at the end of a massless string.When the block accelerates in the upward direction or decelerates in the downward direction, the tension in the string is T = m(g + a).

•

When the block accelerates in the downward direction or decelerates in the upward direction T = m (g - a).

•

When the block is moved up or down with a uniform speed, the tension T = mg.

3.5.3 Tension force Pulley (Atwood's Machine): The Atwood machine is used to determine the acceleration due to gravity at a place. The simple Atwood machine consists of two masses, 'm1' and 'm2', connected by a light string passing over a smooth, frictionless pulley fixed to the ceiling. When the system is released, both the masses move with the same acceleration, say 'a'. As the pulley is smooth, the tension of the string is the same on either side of the pulley.

Let 'm1' be greater than 'm2'.

The mass 'm1' moves down, and 'm2' moves up with the same acceleration 'a'.

Free body diagram of 'm1':

T

T

T a

m1

m2 m g 1

Fig. 3.11 Free body diagram of 'm1'

The forces acting on m1 are: (1) its weight m1g acting down and (2) the tension T acting upward ∴ m1 g − T = m1a ...(1) Free body diagram of 'm2' is

52

IL Foundation Series Class 9

T

a

m2g Fig. 3.12 Free body diagram of 'm2'

The forces acting on m2 are: (1) Its weight acting downward, and (2) The tension T acting upward. ∴ T − m2 g = m 2 a …....................................................(2) Adding (1) & (2), m1 g − m 2 g =( m1 + m 2 ) a

( m1 − m2 ) g ................................(3) ∴ acceleration = m1 + m 2 Substituting the value of 'a' in equation (1) m1 g − T = m1

( m1 − m 2 ) g m1 + m 2

= T m1 g − m1 g

∴

( m1 − m 2 ) m1 + m 2

 ( m + m 2 ) − ( m1 − m 2 )  = m1 g  1  m1 + m 2    2m1m2  T  =  g …..................................(4)  m1 + m 2 

SOLVED EXAMPLES Example 1: A machine gun has a mass of 20 kg. It fires 35 g bullets at the rate of 4 bullets per second, with a speed of 400 ms-1. What force must be applied to the gun to keep it in position?

Solution: Mass of the gun, M = 20 kg Velocity of the bullet, v = 400 ms-1 If V is the velocity of the recoil, then 53

FORCE AND LAWS OF MOTION

V =

mv 0.035 × 400 −1 ms= 0.7 ms−1 = M 20

1 second 4 If F is the average force required to hold the gun in position, then Time taken to fire one bullet, t =

∴ F=

MV 20 × 0.7 N= 56N = t 1/ 4

Example 2: Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 ms-1 before the collision during which they stick together. What will be the velocity of the combined object after the collision?

Solution: Mass of the first object, m1 = 1.5 kg Mass of the second object, m2 = 1.5 kg Velocity of the first object u1 = + 2.5 m/s Velocity of the second object u2 = -2.5 m/s (The '-ve' sign indicates movement in the opposite direction) Velocity of the combined mass, v = ?

From the law of conservation of momentum,

( m1 + m2 ) × v= m1u1 + m2u2  m u + m2 u2  = 1 1   m1 + m2   1.5 × 2.5 + 1.5 × (−2.5)  = v =  0 m/s 1.5 + 1.5   v

The velocity of the combined object is zero.

Example 3: A bullet of mass 10 g travelling horizontally with a velocity of 150 ms -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.

Solution: Mass of bullet, m=10 g = 0.01 kg Initial velocity of the bullet, u = 150 m/s Final velocity of the bullet, v = 0 m/s Time taken t = 0.03 s Force exerted by the block on the bullet F = ?

54

IL Foundation Series Class 9

m(v − u) 0.01 × (0 − 150) = = −50kgms−2 t 0.03 force Acceleration = mass −50 = −5000 m / s2 a= 0.01 F=

Distance of penetration, S = ?

= S

v2 − u2 02 − (+150)2 150 × 150 = ⇒ = S = 2.25m 2a 2 × (−5000) 2 × 5000

Example 4: Two blocks of masses 4 kg and 6 kg are connected by a string, as shown in the figure.

They are initially at rest on the floor. Calculate their acceleration when a force of (a) 50 N (b) 100 N is applied on the pulley.

Solution: Case (i): F = 25 N. 2 The applied force on each block is less than the weight of each block. So, the blocks will not move.

Force on the 6 kg and 4 kg objects =

F

50N

4 kg 6 kg

4 kg

40N

Case (ii):

100 = 50 N 2 The 6 kg block will not move because the applied force is less than its weight.

Force on each block = On 4 kg mass Resultant force = mass × acceleration 50 - 40 = 4a; 10 = 4a; a = 2.5 m/s2

55

FORCE AND LAWS OF MOTION

QUICK REVIEW •

Force is an external agent or push or pull that changes or tends to change the state of rest or the state of uniform motion of a body.

•

Newton has given three laws to describe the motion of bodies. These laws are known as Newton's laws of motion.

•

Newton's first law of motion: Every body continues to remain in its state of rest or of uniform motion unless it is compelled to change its state of rest or of uniform motion by some external force. It is also called the law of inertia.

•

The natural tendency of an objects to resist a change in their state of rest or of uniform motion is called inertia.

•

The product of the mass of a body and its velocity is called linear momentum or momentum.   Momentum = mass × velocity = p = mv

• •

Newton's second law of motion: The rate of change of momentum is directly proportional to the force applied and acts along the direction of the applied force.

•

Total momentum of system before collision = Total momentum of the system after collision. m1u1 + m2u2 = m1v1 + m2v2

• • • •

Newton's third law of motion: For every action, there is an equal and opposite reaction. Action and reaction forces are equal in magnitude and opposite in direction.  change in momentum  m(v − u) Rate of change of momentum =  =  = ma time t   The product of magnitude of force applied on a body within a short interval of time is called impulse.

Tension force is a mutual force between two objects connected by a string or rope, and the magnitudes of these forces are equal.

WORKSHEET - 1 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER I.

Types of forces and Newton's laws of motion

1. When a body is stationary, a. There is no force acting on it b. The forces acting on it are not in contact with it c. The combination of forces acting on it balance each other d. The body is in a vacuum

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IL Foundation Series Class 9

2. Inertia is the property by virtue of which the body is a. Unable to change by itself the state of rest only b. Unable to change by itself the state of uniform motion only c. Unable to change by itself the direction of motion only d. Unable to change by itself the state of rest or uniform motion 3. Inertia of a body has a direct dependency on a. Velocity

b. Mass

c. Area

d. Volume

4. To keep a particle (body) moving with constant velocity on a frictionless horizontal surface, an external force a. Should act continuously

b. Should be a variable force

c. Should not act

d. Should act opposite to the direction of motion

5. Newton’s first law of motion describes the following a. Energy

b. Work

c. Inertia

d. Moment of inertia

6. A person in a car tends to fall back when it suddenly starts. It is due to a. Inertia of rest

b. Inertia of motion

c. Inertia of direction

d. None

7. A rider on a horse falls forward when the horse suddenly stops. This is due to a. Inertia of horse

b. Inertia of rider

c. Large weight of the horse

d. Losing of the balance

8. When a bus takes a turn suddenly, the passengers are thrown outward because of a. Inertia of motion

b. Inertia of rest

c. Inertia of direction

d. Speed of motion

9. Which of the following has more inertia: a rubber ball or a stone of the same size? a. Rubber ball

b. Stone

c. Both

d. Neither ball nor stone

10. Which among the following statements is false? a. The mass of two bodies will be the same, if they exhibit the same inertia b. The mass of a body at the centre of the Earth is not equal to zero c. If a ball is thrown vertically upward by a man sitting in a uniformly moving car. Then the ball will fall behind the man. d. The mass of a body remains constant at all places.

57

FORCE AND LAWS OF MOTION

11. When a wet dog shakes itself, people standing nearby tend to get wet. The water flies outward from the dog. This is due to the a. Inertia of rest

b. Inertia of motion

c. Inertia of direction

d. Force

12. An athlete runs some distance before taking a long jump because a. He gains the energy to take him through long-distance b. It helps to apply a large force c. By the running action, the reaction force increases d. By running, he gives himself a large inertia of motion 13. Bus passengers bend forward when the bus is suddenly stopped due to the a. Inertia of bus

b. Inertia of passenger

c. External force

d. Internal force

14. The water drops striking a cycle tyre are found to leave it tangentially because of the a. Inertia of rest

b. Inertia of motion

c. Inertia of direction

d. None

15. Statement A: The mass of a body is a measure of the quantity of the matter in it.

Statement B: Both mass and inertia have the same SI units. a. Both A and B are true

b. Both A and B are false

c. A is true, B is false

d. A is false, B is true

16. Assertion (A): A tablecloth can be pulled from a table without dislodging the dishes.

Reason (R): The inability of the body to change the state of rest by itself is called the inertia of rest. 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 and R is false. d. A is false and R is true.

17. Assertion (A): Mud-guards are used over the rotating wheel of the vehicles.

Reason (R): Due to the inertia of direction, the mud sticking to the wheel flies off tangentially. This can spoil the vehicle and also the people just behind it. To avoid this, mudguards are used. 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 and R is false. d. A is false and R is true.

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IL Foundation Series Class 9

II.

Second law of motion

1. Newton’s 2nd law explains the a. Definition of force

b. Measure of force

c. Nature of force

d. Types of force

2. When a constant force is applied to a body, it moves with a uniform a. Acceleration

b. Velocity

c. Speed

d. Momentum

3. An object will continue to have an acceleration until a. The resultant force on it begins to decrease b. Its velocity changes direction c. The resultant force on it is zero d. The resultant force is at the right angle to its direction of motion 4. An unbalanced force acts on a body. Then, the body a. Must remain at rest

b. Must move with uniform velocity

c. Must be accelerated

d. Must move along a circle

5. A constant force of 12 N acts on a body for 4s. The change in the linear momentum of the body is a. 48 kgms-1

b. 0.48 kgms-1

c. 4.8 kgms-1

d. 480 kgms-1

6. A force produces an acceleration of 5 cms-2 when it acts on a body of mass 20 g. The force is a. 103 N

b. -103 N

c. 10-3 N

d. 104 N

7. The force that is acting on 1 Kg of mass developing an acceleration 1m/s-2 in its own direction is known as a. 1 kgwt

b. 1 dyne

c. 1 pound

d. 1 newton

8. If a body of mass 1 kg is acted upon by a constant force of 3 N continuously, then the acceleration of the body is 1 m /s 2 b. 3 m/s2 3 9. 1 newton = _________ dynes

a.

a. 105

b. 10-5

c.

2 m/s 2 3

c. 104

d.

3 m/s 2 2

d. 10-4

10. If a body of mass 2 kg is moving with a velocity of 3 ms−1 for 4s, then the force acting on the body is a.1.5 N

b. 2 N

c. 3 N

d. Zero

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FORCE AND LAWS OF MOTION

11. A force acts on a particle of mass 200g. The velocity of the particle changes from 15 ms-1 to 25 ms-1 in 2.5 s. Assuming the force to be constant, the magnitude of the force is a. 8 N

b. 0.8 N

c. 0.08 N

d. 80 N

12. A force of 0.6 N acting on a particle increases its velocity from 5 ms-1 to 6 ms-1 in 2 s. The mass of the particle is a. 12 kg

b. 0.12 kg

c. 1.2 kg

d. 120 kg

13. A body of mass 5 kg at rest is acted upon by a force. Its velocity changes to 10 m/s . Find its initial and final momentum. a. 40 kgm/s, 0

b. 0, 25 kgm/s

c. 0, 50 kgm/s

d. 20 kgm/s, 0

14. A force F1 acting on a 2 kg body produces an acceleration of 2.5 ms-2 . Another force F2 acting on a 5 kg body produces an acceleration of 2 ms-2. The forces F1 and F2 are a. 5 N, 10 N

b. 10 N, 5 N

c. 1 N, 2 N

d. 2 N, 1 N

15. An object of mass 2 kg is sliding with a constant velocity of 4 m/s on a smooth horizontal table. The force required to keep this object moving with the same velocity is a. 32 N

b. 0 N

c. 2 N

d. 8 N

16. If a constant force acts on a body initially kept at rest, the distance moved by the body in time 't' is proportional to a. t

b. t2

c. t3

d. t4

17. A force acting on a particle of mass 200 g displaces it through 400 cm in 2 s. If the initial velocity of the particle is zero, then the magnitude of the force is a. 0.4 N

b. 4 N

c. 40 N

d. 400 N

18. A force of 50 N acts on a mass of 10 kg at rest. What is its acceleration, and what is its velocity after 2s if the same force is acting? a. 2 m / s 2 , 10 m / s

b. 4 m / s 2 , 5 m / s

c. 5 m/s2, 10 m/s

d. 6 m / s 2 , 1 m / s

III. Third law of motion

1. A body of mass 3 kg is lying on the ground, then the normal reaction exerted by the ground on the body is 3kg

a. 15 N

60

b. 25 N

c. 30 N

d. 40 N

IL Foundation Series Class 9

2. A light string is subjected to equal forces at both ends, as shown. What is the tension developed in the string? a. F

F

b. 2 F

F

F 2 d. zero c.

3. A block of mass 3 kg shown in the figure is being pulled by the light horizontal string at an acceleration of 0.5 m/s 2. What is the tension developed in the string? a. 8 N

3kg

b. 6 N c. 5 N d. 1.5 N

4. A block of mass 25 kg is suspended to the ceiling by means of a light string, as shown in the figure. What is the force with which the string pulls the ceiling? a. 125 N b. 150 N c. 250 N

25kg

d. 500 N

5. Two bodies of 5 kg and 4 kg are tied to a string, as shown in the fig. If both the table and the pulley are smooth, the acceleration of the 5 kg body is a. g

4 kg

b. g/4

T T

4 g 5 kg 9 5 d. g 9 6. Two blocks of masses of 1 kg and 2 kg rest on a smooth horizontal table. When the 2 kg block is pulled by a certain force F the tension T in the string is c.

a. F/2 N

b. F/3 N

c. F/4 N

d. F/5 N

1 kg

T

2 kg

F

7. A machine gun of mass 10 kg fires 20 g bullets at the rate of 180 bullets per minute with a speed of 100 ms -1 each. The recoil velocity of the gun is _________________. a. 0.6 m/s

b. 1.2 m/s

c. 0.8 m/s

d. 1.4 m/s

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FORCE AND LAWS OF MOTION

8. Two objects of masses 1 kg and 2 kg are moving with velocities 2m/s and 4m/s, respectively.

They collide, and after the collision, the first object moves at a velocity of 3m/s. What will be the velocity of the second object? a. 2.8 m/s

b. 3.5 m/s

c. 4 m/s

d. 2.4 m/s

9. A 2 kg shot is fired from a canon of mass 198 kg with a velocity of 50 ms-1 with respect to the canon. The KE of recoil of the canon is _____________. a. 28.45 J

b. 36.2 J

c. 25.8 J

d. 24.75 J

WORKSHEET - 2 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Calculate the mass of the body when a force of 525 N produces an acceleration of 3.5 m/s2. a. 160 kg

b. 150 kg

c. 155 kg

d. 175 kg

2. A driver accelerates his car, first at the rate of 1.8 m/s2 and then at the rate of 1.2 m/s2.

Calculate the ratio of the force exerted by the engine in the two cases. a. 4:3

b. 2:3

c. 3:2

d. 3:42

3. A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision, and both of them come to a halt after that. If the collision lasts for one second, which vehicle experiences the greater force? a. The truck b. The car c. Both will experience the same force d. The car will experience a greater force 4. A bullet of mass 0.01 kg is fired with a velocity of 100 m/s from a rifle of mass of 20 kg. The recoil of the rifle is a. 0.05 m/s

b. 20 m/s

c. 10 m/s

d. 1 m/s

5. A force of 15 N acts separately on two bodies of masses 3 kg and 5 kg. The ratio of the acceleration produced in the two cases will be a. 5:3

b. 3:5

c. 8:15

d. 15:8

6. When the force of one newton acts on a mass of 1 kg that is able to move freely, the object moves with a a. Speed of 1 ms-1

b. Speed of 1 kms-1

c. Acceleration of 10 ms-2

d. Acceleration of 1 ms-2

7. A and B are two objects with masses 6 kg and 34 kg, respectively

62

a. A has more inertia than B

b. B has more inertia than A

c. A and B have the same inertia

d. None of the above is true

IL Foundation Series Class 9

8. A rocket or a jet engine works on the principle of a. Conservation of energy

b. Conservation of momentum

c. Conservation of mass

d. Newton's second law of motion

9. Convert 1 kg-wt into dynes a. 9.81×104 dyne

b. 98.1×105 dyne

c. 9.81×105 dyne

d. 98.1×104 dyne

10. If two bodies collide in the absence of any external force, what will be the total change in the momentum of the system? a. One

b. Zero

c. > 1

d. Both (a) and (b)

11. Find the magnitude of the net force on a 20 kg mass if it accelerates uniformly from rest to 5.8 m/s in 3 s. a. 44.4 N

b. 38.6 N

c. 22.2 N

d. 55.5 N

12. The recoil velocity of the gun is a. Equal to the velocity of a bullet

c. Much smaller than the velocity of a bullet

b. Much greater than the velocity of a bullet

d. None

13. Action and reaction a. Always act on two different bodies

b. Are equal in magnitude

c. Always act in opposite directions

d. All the above are true

14. Statement (A): For every action, there exists an equal and opposite reaction.

Statement (B): Action and reaction do not occur on the same body. a. A is true, B is false

b. A is false, B is true

c. Both A and B are true

d. Both A and B are false

15. Statement (A): An action and reaction act at the same instant of time.

Statement (B): Force always exists in an action-reaction pair. a. A is true B is false

b. A is false, B is true

c. Both A and B are true

d. Both A and B are false

16. Assertion (A): If action and reaction act on different bodies, they do not cancel each other.

Reason (R): Action and reactions are forces that are equal in magnitude but opposite in direction. 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 63

FORCE AND LAWS OF MOTION

17. Swimming is possible because of the a. First law of motion

b. Second law of motion

c. Third law of motion

d. Newton's law of gravitation

18. When we jump out of a boat standing in water, it a. Moves backward

b. Moves forward

c. Moves sideways

d. Sinks

19. A weighing machine records 35 kg when a boy stands on it. The reaction of the machine on the boy is a. 0 kg-wt

b. 35 kg-wt

c. 70 kg-wt

d. Data is insufficient

c. Inertia

d. Reaction

20. When we kick a stone, we get hurt. It is due to a. Velocity

b. Momentum

21. A man is at rest beside a space station (there is no rope between the station and the man). He can get himself to the space station by making use of Newton's a. First law

b. Second law

c. Third law

d. All the laws

22. A rocket can go vertically upwards in the Earth's atmosphere because a. It is lighter than air b. Of the gravitational pull of the sun c. It has a fan that displaces more air per unit time than the weight of the rocket d. Of the force exerted on the rocket by gases ejected by it 23. Assertion (A): When a stationary bomb explodes into two pieces, their speeds are in the inverse ratio of mass.

Reason (R): Explosion does not violate the law of conservation of linear momentum. a. Both A and R are correct, and R is the correct explanation of A. b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, and R is incorrect. d. A is incorrect, and R is correct.

24. Assertion (A): When bullets are fired from a gun, the gunner should exert force on the gun in the direction of motion.

Reason (R): The gun recoils when the bullet is fired. a. Both A and R are correct, and R is the correct explanation of A. b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, and R is incorrect. d. A is incorrect, and R is correct.

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IL Foundation Series Class 9

25. Assertion (A): A rocket works on the principle of conservation of linear momentum.

Reason (R): Whenever there is a change in the momentum of one body, the same change occurs in the momentum of the second body of the same system but in the opposite direction. a. Both A and R are correct, and R is the correct explanation of A b. Both A and R are correct, but R is not the correct explanation of A c. A is correct, and R is incorrect. d. A is incorrect, and R is correct.

26. Assertion (A): No force is required by the body to remain in a state of uniform motion.

Reason (R): In uniform linear motion, acceleration has a finite value a. Both A and R are correct, and R is the correct explanation of A b. Both A and R are correct, but R is not the correct explanation of A c. A is correct, R is incorrect d. A is incorrect, R is correct 27. Assertion (A): In the case of a free fall of the lift, the man will feel weightlessness.

Reason (R): In free falling, acceleration of lift is equal to acceleration due to gravity. a) Both A and R are correct, and R is the correct explanation of A. b) Both A and R are correct, but R is not the correct explanation of A. c) A is correct, and R is incorrect. d) A is incorrect, and R is correct.

28. When force is applied to a body, and the force is exerted in the direction of the velocity, in which of the following directions its momentum should be increased? a. Direction of velocity

b. Perpendicular to velocity

c. Opposite to velocity

d. Momentum remains the same

29. Which of the following has the largest inertia? a. A pin

b. An inkpot

c. Your physics book d. Your body

30. A body of weight w1 is suspended from the ceiling of a room through a chain of weight w2. The ceiling pulls the chain by a force w1 + w 2 2 31. When a horse pulls a cart, the force that helps the horse to move forward is the force exerted by a. wl

b. w2

c. w1 + w2 d.

a. The cart on the horse

c. The ground on the cart

b. The ground on the horse

d. The horse on the ground 65

FORCE AND LAWS OF MOTION

32. A car accelerates on a horizontal road due to the force exerted by a. The engine of the car

b. The driver of the car

c. The earth

d. The road

33. A person standing on the floor of an elevator drops a coin. The coin reaches the floor of the elevator in time t1 if the elevator is stationary and in time t2 if it is moving uniformly. Then (depending on whether the lift is going up or down.) a. t1 = t2

b. t1 < t2

c. t1 > t2

d. tf < t2 or t1 > t2

34. The kinetic energy of a freely falling body a. Is directly proportional to the height of its fall. b. Is directly proportional to the square of the time of its fall c. Both a and b d. None of these 35. A long piece of rubber is wider than it is thick. When it is stretched in length by some (finite) amount: a. Its thickness decreases, but its width increases b. Its thickness decreases, but its width remains constant c. Its thickness increases, but its width decreases d. Both its thickness and width decrease 36. The length of a metal wire is l1 when the tension in it is T1 and is l2 when the tension is T2. The natural length of the wire is l1 + l 2

b. l1l 2 2 l T −l T l T −l T c. 1 2 2 2 d. 1 2 2 1 T2 − T1 T2 − T1 37. If a rocket of mass 1000 kg exhausts gases at a rate of 4 kg/sec with a velocity of 3000 m/s, the thrust developed on the rocket is a.

a. 120 N

b. 800 N

c. 12000 N

d. 200 N

38. A man is at rest in the middle of a pond on perfectly smooth ice. He can get himself to the shore by making use of Newton's a. First law

b. Second law

c. Third law

d. All of these laws

39. Which of the following facts is/are direct evidence(s) supporting Newton's first law of motion? a. A feather and a coin take equal amounts of time to reach the ground when dropped from the same height on the Moon's surface. b. A satellite orbits around the Earth with uniform speed without a supply of fuel.

66

IL Foundation Series Class 9

c. A man is thrown forward on a bus which stops suddenly. d. A man walks on a horizontal road.

40. Two bodies of mass M and m, where (M > m), are released freely in the air, then the acceleration of the a. Body 'M' is more

b. Body 'm' is more

c. Bodies is equal

d. Bodies depends on height

41. The net force acting on a body of mass 1 Kg moving with a uniform velocity of 5 m/s is _____. a. 5 N

b. 0.2 N

c. 0 N

d. 1 N

42 A stone, when thrown on a glass window, smashes the window pane to pieces. But a bullet fired from a gun passes through, making a hole. Why? a. Inertia of motion

b. Inertia of direction

c. Inertia of rest

d. Less force applied by the bullet

43 By applying a force of 1 N, one can hold a body whose mass is approximately equal to a. 100 mg

b. 100 g

c. 1 kg

d. 10 kg

44. A coin flicked across a table stops because a. No force acts on it

b. It is very heavy

c. The table exerts a frictional force on it

d. The Earth attracts it

45. The principle of conservation of linear momentum states that the linear momentum of a system: a. Never stays the same. b. Changes only when an internal force acts. c. Remains constant if no external force acts. d. Always stays the same, no matter what.

67

GRAVITATION

4

4.1 INTRODUCTION 4.1.1 What is gravitation?

The force of attraction between any two masses by virtue of their mass is called gravitational force.

•

It is one of the four fundamental forces of nature.

•

It is the weakest force with the longest range.

Earth Moon Gravitational force of the moon

High tide

High tide

Fig. 4.1 Gravitational force (A)

(B)

4.1.2 Gravitational force providing centripetal acceleration We know that the moon moves around the Earth. When we throw something up, it goes up and then comes back down. One day, Newton was sitting under a tree when an apple fell on him. This made him think: if the Earth can pull the apple down, can it also pull the moon? Is the force the same in both cases? Newton guessed that the same force is responsible for both the apple falling and the moon moving around the Earth. He thought that at each point in its orbit, the moon falls toward the Earth instead of going straight. So, there must be a force pulling it. But we don't really see the moon falling towards the Earth.

To understand this, let's do an activity:

68

1.

Take a piece of thread.

2.

Tie a small stone to one end and whirl it around.

3.

Notice the motion of the stone.

4.

Release the thread.

5.

Note the direction of the stone's motion before and after releasing it.

IL Foundation Series Class 9

Fig. 4.2 Centripetal force

Observation: Before releasing the thread, the stone moves in a circle with a certain speed, changing direction at every point. The force causing this change in direction is called the centripetal force, pulling towards the centre. Without this force, the stone would fly off in a straight line tangent to the circular path.

The moon's motion around the Earth is also due to a centripetal force provided by Earth's attraction. Without this force, the moon would move in a straight line.

Interestingly, when an apple falls, it is attracted to Earth. But does the apple attract the Earth?

According to Newton's third law of motion, yes. However, according to the second law, for a given force, acceleration is inversely proportional to the mass of an object. The Earth's mass is much larger than the apple's, so we don't see the Earth moving toward the apple. The same argument applies to why Earth doesn't move towards the moon.

In our solar system, all planets orbit the Sun. Applying the same reasoning, there must be a force between the Sun and the planets. Newton concluded that not only does Earth attract an apple and the moon, but all objects in the universe attract each other. This force is called gravitational force.

4.1.3 Kepler's laws 1st Law (Law of orbits): Every planet revolves around the sun in an elliptical orbit, with the Sun at one of the foci.

Planet

Sun

Fig. 4.3 Law of orbits

•

The nearest point of the path of the planet to the sun is called perihelion.

•

The farthest point is called aphelion.

2nd Law (Law of areas): The areal velocity of the radius vector of a planet originating from the Sun is constant. (or) The radius vector of the planet from the Sun sweeps out equal areas in equal intervals of time.

69

GRAVITATION

Δt

A2 A1 A1= A2

Fig. 4.4 Law of areas

The validity of Kepler’s second law means a planet must have a higher than average velocity near perihelion (when it is closest to the Sun) and a lower than average velocity near aphelion (when it is farthest away from the Sun).

According to Kepler's second law, the angular momentum of a planet remains constant.

3rd Law (Law of Periods): The square of the time period of revolution of a planet around the sun is proportional to the cube of the mean distance of the planet from the sun.

If T is the time period of a planet and R is the mean radius of the path of a planet, according to Kepler's third law, T 2 4 π2 T ∝ R ⇒ 3 = (constant) R GM 2

3

4 π2 is a proportionality constant.

GM 4.1.4 Universal law of gravitation Newton's law of universal gravitation can be stated as follows: Every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them, and this force acts along the line joining the two particles.

Consider two particles A and B of masses m1 and m2. Let r be the distance between their centres and F be the force of attraction between them. m1 A

F12

F21

m2 B

r

Fig. 4.5 Universal law of gravitation

According to the universal law of gravitation, F ∝ m1. m 2 , F ∝

70

1 r2

IL Foundation Series Class 9

∴F ∝

m1 m 2 r2

m1 m 2 r2 where G is the constant of proportionality and is called the universal gravitational constant.

G Or ∴ F =

4.1.5 The gravitational constant We have, F = G

m1 m 2 r2

... (1)

Fr 2 m1 m 2 SI unit: Nm2 / kg2 , Or G =

CGS unit: dynes cm2 / g 2 The value of G: 6.67 × 10−11 N.m2 / kg2

 MLT−2 ⋅ L2  = M −1 L3 T−2  Dimensional formula of G: [G] =    MM  Definition of G: Let m1 = m2 =1and r = 1

G × 1× 1 = G 12 "Thus universal gravitational constant is equal to the force of attraction acting between two bodies each of unit mass, whose centers are placed unit distance apart."

From Eq= (1), F

It is a scalar quantity. Its value is the same throughout the universe and is independent of nature and the bodies as well as the nature of the medium between the bodies.

Note: Newton's law of gravitation holds good for objects lying at very large distances and also very short distances. It falls when the distance between the objects is less than 10−9 m. (i.e., of the order of intermolecular distances) SOLVED EXAMPLES Example 1: 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. (G = 6.67 × 10-11 N.m2/kg2) Solution: Mass of sun, m1= 2 ×1030 kg Mass of earth, m 2= 6 ×1024 kg Distance between them, = r 1.5 ×1011 m then F =

Gm1m 2 r2 71

GRAVITATION

= = = F

6.67 × 10-11 × 2 × 1030 × 6 × 1024

(1.5 × 10 )

11 2

6.67 × 2 × 6 × 1043 1.5 × 1.5 × 1022 3.56 × 1022 N

Example 2: The mass of the earth is 6 × 1024 kg and that of the 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 the Earth on the moon. (Take = G 6.7 ×10−11 Nm 2 kg −2 ) Solution: The mass of the earth, M= 6 ×1024 kg. The mass of the moon, m = 7.4 ×1022 kg. The distance between the earth and the moon,

= d 3.84 ×105 km = 3.84 ×105 ×1000 m = 3.84 ×108 m = G 6.7 ×10−11 N m 2 kg −2 The force exerted by the earth on the moon is = F G

M × m 6.7 ×10−11 Nm 2 kg −2 × 6 ×1024 kg × 7.4 ×1022 kg = = 2.02 ×1020 N 2 2 8 d ( 3.84 ×10 m )

Thus, the force exerted by the earth on the moon is 2.02 × 1020 N. Example 3: Two particles of masses 1.0 kg and 2.0 kg are placed at a seperation of 50 cm. Assuming that only gravitational forces act on the particles mutually, the initial acceleration of the first particle.

Solution: Gravitational force between the two particles

F=

Gm1 m 2 6.67 ×10−11 ×1× 2 ⇒ = = 5.3 ×10−10 N F 2 2 r (0.5)

The acceleration of 1.0 kg particle is

a= 1

F 5.3 ×10−10 = = 5.3 ×10−10 m / s 2 m1 1 It is towards the 2.0 kg particle.

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4.1.6 Inverse square law In establishing the inverse square law for gravitational forces, Newton based his calculations on certain assumptions related to planetary motion. He considered planetary orbits to be circular and accounted for the gravitational force between the sun and a planet as the centripetal force responsible for the planet's motion in its orbit. The magnitude of the centripetal force (F) acting on a planet in its orbit is expressed as F = mv2/r, where 'm' is the mass of the planet, 'v' is the orbital velocity, and 'r' is the radius of the orbit. The orbital velocity (v) of the planet is given by v = 2πr / T , where ' T ' is the period of revolution, and 2πr is the circumference of the orbit.

Substituting this into the centripetal force equation, we get: = F

m(2πr / T)2 m × 4π2 r 2 4π2 mr3 = = r T2r T2r2

Now, according to Kepler's third law of planetary motion, r3/T2 is constant. Assuming the mass of the planet is constant, we obtain:

F =( 4π2 m ) × ⇒F∝

1 r3 / T 2 =Constant × 2 , 2 r r

1 which is the formulation of the inverse square law. r2

Example: How does the force of gravitation between two objects change when the distance between them is reduced to half?

Solution: 1 [inverse square law ] r2 F2 r12 r12 = = = 4 F1 r22  r1 2   2

As F ∝

Force becomes four times when the distance between the objects is reduced to half.

4.1.7 Importance of the universal law of gravitation The law of gravity helps us to understand (i) why we stay on Earth, (ii) how the moon moves around the Earth, (iii) how planets move around the Sun, and (iv) why we have tides caused by the moon and the Sun.

73

GRAVITATION

4.2

ACCELERATION DUE TO GRAVITY

4.2.1 Free fall Whenever objects fall towards the earth under gravitational force alone, we say that the objects are in free fall.

When a stone is dropped from a cliff, in addition to gravity, it experiences air resistance.

Consequently, not all objects fall freely. A body is considered to be in free fall when only gravity influences it, without any other forces acting on it.

During free fall, objects experience the same acceleration due to gravity, irrespective of their masses. For instance, if a small iron ball and a feather of the same mass are dropped, they both have the same acceleration, 'g,' but the feather takes longer to reach the ground due to increased air resistance.

Now, consider dropping an iron ball and a wooden ball of the same size simultaneously. Because they experience similar air resistance, the effect of air resistance can be neglected, and both balls can be treated as if they are in free fall, reaching the ground at the same time.

Galileo demonstrated this by dropping an iron ball and a wooden ball from the leaning tower of Pisa, disproving Aristotle's theory that heavier bodies fall faster than lighter ones. This experiment highlighted that mass does not influence the acceleration due to gravity during free fall.

Sir Isaac Newton further confirmed this idea by experimenting with an evacuated tube, where a coin and a feather fell simultaneously for a distance of 6 meters. In the absence of air resistance, both objects reached the bottom of the tube at the same time, providing additional evidence that mass does not affect the acceleration due to gravity during free fall.

4.2.2 Acceleration due to gravity (g) If the force on the body is due to gravity of the earth, then acceleration produced in that body is called acceleration due to gravity.

•

SI unit of acceleration due to gravity (g) is ms −2

•

The dimensional formula of acceleration due to gravity is M 0 L1 T−2    Acceleration due to gravity independent of shape, size, and mass of the body

•

Relation between g and G

Consider a body of mass 'm' located on the surface of the earth. Let the mass and radius of the Earth be given by M and R, respectively.

According to Newton's law of universal gravitation, F = G Since this gravitational force is the weight of the body

74

Mm .

R2

IL Foundation Series Class 9

F = mg GMm ⇒ mg = R2 GM g= ⇒ R2 where M is the mass of the Earth, and R is the distance between the object and the Earth.

To calculate the value of g

In order to calculate the value of g, we should put the values of G, M, and R in g = GM .

2 R We have, Mass of the Earth, M= 6 ×1024 kg Radius of the Earth, R = 6400 km = 64 ×105 m Universal gravitational constant,= G 6.67 ×10−11 N m 2 kg −2 Acceleration due to gravity on the surface of the Earth, g =

= g

GM R2

GM 6.67 ×10−11 × 6 × 2024 = = 9.8 m s −2 5 r2 64 × 10 ( )

Note: The Earth is not a perfect sphere. As the radius of the Earth increases from the poles to the equator, the value of g becomes greater at the poles than at the equator, unlike the universal gravitational constant. It fluctuates depending on the location, leading to variations in the weight of objects. The standard value for acceleration due to gravity at the Earth's equator is 9.82 m/s2. Various factors influence this acceleration, and some of them are explained below.

4.2.3 Variation in acceleration due to gravity (g) Variation of 'g' with altitude

The acceleration due to gravity at a height 'h' from the surface

is g h =

GM .

(R + h)2

 2h  If h  R, then g h  g 1 −  by applying binomial expansion.

R  At smaller altitudes, ∆g 2h ∆w 2h = = and (where 'w' is the weight on the surface of the Earth). g R w R Variation of 'g' with depth

The acceleration due to gravity at a depth 'd' from the surface of the Earth gd is

 d ⇒ gd = g 1 −   R

75

GRAVITATION

∆g ∆w d = = g w R If gh = gd when h << R and d << R, then 2h = d The variation between r and g is as shown (where 'r' is the distance from the centre of the Earth).

gmax g

g∝r

r=o

r

g ∝ 12 r

r=R

Fig. 4.6 Variation in g with altitude and depth Variation of 'g' according to the shape of the earth

1 As g ∝ 2 R a) g is minimum at the equator b) g is maximum at the poles Variation of g according to local conditions: The value of g is slightly more at the location of mineral deposits.

SOLVED EXAMPLES Example 1: Suppose a planet exists whose mass and radius both are half of those of the Earth.

Calculate the acceleration due to gravity on the surface of this planet.

Solution: Take the value of 'g' on the surface of the Earth as 9.8 m/s2.

M g = G 2M g R G 2 R = g M 9.8 m / s2 2 gP ? g M 9.824m / s g P ? 24 6 ×10 kg 3 ×10 kg MM = MP = 24 M M 6 106 kg M P 3 10624 kg 6.4 ×10 m RP = 3.2 ×10 m6 RH = RH 6.4 106 m RP 3.2 2 10 m

g P GM P RM2 3 ×1024 × 6.4 × 6.4 ×1012 = 2 × = gM RP GM m 3.2 × 3.2 ×1012 × 6 ×1024 = g p 2= = 19.6 m / s 2 ( gM ) 2(9.8)

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IL Foundation Series Class 9

Example 2: At what height above the Earth's surface would the value of acceleration due to gravity be half of what is on the surface? Calculate.

Solution: GM R2 As height increases above the surface of the earth, the value of g decreases. Its formula is We know the value of g on the surface of the earth is g =

gh =

GM ( R + h )2

rom the statement 'g' value at a height 'h' metre above the surface earth = half of the 'g' value on F the earth. g GM GM gh = ⇒ = 2 2 (R + h) R2 × 2 2R 2 = ( R + h )2 ⇒ R + h =

2R

h= ( 2 − 1) × R h= ( 2 − 1) × R.

Example 3: Calculate the height at which the value of acceleration due to gravity becomes 50% of that at the surface of the Earth. (Radius of the Earth = 6400 km ) Solution: g at height, g h =

= gh Here,

g h  1 +   R

2

50 = g 0.5 g 100 2

h g  ⇒ 1 + = = 2  R  0.5 g h  1 +   R h h 1 + = 2; = 2 − 1 ⇒ the height, R R h = 0.414R = 0.414 × 6400 = 2650 km.

0.5 = g

g

2

Example 4: Find the depth at which the value of g becomes 25% of that at the surface of the Earth.

(Radius of the earth =6400 km ) Solution:

 d g at a depth, = gd g 1 −   R 25 In this P, = gd = g 0.25 g 100

77

GRAVITATION

substituting = gd 0.25 = g and R 6400 km we get d = 4800 km.

4.3

M OTION OF OBJECTS UNDER THE INFLUENCE OF GRAVITATIONAL FORCE OF THE EARTH

4.3.1 Dropping of an object •

If a person at rest drops an object, its initial velocity will be zero, and then it starts accelerating downwards with an acceleration a = g downwards.

•

If a person, moving with a velocity 'v' in the vertically upward direction drops an object, then its initial velocity, u = v and acceleration, a = –g (downwards).

•

If a person, moving with a velocity ' v ' in a downward direction drops an object, then its initial velocity u = +v, and it starts accelerating downwards with an acceleration a = +g (downward).

4.3.2 Free falling body Consider an object dropped from the top of a building of height ' h '.

h

Fig. 4.7 Object dropped from a height 'h'

•

Its initial velocity, u = 0

•

Its acceleration, a = g downwards

•

Its displacement on reaching the ground, S = h downwards

If we assume vertically downward direction as '+ve' direction, then u = 0, a = +g, S = +h Time taken by the object to reach the ground is given by 1 = S ut + at 2 .

2

1 = S 0(t ) + gt 2 2 1 2 h = gt 2 2h t= g 78

IL Foundation Series Class 9

Final velocity (v) of the object on reaching the ground is given by v 2 − u2 = 2a S. v 2 − 02 = 2gh ⇒ v 2 = 2gh;

v = 2gh 4.3.3 Projection Consider an object projected vertically upwards with an initial velocity ' u ' from the ground.

•

Its velocity decreases continuously while it is moving vertically upwards, and finally, it will become zero.

•

Its velocity at the maximum height is zero.

•

fter reaching maximum height, it starts falling. Its velocity increases in the downward A direction, and finally, it reaches the ground with the velocity u, which is the magnitude of the initial velocity of the projection.

•

Throughout the upward motion, at the maximum height and downward motion, its acceleration is a (= g) downwards.

If we assume vertically upward direction as the positive direction, the initial velocity, u = +u Final velocity during upward motion, v = 0, acceleration a = -g The time taken by the object to reach the maximum height, which is called the time of ascent ( ta ) , is given by, v = u + at ⇒ 0 = u − gta ⇒ ta =

u g

2aS The maximum height (H) reached by the body is given by v 2 − u2 =

u2 2g During the downward journey (descent) Thus, H =

Initial velocity u = 0 Acceleration a = -g Final velocity v = -u The time taken by the object to reach the ground from the maximum height, which is called the time of descent (td), is given by v = u + at Thus, td = u/g Time of ascent is equal to time of descent ta = td

79

GRAVITATION

The total time taken by the vertically projected object to reach the maximum height and then fall freely to touch the initial position is called time of flight. Time of flight is the sum of time of ascent and time of descent t = ta + td Thus, t = 2u g SOLVED EXAMPLES Example 1: A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate: i.

The maximum height to which it rises.

ii. The total time it takes to return to the surface of the earth.

Solution: Initial velocity, u=49 m/s; Final velocity, v = 0 m/s Acceleration of the ball, a = g = −9.8 m / s 2 (for upward motion) Maximum height reached by the ball, s= h= ? v − u2 = 2as changes to v 2 − u2 = −2gh 2

⇒= h

v 2 − u2 0 − (49 × 49) = ⇒ = h 122.5 m −2g −2 × 9.8

v − u 0 − 49 = ⇒= t 5s −g −9.8 Total time for journey, T = 2t = 10 s.

Time of ascent= t

Example 2: A stone is released from the top of a tower of 19.6 m in height. Calculate its final velocity just before touching the ground.

Solution: Initial velocity, u = 0 m/s Acceleration of the stone, a= g= 9.8 m / s 2 Height covered, s= h= 19.6 m v 2 = u2 + 2gh = 0 + 2gh 2 v =2 × 9.8 ×19.6 =(19.6)2 ⇒v = 19.6 m / s

80

IL Foundation Series Class 9

Example 3: 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?

Solution: Initial velocity, u = 40 m/sec Acceleration of the stone, a = g = -10 m/s2 Final velocity at the highest point, v = 0 m / s

v 2 − u2 2(−g ) 0 − (40 × 40) = 2 × −10 h = 80 m h=

As the stone reaches the ground Net displacement = 80 - 80 = 0 m Total distance = 80 + 80 = 160 m.

Example 4: 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.

Solution: Let the stones meet after t sec at x m below the top of tower.

1

u=0 x

1 h

2 100 - x 2

During the downward journey for the first stone distance, 1 x = gt 2 …(1) 2 Upward journey for the second stone

81

GRAVITATION

1 100 − x= 25t − gt 2 2

...(2)

From (1) and (2) 1 1 100 − gt 2 =25t − gt 2 2 2 100 = 25t t = 4s

When the stones meet after 4 seconds, the distance covered,

1 x = gt 2 2 1 = ×10 × 42 2 x = 80 m The stones meet after 4 seconds, 80 meter below the top of the tower.

4.4

MASS AND WEIGHT

4.4.1 Mass Mass is a measure of the amount of matter present in a body. It remains the same whether the object is on the Earth, the Moon, or even in outer space. It is constant everywhere.

The C.G.S. unit for mass is the gram (g), while its S.I. unit is the kilogram (kg).

4.4.2 Weight Consider a spherical body with mass 'm' near the Earth's surface. If the Earth's mass is 'M', a gravitational force of attraction exists between the two bodies, and this force acts along the line joining their centres. Due to the Earth's much larger mass, the sphere is pulled towards its centre.

So, the weight of an object is the force with which it is attracted towards the Earth. It is denoted by 'W'.

When a body is dropped from a certain height on Earth, it falls to the ground due to the gravitational pull, or in other words, its weight is directed towards the centre of the Earth. Newton's second law of motion (F = ma) can be applied in this context, where 'a' is the acceleration, 'm' is the mass, and 'F' is the force acting on the body.

For a falling body, the force is its weight (W), and the equation becomes W = mg, where 'g' is the acceleration due to gravity.

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IL Foundation Series Class 9

The magnitude of 'g' varies from place to place, making the weight of the body non-constant in the universe. However, the mass of the body remains the same everywhere.

In the C.G.S. system, the unit of weight is the dyne, while in the S.I. system, it is the newton (N).

4.4.3 Weightlessness When a load is attached to a spring balance, and the reading is noted, the spring balance registers the weight of the load. However, when the spring balance, along with the load, is allowed to fall freely, it shows a zero reading, giving the impression that no weight is acting on it.

Similarly, when an individual jumps from a height, there is a sensation of having no weight. This phenomenon is known as weightlessness.

Weightlessness occurs when a body is in free fall, experiencing a gravitational force that accelerates it towards the centre of the Earth. During free fall, there is no reaction force acting on the body, resulting in a feeling of weightlessness. This concept is exemplified by the scenario when a lift starts descending, and for a brief moment, individuals inside the lift experience a decrease in their perceived weight.

For example, astronauts aboard spaceships orbiting the Earth are within the Earth's gravitational field, yet they, too, experience weightlessness. In this situation, the force of gravity serves as the centripetal force required to keep them in orbit. In the reference frame of the spaceship, every part of the ship is subject to the same gravitational force, creating an equivalent scenario to free fall.

Consequently, astronauts feel weightless despite being under the influence of gravity.

4.4.4 Weight of an object on the Moon The weight of an object on the Moon is the force with which the Moon attracts that object. The Moon's mass is less than the Earth's, which means it pulls things with less force.

Let’s see if we can calculate an object's weight on the Moon (Wm) using a formula when it has mass ' m': G ⋅ ( Mm ⋅ m ) … ...(1) Rm2 where Mm is the mass of the Moon and Rm is its radius.

Wm =

Now, let's compare that with the weight of the same object on Earth (We), using a similar formula: G ⋅ (M ⋅ m ) ...(2) R2 where M is the mass of the Earth and R its radius.

We =

We have, Celestial body

Mass (kg)

Radius (m)

Earth

5.98 ×1024

6.37 ×106

Moon

7.36 ×1022

1.74 ×106

Table 4.1 Mass and Radius of Earth and Moon 83

GRAVITATION

Substituting the values from the above table in equations 1 and 2, we get Wm = G

7.36 ×1022 kg × m

(1.74 ×10 m ) 6

2

Wm = 2.431×1010 G × m and We = 1.474 ×1011 G × m Dividing Wm by We, we get

Wm 2.431×1010 = We 1.474 ×1011 or Wm 1 = 0.165 ≈ We 6 Weight of the object on the Moon 1 = Weight of the object on the Earth 6 Weight of the object on the Moon = (1/ 6) × its weight on the Earth.

SOLVED EXAMPLES Example 1: The mass of an object is 10 kg. What is its weight on the Earth?

Solution: Mass, m=10 kg Acceleration due to gravity, g = 9.8 ms-2 W= m × g

W= 10 kg × 9.8 ms −2 = 98 N Thus, the weight of the object is 98 N. Example 2: A n object weighs 10 N when measured on the surface of the Earth. What would be its weight when measured on the surface of the Moon?

Solution: We know, = (1/ 6) × its weight on the Earth. weight of object on the Moon

That is,

We 10 = N 6 6 = 1.67 N.

= W m

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IL Foundation Series Class 9

Thus, the weight of the object on the surface of the moon would be1.67 N .

1 Example 3: Gravitational force on the surface of the Moon is only as strong as gravitational force 6 on the Earth. What is the weight in newtons of a 10 kg object on the Moon and on the Earth?

Solution: Fmoon Wmoon m × g moon = = Fearth Wearth m × g earth =

g moon 1 = g earth 6

g earth = 9.8 m / s 2 g moon =

9.8 = 1.63 m / s 2 6

Mass, m = 10 kg Weight on the Moon, Wmoon= m × g moon

= 10 ×1.63 = 16.3 N Weight on the Earth, Wearth= m × g earth

= 10 × 9.8 Wearth = 98 N

4.5

PRESSURE AND THRUST

4.5.1 Thrust A force acting normally on a surface is called thrust.

•

Its unit in CGS system is dyne.

•

Its unit in SI system is Newton (N).

•

The gravitational unit of thrust is kilogram-force (kgf).

•

Its Dimensional formula is M1L1T−2  .

•

It is a vector quantity.

4.5.2 Pressure The force (thrust) acting normally on the unit surface area is called pressure.

Thrust .

Area The unit of pressure in CGS system is dyne/cm2 .

Mathematically, Pressure(P) = •

85

GRAVITATION

• • • •

newton (N) (or) Pascal (Pa). m2 Its Dimensional formula M1 L−1 T−2  .

Its unit in SI system is

It is a scalar quantity.

Nm −2 is also called pascal (Pa) in the honour of physicist Pascal who discovered the law for the transmission of pressure in fluids.

1 Pascal: When a force of 1 N (thrust) acts normally on an area of 1 m2, then the pressure acting on the surface is called 1 pascal.

SOLVED EXAMPLES Example 1: Calculate the pressure produced by a force of 2000 N acting on an area of 2.0 m2.

Solution: Force = 2000 N Area = 2m2 Pressure = ?

The formula of pressure is given by:

Force Area 2000 N 1000 N / m 2 ∴ Pressure = = 2 2m

Pressure =

2

Therefore, the pressure exerted on the given area will be 1000 N / m .

Example 2: A block of wood is kept on a tabletop. The mass of the wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm.

40cm 20cm

10cm 10cm

20cm 40cm

ind the pressure exerted by the wooden block on the tabletop if it is made to lie on the tabletop F with its sides of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.

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IL Foundation Series Class 9

Solution: a) 20 cm ×10 cm.

40 cm 10 cm 20 cm

The mass of the wooden block = 5 kg The dimensions = 40 cm × 20 cm ×10 cm Here, the weight of the wooden block applies a thrust on the tabletop.

That is , Thrust= F= m × g = 5 kg × 9.8 ms −2 = 49 N Area of a side = length × breath

= 20 cm × 20 cm = 200 = cm 2 0.02 m 2

= Pressure

49 N = 2450Nm −2 2 0.02 m

b) 40 cm × 20 cm.

10 cm 20 cm 40 cm

When the block lies on its side of dimensions 40 cm × 20 cm, it exerts the same thrust.

Area length × breadth = = 40 cm × 20 cm = 800 cm 2 = 0.08 m 2 49 N = = 612.5 Nm −2 Pressure 2 0.08 m

87

GRAVITATION

The pressure exerted by the side 20 cm ×10 cm is 2450 Nm −2 and by the side 40 cm × 20 cm is 612.5 N m −2 .

4.5.3 Density Density refers to the mass per unit volume of a substance.

Mathematically, density (ρ) is calculated using the formula: density = (ρ) •

SI unit → kg / m3 .

•

CGS unit → g / cm3 .

•

Density of water is 1000 kg/m3 or 1g/cm3.

mass m = volume v

4.5.4 Relative density It is defined as the ratio of the density of the substance to the density of water at 4° C .

Density of Substance Density of water at 4C It is a unitless and dimensionless physical quantity. = Relative Density =

4.6 PRESSURE EXERTED BY LIQUIDS AND GASES 4.6.1 Pressure in a fluid When a fluid (either liquid or gas) is at rest, it exerts a force perpendicular to any surface in contact with it, such as a container wall or a body immersed in the fluid.

While the fluid as a whole is at rest, the molecules that make up the fluid are in motion; the force exerted by the fluid is due to molecules colliding with their surroundings.

If we think of an imaginary surface within the fluid, the fluid on the two sides of the surface exerts equal and opposite forces on the surface. Otherwise, the surface would accelerate, and the fluid would not remain at rest. dF

dA

dF

Fig. 4.8 Small surface of area dA

Consider a small surface of area dA centred on a point on the fluid; the normal force exerted by the fluid on each side is dF⊥ . The pressure P is defined at that point as the normal force per unit area, i.e.,

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IL Foundation Series Class 9

P=

dF⊥ dA

If the pressure is the same at all points of a finite plane surface with area A, then P=

F⊥ dA

Where F⊥ is the normal force on one side of the surface.

4.6.2 Pascal's law "The pressure in a fluid in equilibrium is the same everywhere if the effect of gravity can be neglected."

It was formulated by Blaise Pascal. This law constitutes one of the basic principles of hydrostatics.

Mathematical Expression for pressure in fluids with pressure of gravity

Consider a liquid of density ' ρ' contained in a beaker. Consider a liquid column of height 'h' and area of cross-section 'a'. Let A and B be the top and bottom surfaces of the liquid column, respectively. a

A Beaker

h

Liquid

B Fig. 4.9 Beaker containing liquid of density ρ

∴ Volume of liquid column (V) = area of cross-section × length = ah Mass of liquid column = volume × density Mass of liquid column = V × ρ Mass of liquid column = ahρ ∴ Weight of liquid column = mass × acceleration due to gravity = mg = ahρg ∴ Thrust exerted by liquid column on the base of the beaker = ahρg ∴ Pressure due to liquid column: Thrust ahρg = Area a So, P= hρg

= P

89

GRAVITATION

∴ Pressure (P) = height of the liquid column (h) × density of the liquid (ρ) × acceleration due to gravity (g) Factors on which the pressure at a point in fluids depends

a) Pressure in a fluid is directly proportional to its height (or) depth. b) Pressure in a fluid is directly proportional to its density. c) Pressure in a fluid is directly proportional to acceleration due to gravity. d) Pressure in a fluid is independent of the area of cross - section. e) If PA is pressure due to the atmosphere, then total pressure at point B is PB = PA + hρg.

Laws of liquid pressure

a) Pressure at a point inside the liquid increases with the depth from the free surface of the liquid. b) Pressure at a point inside the liquid at a given depth increases with an increase in the density of the liquid. c) Pressure is the same in all directions, about a given point within the liquid. d) Pressure is the same at all points in a horizontal plane at a given depth in stationary liquid. e) A liquid seeks its own level.

SOLVED EXAMPLES Example 1: A submarine is cruising at a depth of 1000 m below sea level. The density of seawater is 1025 kg cubic meters. What is the pressure exerted by the seawater on the submarine?

Solution: Density of the seawater = 1025 kg/m3, Depth of the seawater = 1000 m, Acceleration due to gravity = 10 m/s2.

So, the pressure exerted by the seawater on the submarine is: Pressure = Density × Depth × Acceleration due to gravity

= 1025 kg / m3 ×1000 m ×10 m = / s 2 1, 02,50, 000 Pa

Therefore, the pressure exerted by the seawater on the submarine is 1,02,50,000 Pa. Example 2: A dam is filled with water till a height of 127 meters. If the mass of water per cubic centimeter is one gram, then find the difference in pressures acting at the following two points. (a) Point exactly at a depth half that of the dam.

90

IL Foundation Series Class 9

(b) Point at the bottom of the dam.

Solution: Height/depth of dam =127 metres.

Density = (ρ)

1 gram Mass = = 1g = cm −3 1000 kg m −3 3 Volume 1 cm

depth 127 = = 63.5 m 2 2 Pressure at a point ( Pa ) = hρg = 63.5 ×103 ×10 = 635 ×103 pascals

= (a) Height of the point

(b) Depth of the point = 127 m Pressure at a point ( Pb ) = 127 ×103 ×10 = 1270 ×103 pascals

(1270 − 635) ×103 Therefore, difference in pressures = Pb − P= a = 635 ×103 Pa 4.6.3 Viscosity Imagine three tubes connected to a horizontal rod that you can rotate using a handle at one end, like in the figure given below.

Crude Oil

Mercury

Water

Fig. 4.10 Viscosity

These tubes are closed at both ends and filled halfway with crude oil, mercury, and water. Look at Fig. 4.14. You will see that they are in the lower half of the tubes. Now, turn the handle to invert all the tubes. The liquids start flowing down, but they take different times to reach the bottom.

Apart from the friction between the glass surface and the liquids, there's also friction between the different layers of liquid, which resists their movement. This friction is called viscous force. Since the liquids take varying times to reach the bottom, it means the viscous force is different for each liquid. The less viscous force there is, the more mobile the liquid is. This special property of liquids is called viscosity.

Also, note that the viscosity of a liquid decreases as its temperature increases.

91

GRAVITATION

4.6.4 Stoke’s law–viscous drag Stoke’s law describes when a body is in motion in the fluid, an opposing force develops, which opposes the motion of the body by the fluid. This force is called viscous drag. In most cases, when the speed is low, viscous drag is proportional to the speed of the body.

Note: Viscous drag in the fluid is analogous to frictional force in the case of solids. The origin of all the forces except gravitational pull is the electromagnetic force.

4.6.5 Atmospheric pressure (P­0) It is the pressure of the Earth's atmosphere. This changes with weather and elevation. Normal atmospheric pressure at sea level is 1.013 × 105 pa

1 atm = 1.013 ×105 Pa Activity

Empty a can and fill it with water. Set the can on a tripod stand and apply heat using a burner. Once the water reaches a boiling point, steam will emerge, displacing the air inside the can. Keep heating until steam escapes freely from the can's mouth. Cease heating and promptly seal the can tightly.

Detach the can from the tripod stand and cool it by pouring cold water. The observation reveals that the can collapse due to the atmospheric pressure exerted on it.

Steam pressure Air pressure Water

Fig. 4.11 Activity to show atmospheric pressure

When cold water is applied to the can, the steam within condenses back into the water, leading to a significant decrease in pressure. Simultaneously, the external atmospheric pressure, being substantial, presses upon the can, causing it to collapse.

At sea level, atmospheric pressure is commonly measured in atmospheres (atm). The standard atmospheric pressure at sea level is 1 atm.

Experimental findings indicate that at sea level, a mercury column of approximately 76 cm exerts an equivalent pressure to that of the atmospheric air.

The relationship between pressure (P), the height of the mercury column (h), and the density of mercury (ρ) is represented by the equation P = hρg.

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IL Foundation Series Class 9

Where h= height of mercury = column 76 = cm of Hg 0.76 m of Hg

d density of mercury 13.6 g cm −3 (or ) 13.6 ×103 kg m −3 = = g = 9.8 m s −2 ∴1 atm. = 0.76 ×13.6 ×103 × 9.8 Pa =1.013 ×105 Pa ≅ 105 Pa. Absolute pressure and gauge pressure

The excess pressure above atmospheric pressure is usually called gauge pressure and the total pressure is called absolute pressure. Thus, • Gauge pressure = absolute pressure - atmospheric pressure • Absolute pressure is always greater than or equal to zero.

• While gauge pressure can be negative also.

Different units of pressure

• 1 Pa = 1 N m −2 • 1 kg f m −2 = 9.8 Pa 105 dyne 104 cm 2 • 1 Pa = 10 dyne cm −2 • 1 N m −2 =

• 1 bar ≈ 105 Pa • 1 torr = 1 mm of Hg • 1 atm ≅ 1.013 bar

4.7

BUOYANCY

4.7.1 Buoyant force Consider a part ABCD of the fluid. Let the mass of the part of the fluid be (mf). When the part of the fluid ABCD is at rest: B A

B

D

C mfg

Fig. 4.12 Part of fluid ABCD and its balancing forces

The weight of ABCD(mfg) is balanced by an upward force exerted by the remaining fluid. This upward force exerted by the remaining fluid is called buoyant force (B).

93

GRAVITATION

B = mfg ⇒ B = Vρfg where V is the volume of the fluid displaced and ρf is the density of the fluid.

Note: If the part of the fluid is replaced by another object of the same volume, irrespective of the material, the buoyant force remains the same. When the fluid is at rest, the buoyant force is equal to the weight of the fluid displaced.

B

m0 is the mass of the object m0g Fig. 4.13 Buoyant force

This phenomenon is known as buoyancy.

4.7.2 Flotations - why objects float or sink when placed on the surface of water?

When a solid body is dipped into a fluid, the fluid exerts an upward force of buoyancy on the solid.

If the force of buoyancy equals the weight of the solid, the solid will remain in equilibrium. This is called flotation.

When the overall density of the solid is smaller than the density of the fluid, the solid floats with a part of it in the fluid. The fraction dipped is such that the weight of the displaced fluid equals the weight of the solid. Consider an object of volume V and density ρs floating in a liquid of density ρl .

Let Vi be the volume of the object immersed in the liquid for the equilibrium of the object.

Weight = up thrust Vρs g =Viρl g Vi ρs = V ρl The above equation represents the fraction of volume immersed.

Vi ρs % of volume immersed in liquid = ×100 = ×100 V ρl Case - 1: If ρs < ρl , then only a fraction of the body will be immersed in the liquid, Case - 2: If ρs =ρl , then the whole rigid body will be immersed in the liquid. Hence the body remains floating.

94

IL Foundation Series Class 9

Case -3: If ρs > ρl , then the body will sink.

Apparent weight of body inside a liquid

If a body is completely immersed in a liquid its effective weight decreases. The decrease in its weight is equal to upthrust on the body. = Wapp Wactual − upthrust = Vρs g − Vρl g Wapp = Vg ( ρs − ρl ) where V total volume = = of body, ρs density = of body, and ρl density of liquid 4.7.3 Archimedes' principle When a body is partially or fully dipped into a fluid at rest, the fluid exerts an upward force of buoyancy equal to the weight of the displaced fluid.

Verification of Archimedes' principle

•

Take a clean and dry beaker and weigh it carefully on a physical balance. Let the weight of the empty beaker be mg.

•

Take a spring balance and suspend it from an iron stand. From its hook, suspend a piece of some metal, say iron, and record the weight. Let it be wg.

•

Take an empty jar and fill it with water till the water is on the verge of overflowing. Place the beaker, which was previously weighed, under its spout. Gently lower the solid, suspended from spring balance, into the water till the solid is completely immersed.

•

Then, note down the reading of spring balance as w1g. ∴ Apparent loss of weight of the solid in water = ( w − w1 ) g

•

Weigh the beaker along with the displaced liquid. Let it be m1g ∴ Weight of water displaced by the solid = ( m1 − m ) g

•

If we compare the weight of the water displaced by the solid ( m1 − m ) with the apparent loss of weight in water, it is found that these are equal.

Applications of Archimedes' principle

It is employed for the determination of the relative density of solids and liquids (heavier or lighter than water). = Re lative density of solid

Weight of solid in air wa = Loss of weight of solid in water wa − w w

Where, wa = weight of solid in air

ww = weight of solid in water

95

GRAVITATION

= Re lative density of liquid

Loss of weight of solid in liquid wa − wl = Loss of weight of solid in water wa − w w

Where, wa = weight of solid in air

wl = weight of solid in liquid

ww = weight of solid in water.

QUICK REVIEW •

The force of attraction between any two masses by virtue of their mass is called gravitational force.

•

1st Law (Law of Orbits): Every planet revolves around the Sun in an elliptical orbit, with the Sun at one of the foci.

•

2nd Law (Law of Areas): The radius vector of a planet from the Sun sweeps out equal areas in equal intervals of time.

•

3rd Law (Law of Periods): The square of the time period of revolution of a planet around the Sun is proportional to the cube of the mean distance of the planet from the sun.

•

Universal law of gravitation: F = G

•

The value of G : 6.67 ×10−11 N.m 2 / kg 2

•

According to the inverse square law, F ∝ 2 .

•

Whenever objects fall towards the Earth under gravitational force alone, we say that the objects are in free fall.

•

If the force on the body is due to the gravity of the Earth, then acceleration produced in that body is called acceleration due to gravity.

GM Relationship between g and G : g = 2 .

R -2 g = 9.8 ms GM Variation of 'g' with altitude: g h = . ( R + h )2

• • • • •

96

m1 m 2 . r2 1 r

 d Variation of 'g' with depth: = gd g 1 −   R Variation of 'g' according to the shape of the Earth: g is minimum at the equator and g is maximum at the poles.

•

Variation of 'g' according to local conditions: The value of g is slightly more at the location of mineral deposits.

•

Mass is a measure of the amount of matter present in a body.

•

The weight of an object is the force with which it is attracted towards the earth. It is denoted by 'W'.

IL Foundation Series Class 9

• •

When an individual jumps from a height, there is a sensation of having no weight. This phenomenon is known as weightlessness.

Weight of the object on the Moon = (1/ 6) × its weight on the earth.

•

A force acting normally on a surface is called thrust.

•

The force (thrust) acting normally on a unit surface area is called pressure. mass Density (ρ) = volume Density of Substance Relative Density = Density of water at 4C When a fluid (either liquid or gas) is at rest, it exerts a force perpendicular to any surface in contact with it.

• • • • • •

P= hρg When a body is in motion in the fluid, then there develops an opposing force, which opposes the motion of the body by the fluid. This force is called the viscous drag.

Atmospheric pressure is the pressure of the Earth's atmosphere. This changes with weather and elevation. Normal atmospheric pressure at sea level is 1.013 ×105 Pa.

•

The upward force exerted by the fluid is called buoyant force (B).

Weight of solid in air wa = • Re lative density of solid = Loss of weight of solid in water wa − w w = • Re lative density of liquid

Loss of weight of solid in liquid wa − wl = Loss of weight of solid in water wa − w w

WORKSHEET - 1 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER I.

Gravitation

1. The force of gravitation is a. Repulsive

b. Electrostatic

c. Conservative

d. Non-conservative

2. The gravitational constant depends upon a. Mass of the bodies

b. Gravitational force

c. Distance between the bodies

d. None of the above factors

3. If the distance between two masses is doubled, then the gravitational attraction between them a. Is doubled

b. Becomes 4 times

c. Is reduced to half

d. Is reduced to a quarter

97

GRAVITATION

4. The gravitational force between two point masses, m1 and m2 ,at separation 'r' is given by km1m 2 F= . r2 The constant 'k' depends on a. The system of units only

b. The medium between masses only

c. Both a and b

d. Neither a nor b

5. Gravitational force is universal in nature. However, when we stand near a wall, we are not pulled by the wall because a. The wall has a large area b. The wall is at a small distance c. Flat surfaces do not exert gravitational force d. The gravitational force due to the wall is negligible as compared to that due to the Earth 6. Statement (A): A person sitting in an artificial satellite revolving around the Earth feels weightless.

Statement (B): There is no gravitational force on the satellite. a. Both statements are true

b. Both statements are false

c. A is true, B is false

d. A is false, B is true

7. The gravitational force between two point masses m and M separated by a distance d is F. Now a point mass ‘3m’ is placed next to m so that the total force on M is XF. What must be the value of X?________. a. 4

b. 3

c. 6

d. 7

8. The gravitational force between two stones of mass 1 kg each, separated by a distance of 1 m in vacuum, is a. Zero

b. 6.675 ×10−5 N

c. 6.675 ×10−8 N

d. 6.675 ×10

−11

N

9. Two spheres of mass m and M are situated in the air, and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3 . The gravitational force will now be F F c. d. 3 F 3 9 10. Two identical spheres are placed in contact with each other. The force of gravitation between the two spheres will be proportional to (R = radius of each sphere)

98

a. F

b.

a. R2

b. R-2

c. R4

d. R-4

IL Foundation Series Class 9

11. Two particles of masses 1 kg and 2 kg are placed at a separation of 50 cm. Assuming that the only forces acting on the particles are their mutual gravitation, the initial accelerations of the two particles are a. 6 × 10−10 m / s 2 and 8.7 × 10−10 m / s 2

b. 5.3 ×10−10 m / s 2 and 2.6 × 10−10 m / s 2

c. 1.57 × 10−10 m / s 2 and 3.4 × 10−10 m / s 2

d. 1.67 ×10−10 m / s 2 and 1.08 × 10−10 m / s 2

12. The orbit of a planet around a star is a. A circle

b. An ellipse

c. A parabola

d.A straight line

13. Kepler's second law regarding the constancy of the areal velocity of a planet is a consequence of the law of conservation of a. Energy

b. Angular momentum

c. Linear momentum

d. None of these

14. Kepler's 2nd law of motion indicates a. How the time period of a planet varies with the radius of its orbit b. The conservation of angular momentum principle c. The nature of energy principle d. The nature of Earth's orbit 15. The orbital speed of Jupiter is a. Greater than the orbital speed of earth

b. Less than the orbital speed of earth

c. Equal to the orbital speed of earth

d. Zero

16. If the Earth is at one-fourth of its present distance from the Sun, then the duration of the year will be a. Half of the present year

b. One-eight of the present year

c. One-fourth of the present year

d. One-sixth of the present year

17. The planet Neptune travels around the sun with a period of 165 years. The ratio of its mean radius of the orbit to that of the earth is approximately a. 10:1

b. 20:1

c. 30:1

d. 40:1

99

GRAVITATION

18. The Earth 'E' moves in an elliptical orbit with the Sun 's' at one of the foci as shown in the figure. Its speed of motion will be maximum at the point.

C

E A

a. C

B

s D

b. A

c. B

d. D

19. The figure shows the motion of a planet around the Sun in an elliptical orbit with the Sun at the focus. The shaded areas A and B are assumed to be equal. If t1 and t2 represent the time for the planet to move from a to b and c to d, respectively, then

b

t1 a

A s B c t d 2 a. t1 < t2 II.

b. t1 > t2

c. t1 = t2

d. t1 ≤ t2

Acceleration due to gravity

1. The acceleration due to gravity (on the Earth) depends upon a. Size of the body

b. Gravitational mass of the body

c. Gravitational mass of the Earth

d. Gravitational force of the Sun

2. The acceleration due to gravity near the surface of a planet of radius R and density d is proportional to a. d/R2

b. dR2

c. dR

d. d/R

3. The acceleration due to gravity would be half its value at sea level at an altitude a. R

b.

2R

c. ( 2 − 1)R

d. ( 2 + 1)R

4. If the Earth stops revolving in its orbit about the Sun, there will be variation in the weight of the bodies at a. Equator 100

b. Latitude

c. Poles

d. Nowhere

IL Foundation Series Class 9

5. An iron ball and a cork ball of the same radius are released from the same height in a vacuum.

Both of them reach the ground simultaneously. It is because i. Acceleration due to gravity is independent of the mass of falling bodies ii. Acceleration due to gravity in vacuum is independent of the size of the bodies iii. In a vacuum, the acceleration due to gravity is zero iv. In a vacuum, there is no resistance to the motion of the balls a. i, ii

b. ii, iii

c. i, iv

d. i, ii, iv

6. The acceleration due to gravity decreases if i. We go down from the surface of the earth towards its center. ii. We go up from the surface of the earth iii. The rotational speed of the earth is increased. iv. We go from the equator towards the poles. a. i, ii, iii

b. i, iii

c. iii, iv

d. i, ii

7. Among the following statements, the true statement is a. g is less at the earth's surface than at a height above or at a depth below b. g is the same at all places on the surface of the earth c. g has its maximum value at the equator d. g is greater at the poles than at the equator 8. Two planets of radii R1 & R2 are made up of the same material. The ratio of acceleration due to gravity at the surfaces of the two planets is 2

2

c. ge = 4gp

d. ge = 2gp

 R1   R2  c.  d.     R2   R1  9. Let ge and gp be the accelerations due to gravity on the Earth and on a different planet, respectively. The radius of the planet as well as its mass are twice that of the Earth. The correct relation among the following is

R1 a.

R2

R2 b.

R1

a. gp = 4ge

b. gp = 2ge

10. If the density of the Earth is doubled, keeping its radius constant, then acceleration due to gravity will be a. 2.45 ms −2

b. 4.9 ms

−2

c. 9.8 ms

−2

d. 19.6 ms

−2

11. Planet x is twice the radius of planet y and is made of a material with the same density. The ratio of acceleration due to gravity at the surface of x to that at the surface of y is a. 1:4

b. 1:2

c. 2:1

d. 4:1

101

GRAVITATION

12. If the value of g at the surface of the Earth is 9.8 ms −2 , then the value of g at a place 480 km above the surface of the Earth will be (radius of earth R = 6400 km ) a. 4.2 ms −2

b. 7.2 ms

−2

c. 8.5 ms

−2

d. 9.8 ms

−2

13. The diameter of a planet is 4 times that of the Earth and its mean density is equal to that −2 of the Earth. If the acceleration due to gravity on the Earth's surface is 9.8 ms , then the acceleration due to gravity on the other planet's surface is a. 4.9 ms −2

b. 9.8 ms

−2

c. 19.6 ms

−2

d. 39.2 ms

−2

14. If the mass of Earth is 80 times that of the Moon and the diameter is double that of the Moon −2 and g on the surface of the earth is 9.8 ms , then the value of g on the Moon is a. 0.49 ms

−2

b. 0.98 ms

−2

c. 4.9 ms

−2

d. 49 ms

−2

III. Motion of objects

1. The final velocity of the freely falling body, when it reaches the ground in 't' seconds is. a. v = gt

b. v = u + gt

c. v = u - gt

d. v = g/t

2. An iron ball and a wooden ball of different masses and the same size are released from the same height in a vacuum. They take the same time to reach the ground. The reason for this is a. Acceleration due to gravity in vacuum is same irrespective of the size and mass of body b. Acceleration due to gravity in vacuum depends upon the mass of the body c. There is no acceleration due to gravity in vacuum d. In vacuum, there is a resistance offered to the motion of the body and this resistance depends upon the mass of the body 3. A lift is moving up with a constant velocity of 5 cms-1. A bolt is detached from the lift then its initial velocity is a. 0 cm s −1

b. 5 cm s-1 upward

c. 5 cm s-1 downward

d. 9 cm s-1

4. A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is a. Equal to the time of fall

b. Less than the time of fall

c. Greater than the time of fall

d. Twice the time of fall

5. A stone dropped from the top of a tower touches the ground in 4 seconds. The height of the tower is (g = 10 ms −2

)

a. 80 m

102

b. 40 m

c. 20 m

d. 160 m

IL Foundation Series Class 9

Passage: Q6-Q8: A ball is dropped from a balloon which is moving up with a constant speed of 10 ms-1. When it is at a height 40 m from the ground, then answer the following: 6. The time taken by the ball to reach the ground a. 4 s

b. 6 s

c. 2 s

d. 5 s

7. The maximum height reached by the ball from the ground a. 40 m

b. 45 m

c. 50 m

d. 55 m

8. The maximum height reached by the ball with respect to the balloon a. 5 m

b. 10 m

c. 15 m

d. Data inadequate

9. A body is projected up with a speed 'u' and the time taken by it is 'T' to reach the maximum height 'H'. Pick out the correct statement based on this information. u T in a. It reaches H in T seconds b. It acquires velocity seconds 2 2 2 2 u H c. Its velocity is at d. Same velocity at 2 T 2 2 10. A stone falls from a balloon that is descending at a uniform rate of 12 ms-1. The displacement of the stone from the point of release after 10 seconds is a. 490 m

b. 510 m

c. 610 m

d. 725 m

11. When a body is thrown up vertically with an initial velocity, then a. Gradually its velocity decreases while it is moving upwards b. Gradually its velocity increases while it is moving upwards c. Its final velocity becomes zero d. Both a and c 12. When a body is thrown up vertically with initial velocity u a. time of ascent = time of descent

b. time of flight =

c. time of ascent > time of descent

d. both a and b

2u g

13. The distance described during the last second of the upward motion, of a body projected vertically upward is a. Dependent upon the velocity of projection b. Dependent upon the time taken to reach maximum height c. Always constant d. Dependent upon the mass of the body 14. The maximum height reached by the body projected vertically up with an initial velocity u is u a. 2 g

u2 b. 2g

u2 c. g

2u2 d. g 103

GRAVITATION

15. An object is projected upwards with a velocity of 100 ms-1. It will strike the ground after [g = 10 ms-2] a. 10 second b. 20 second c. 15 second d. 5 second 16. A ball is projected vertically upward with a speed of 50 ms −1 . Then the speed at half of the maximum height is g = 10 ms −2

(

a. 100 ms-1

)

b. 125 ms-1

c. 35 ms-1

d. 45 ms-1

17. A stone is thrown vertically upward with a speed of 49 ms-1. Then the velocity of the stone one second before it reaches the maximum height is a. 4.9 ms-1

b. 9.8 ms-1

c. 13.6 ms-1

d. 19.6 ms-1

Passage: Q18-Q20: A ball of mass 2 kg is thrown vertically up with a velocity 20 ms-1. [g = 10 ms-2] 18. The maximum height reached by the ball a. 20 m

b. 25 m

c. 5 m

d. 10 m

19. The time taken by the ball to reach the same point of projection a. 2 s

b. 3 s

c. 4 s

d. 5 s

20. The distance covered by the ball during the first second in the downward journey a. 20 m

b. 25 m

c. 5 m

d. 10 m

21. The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on the planet B. A man jumps to a height of 2 m on the surface of A. What is the height of the jump by the same person on B? 2 a. 18 m b. 6 m c. m d. 2/9 m 3 22. A body 'A' is projected upwards with a velocity of 98 ms-1 and a second body 'B' is projected upwards with the same initial velocity but after 4 seconds. Both the bodies will meet after a. 6 s

b. 8 s

c. 10 s

d. 12 s

23. A ball is dropped from the top of a tower of 100 m height. Simultaneously, another ball was thrown upward from the bottom of the tower with the speed of 50 ms-1. They will cross each other after g = 10ms −2    a. 1 s

b. 2 s

c. 3 s

IV. Mass and weight

1. Which among the following is not a property of mass? a. It is the amount of matter contained in a body b. It can be measured by a physical balance c. It varies as per the location d. Its S.I. unit is kilograms 104

d. 4 s

IL Foundation Series Class 9

3 2. If an object weighs 5 N on the Earth, how much would it weigh on the Moon?

5 a. 93 N b. 0.93 N c. 9.3 N d. 9300N 3. The weight of an object on the surface of the Moon is 60 Newton. What will be its weight on the surface of a celestial body where the gravitational ratio is 1/12 of the Moon's gravity? a. 10 N V.

b. 5 N

c. 14 N

d. 16 N

Pressure, thrust, and pressure exerted by liquids and gases

1. The unit of thrust in the SI system is a. Dyne

b. Joule

c. Newton

d. Pascal

2. The force (thrust) acting normally on a unit surface area is called a. Volume

b. Density

c. Mass

d. Pressure

3. Gauge Pressure = a. absolute pressure - atmospheric pressure

b. atmospheric pressure - absolute pressure

c. absolute pressure + atmospheric pressure

d. absolute pressure × atmospheric pressure

4. Pressure (P)= Area Thrust b. c. Thrust × Area d. Thrust + Area Thrust Area 5. The three vessels shown in the figure have the same base area. Equal volumes of a liquid are poured into the three vessels. The force on the base will be

a.

A

B

C

a. Maximum in vessel A

b. Maximum in vessel B

c. Maximum in vessel C

d. Equal in all the vessels

6. Equal mass of three liquids are kept in three identical cylindrical vessels A, B, and C. The densities are ρA , ρB , ρC with ρA < ρB < ρC . The force on the base will be a. Maximum in vessel A

b. Maximum in vessel B

c. Maximum in vessel C

d. Equal in all the vessels

7. Statement (A): Pressure is a vector quantity.

a. Both A and B are true

F . Here F (Force) is a vector quantity.

A b. Both A and B are false

c. A is true, and B is false

d. A is false, and B is true

Statement (B): Pressure P =

105

GRAVITATION

8. The force experienced by the body is 10 N and the cross-section area is 20 m2, so the pressure is a. 1 N/m2

b. 0.5 N/m2

c. 2 N/m2

d. 2.5 N/m2

9. A force of 60 N is applied on a nail, where the tip has an area of cross-section of 0.001cm2, then the pressure on the tip is a. 6 × 108 Pa

b. 6 × 106 Pa

c. 6 × 1010 Pa

d. 6 × 104 Pa

10. The pressure at the bottom of a lake due to water is 4.9 × 106 N/m2. Then the depth of the lake is a. 400 m

b. 500 m

c. 600 m

d. 300 m

c. P = hρg

d. P =

11. Pressure in a fluid at a depth 'h' is a. P = hg

b. P =

VI. Buoyancy

h ρg

ρ hg

1. If the weight of a body and up thrust on the body in a liquid are equal, then the body a. Sinks

b. Floats

c. Just completely immerses in the liquid

d. Completely lies outside the liquid

2. A bird is at rest on the floor of an air-tight box which is being carried by a boy. If the bird starts flying, he feels the box as: a. Heavier

b. Lighter

c. Of the same weight as before

d. Lighter in the beginning and heavier afterwards

3. A body is floating in a liquid. Weight of the body a. Equal to weight of the body

b. Equal to weight of the liquid displaced by the body

c. Less than the weight of the body

d. Zero

4. A body weighs more in a. Air

b. Hydrogen

c. Water

5. The clouds float in air due to this reason: a. Viscosity of air

b. Clouds are heavier

c. At higher altitudes density of air is more

d. Clouds are lighter

6. The buoyant force is equal to a. Weight of the body b. Weight of the liquid displaced by the body c. Difference of the weights of the body and liquid displaced d. Sum of the weights of the body and the liquid displaced 106

d. Vacuum

IL Foundation Series Class 9

7. A boat in the river enters into the seawater, and then it a. Sinks

b. Rises

c. Remains the same

d. Comes to rest immediately

8. A boat contains leaves. When they are thrown into the water, the level of water in the lake a. Remains same

b. Falls

c. Rises

d. First rises, then falls

9. Statement (A): The force which opposes relative motion between the layers of a fluid is called viscous force.

Statement (B): The force which opposes the relative motion between solid surfaces is called frictional force. a. Both A and B are true

b. Both A and B are false

c. A is true, B is false

d. A is false, B is true

10. When a boat moves slowly on the water of a calm river Statement (A): The water in contact with the boat is dragged.

Statement (B): The water in contact with the bed of the river remains at rest. a. Both A and B are true

b. Both A and B are false

c. A is true, B is false

d. A is false, B is true

11. If a body of weight w1 displaces an amount of water w2, then w1 < w2. This statement is a. true

b. false

c. depends on the situation

d. cannot be determined

Passage: Q12-Q14: When a body is immersed in a fluid, it experiences an upward thrust which is called the buoyant force. Let m0 g be the weight of the body and B be the buoyant force.

12. If the body floats in the fluid, then the relation between B and m0 g is a. m0 g = B

b. m0 g > B

c. m0 g < B

d. m0 g ≥ B

13. If the buoyant force is equal to the weight of the body, then the a. Body floats

b. Body sinks

c. Body must suspend

d. Body will fly out of fluid

14. Buoyant force on a body depends on a. Mass of the body

b. Mass of the fluid

c. Volume of the body

d. Volume of the fluid displaced

15. A body will experience minimum upthrust when it is completely immersed in: a. Turpentine

b. Water

c. Glycerine

d. Mercury

107

GRAVITATION

16. Choose the correct statement: a. Apparent weight = Real wt. - Buoyant force b. Apparent weight = Real wt. + Buoyant force c. Apparent weight = Real wt. × Buoyant force d. Apparent weight = Real wt. ÷ Buoyant force 17. The pressure P1 at a certain depth in river water and P2 at the same depth in sea water are related as: a. P1 > P2

b. P1 = P2

c. P1 < P2

d. P1 = P2 atmospheric pressure

18. A body of mass 716 g and volume 448 cm3 is put in a liquid of density 1.3 g cm-3 . Will it float, sink, or partially sink? a. Partially Float

b. Float

c. Sink

d. Can't determine

19. A piece of metal weighs x newton in air, y newton when completely immersed in water, and z newton when completely immersed in liquid. The relative density of the liquid is x−z y−z y−z x−y b. c. d. x−y x−y x−z x−z 20. The surface of water in a tank on the top of a house is 4 m above the tap level. Then the pressure of water at tap when the tap is closed

a.

a. 40000 Nm −2

b. 4000 Nm −2

c. 50000 Nm −2

d. 500 Nm −2

21. An ornament weighing 36 g in air, weighs only 34 g in water. Assuming that some copper is mixed with gold to prepare an ornament, the amount of copper in it is (specific gravity of gold is 19.3 and copper is 8.9 ) a. 1.2 g

b. 2.2 kg

c. 2.2 g

d. 2 kg

Statement: Q22- Q25: When a body is partially or fully dipped into a fluid at rest, the fluid exerts an upward force of buoyancy equal to the weight of the displaced fluid.

22. A body is lowered into a liquid. Loss of weight of the body depends upon a. Volume of the body

b. Density of liquid

c. Acceleration due to gravity

d. All the above

23. Two balls, one of iron and the other of aluminium experience the same upthrust when dipped in water if a. both have the same mass

b. one has half the volume of the other

c. both have equal volume

d. one has one-fourth of the volume as that of other

24. If an object sinks in water, it means:

108

a. The object's density is less than water.

b. The object's density is equal to water.

c. The object's volume is greater than water's.

d. The object's density is greater than water.

IL Foundation Series Class 9

25. A boy is carrying a bucket of water in one hand and a piece of plastic in the other hand. After transferring the plastic piece to the bucket (in which it floats), the boy will carry a. more load than before

b. less load than before

c. same load as before

d. none of these

26. A boat carrying a large number of stones is floating in a water tank, if the stones are unloaded into the water in the tank, then the level of water a. remains unchanged

b. rises

c. falls

d. falls at first and then rises to the same level as before

27. A piece of ice, with a stone frozen inside it, is floating in water contained in a beaker. When the ice melts, the level of water in the beaker a. rises

b. falls

c. remains unchanged

d. falls at first and then rises to the same height as before

28. Choose the correct statements from the following: a. A body will sink in a liquid if its weight is equal to or greater than the weight of the liquid displaced by it. b. A body will sink in a liquid if its weight is less than the weight of the liquid displaced by it. c. When a body floats in a liquid, the portion of the body above the surface of the liquid is independent of the density of the body relative to that of the liquid. d. In still air, a hydrogen - filled balloon rises up to a certain height and then stops rising.

29. A cube of ice is floating in water contained in a vessel. When the ice melts, the level of water in the vessel a. rises

b. falls

c. remains unchanged

d. falls at first and rises to the same height as before

30. The fraction of a floating object of density d0 above the surface of a liquid of density d. d0 − d d0 − d d − d0 d0 b. c. d. d d d0 d 31. A ferry boat has an internal volume 1 m3 weight 50 kg, neglecting the thickness of wood. If a leak develops in the bottom and water starts coming in, then the fraction of the boat volume that will be filled with water before water starts coming in from the sides is

a.

7 15 d.

48 28 32. Ice has a density relative to seawater of 0.90, the portion of an iceberg is submerged is

a.

11 15

a. 15%

b.

19 20

b. 68%

c.

c. 22%

d. 90%

109

GRAVITATION

WORKSHEET - 2 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. How much force acting perpendicular to a surface having an area of 0.5 m2 will produce a pressure of 500 pa? a. 150 N

b. 350 N

c. 175 N

d. 250 N

2. Atmospheric pressure is 1.01 × 105 pascal. How much force is exerted by air on the inside of a window plane that is 40 cm × 100 cm? a. 2.04 ×105 N

b. 4.04 × 104 N

c. 4.04 ×106 N

d. 2.04 ×104

3. An object weight 10 N when measured on the surface of the earth, what would be its weight when measured on the surface of the moon? a. 1.77 N

b. 1.67 N

c. 2.77 N

d. 2.67 N

4. The weight of a certain object is 6 N on the earth. What will be the weight of the same object on the moon a. 2 kg

b. 1.5 kg

c. 1 kg

d. 2.5 kg

c. 0.98 N

d. 0.098 N

5. How many newtons are there in 1 kg-wt ? a. 980 N

b. 9.8 N

6. What are the SI unit of weight? i. dyne

ii. kg-wt

iii. N

iv. g-wt

a. ii and iii

b. ii only

c. i only

d. i and iii

7. The mass of a boy is 50 kg. What will be his weight on the earth? What is his weight on the moon? (g = 10 m/s2) a. 300 N, 30.33 N

b. 400 N, 88.33 N

c. 500 N, 83.33 N

d. 400 N, 33.83 N

8. 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. a. 2 N

b. 8 N

c. 10 N

d. 12 N

9. The relation between pressure (P), force (F), and area of cross-section (A) is P A 10. According to the principle of flotation

a. P = FA

b. F =

c. P =

F A

a. Weight of the liquid displaced = weight of the floating body b. Weight of the liquid > weight of the floating body c. Weight of the liquid displacement < weight of the floating body d. None of these 110

d. P =

F A2

IL Foundation Series Class 9

11. A ball is thrown up and attains a maximum height of 20 m. Calculate its initial speed. (g =10 ms-1) a. 19.7 m/s

b. 20 m/s

c. 25 m/s

d. 35 m/s

12. Two bodies of mass 10 kg each are separated by 1 metre. The force of attraction between them is a. 6.67 × 10-9 N

b. 6.67 × 105 N

Gm1m 2 is valid for r2 a. Rectangular bodies b. Circular bodies

c. 6.67 × 10-5 N

d. 6.67 × 109 N

c. Elliptical bodies

d. Spherical bodies

13. The equation F =

14. A particle is taken to a height R above the earth's surface where R is the radius of the earth.

The acceleration due to gravity there is a. 2.45 m/s2

b. 4.9 m/s2

c. 9.8 m/s2

d. 19.6 m/s2

15. Where a body is thrown up, the force of gravity is a. In the upward direction

b. In the downward direction

c. Zero

d. In the horizontal direction

16. A coin and a feather are dropped together in a vacuum a. The coin will reach the ground first b. The feather will reach the ground first c. Both the bodies will reach the ground together d. The feather will not fall down 17. A person takes out a coin from his pocket and drops it from a height of 1.6 m. With what speed will it strike the ground? a. 6.6 m/s

b. 5.6 m/s

c. 4.6 m/s

d. 3.6 m/s

18. The mass of the object a. Increases in everywhere

b. Remains the same in everywhere

c. Decreases in everywhere

d. None of the above

19. A stone thrown upwards attains a maximum height of 19.6 m. Find the velocity with it was thrown. a. 20 m/s

b. 18.36 m/s

c. 19.6 m/s

d. 20.6 m/s

20. The value of the gravitational constant is a. 6.67 ×1011 Nm 2 / kg 2

−11 2 2 b. 6.67 ×10 Nm / kg

c. 11.67 ×10−11 Nm 2 / kg 2

d. none

111

GRAVITATION

21. The upward pressure exerted by a static liquid is explained by which of the following principles a. Pascal's law

b. Archimedes principle

c. Bernoulli's theorem

d. Newton's law

22. Pressure exerted by a liquid is a. Directly proportional to its density b. Inversely proportional to its density c. Independent of its density d. Directly proportional to the area of cross-section 23. The force of attraction between the molecules of the same medium is called a. Adhesive force

b. Cohesive force

c. Magnetic force

d. Electric force

24. The force of attraction between the molecules of different mediums is called a. Adhesive force

b. Cohesive force

c. Magnetic force

d. Electric force

25. At a given depth, liquid exerts a. Same pressure in all directions

b. No pressure at all

c. More pressure in a particular direction

d. Less pressure in a particular direction

26. Choose the correct option a. 1kg / m3 = 10−3 g / cm3

b. 1kg / m3 = 103 g / cm3

c. 1kg / m3 = 10−5 g / cm3

d. 1 kg/m3 = 105 g/cm3

27. Which of the following is merely a number and has no units? a. Density

b. Relative density

c. Thrust d. Pressure

28. Choose the correct option a. Thrust at a point is a vector quantity

b. Pressure at a point is a scalar quantity

c. Thrust at a point is a scalar quantity

d. Both a and b

29. When a liquid is taken in a container a. The pressure inside the liquid increases with depth b. The pressure inside a liquid at depth h is given by P = P0 + ρ1gh c. The pressure inside the liquid at a constant depth is constant d. All the above

112

IL Foundation Series Class 9

30. When an object floats on the surface of a liquid, a. It displaces a weight of liquid equal to its own weight. b. It displaces a weight of liquid less than its own weight c. It displaces a weight of liquid greater than its own weight d. None of these 31. Two spheres of equal mass M are separated by a distance 2R. A particle of mass m is placed exactly midway between them. What is the net gravitational force on the particle due to the two spheres? a. Zero

b. GMm/r2

c. 2GMm/r2

d. GMm/(2r2 )

32. The gravitational force between two bodies is a. Always attractive

b. Always repulsive

c. Sometimes attractive d. Neither attractive nor repulsive 33. The mass of the planet is M. If the mass of the planet is reduced to 1/8th of the original without a change in density. Then, what is the new value of acceleration due to gravity of the planet? g g c. 3 g d.

2 3 34. A body of mass 'm' is taken from the earth's surface to the height 'h' equal to the radius of earth. The change in potential energy will be

a. 5 g

b.

1 mgr 2 35. We get less sugar for a weight of 5 kg at

c. 2mgR

a. Equator

b. Centre of the earth

c. Pole

d. None of the given three

a. mgr

b.

d.

1 mgR 4

36. Two planets, 'A' and 'B' have the same radius are shown in the figure. ρ1 and ρ2 are densities of the materials in the planets ρ1 < ρ2. If the accelerations due to gravity on the surfaces of the planets A and B are gA and gB respectively, then

ρ2

ρ1

ρ1

"A"

ρ2

"B"

a. gA < gB

b. gA > gB

c. Given information is not sufficient

d. gA = gB 113

GRAVITATION

37. The time period of a satellite near the earth's surface neglecting the height of the orbit of the satellite from the surface of the ground will be equal to a. T = 2π

R

g

b. T = 2π

g R

R3 g 38. A body weighs 63 N on the surface of the earth. How much will it weigh at a height equal to half the radius of the earth?

c. T = 2π Rg

d. T = 2π

a. 63 N

c. 28 N

b. 32.5 N

d. None of these

39. If gp is the acceleration due to gravity at the poles and gc that at the equator then. a. gp < ge

b. gp > ge

c. gp = ge

d. ge = 0

40. The correct relation between gravitational mass and inertial mass is a. inertial mass > gravitational mass

b. inertial mass = gravitational mass

c. inertial mass < gravitational mass

d. none

41. As the distance of a planet from the Sun increases, then the time period of revolution of the planet a. Increases

b. Decreases

c. Constant

d. None

42. The value of G depends on a. The mass of bodies

b. The medium between the bodies

c. The temperature of bodies

d. It is an independent constant

43. The atmosphere is held to the earth by a. The rotation of the Earth

b. The attraction of the Sun

c. The gravity

d. All

44. Fg, Fe, and Fn represent the gravitational, electromagnetic, and nuclear forces, respectively.

Now, arrange them in increasing order of their strengths. a. Fn,Fe,Fg

b. Fg,Fe,Fn

c. Fe,Fg,Fn

d. Fg, Fn,Fe

b. Repulsive

c. Both a and b

d. None of these

45. The force of gravity is a. Attractive

46. When we try to push an empty plastic bottle into the water with our hand, we feel a. Upward thrust increases as we try to push the bottle deep into a liquid b. Upward thrust decreases as we try to push the bottle deep into a liquid c. Upward thrust remains the same as we try to push the bottle deep into a liquid d. None of these 114

IL Foundation Series Class 9

47. Archimedes made the wonderful observation that when a body is kept immersed in a liquid a. It appears to lose some of its weight b. The apparent loss of weight of the immersed body is equal to the weight of the liquid displaced c. The apparent loss of weight of the immersed body is greater than the weight of the liquid displaced d. Both a and b 48. Our body is not crushed by the atmospheric pressure because a. Our blood exerts a pressure which is slightly more than the atmospheric pressure b. Our blood exerts pressure equal to the atmospheric pressure. c. Our blood exerts pressure that is slightly less than atmospheric pressure. d. None of these 49. A man can easily float on the water of the Dead Sea without sinking because a. Density of human body > Density of Dead Sea b. Density of human body < Density of Dead Sea c. Density of human body = Density of Dead Sea d. None of these 50. Various applications of the Archimedes Principle are a. Designing of submarines and ships b. Lactometers which are used to determine the purity of sample of milk. c. Hydrometers which are used to determine the density of liquids d. All the above 51. Assertion (A): A solid body of density half that of water, falls from a height of 10 m and then enters into water. The depth to which it will go in water is 10 m.

Reason (R): Inside the water, the body experiences buoyant force a. Both A and R are correct and R is the correct explanation of A b. Both A and R are correct but R is not the correct explanation of A c. A is correct, R is incorrect d. A is incorrect, R is correct 52. Assertion (A): A man sitting in a boat which is floating on a pond. If the man drinks some water from the pond, the level of water in the pond decreases.

Reason (R): According to Archimedes’ principle, the weight of the displaced fluid by the body is equal to the weight of the body.

115

GRAVITATION

a. Both A and R are correct, and R is the correct explanation of A b. Both A and R are correct, but R is not the correct explanation of A c. A is correct, R is incorrect d. A is incorrect, R is correct 53. Assertion (A): Graph between pressure 'p' and depth ' h ' below the surface of a liquid open to the atmosphere as shown.

Reason (R): Static pressure increases linearly with depth. p

h

a. Both A and R are correct, and R is the correct explanation of A b. Both A and R are correct, but R is not the correct explanation of A c. A is correct, R is incorrect d. A is incorrect, R is correct 54. Assertion (A): The universal gravitational constant is the same as acceleration due to gravity.

Reason (R): The electrical force is experienced by charged particles only. a. Both A and R are correct, and R is the correct explanation of A b. Both A and R are correct, but R is not the correct explanation of A c. A is correct, R is incorrect d. A is incorrect, R is correct 55. Assertion (A): The universal gravitational constant is the same as acceleration due to gravity.

Reason (R): Gravitational constant and acceleration due to gravity have different dimensional formula. a. Both A and R are correct, and R is the correct explanation of A b. Both A and R are correct, but R is not the correct explanation of A c. A is correct, R is incorrect d. A is incorrect, R is correct 116

WORK AND ENERGY

5

5.1

INTRODUCTION

Work and energy are closely associated with each other. Living beings, like humans and animals, get energy from food for vital processes and other tasks like playing, thinking, and physical actions.

Machines also require energy from sources like fuel or electricity to function and perform specific tasks. In our daily lives, the activity we do is termed as work done. But, there is a difference between the work in day-to-day life and the 'work' that we refer to in science.

In everyday life, any physical or mental effort is 'work.' For example, studying, reading, and discussing require hard work. However, in science, 'work' has a specific meaning. According to science, work is only done when an object is moved by a force. So, if we push a rock but it doesn't move, we haven't done any work on the rock. Similarly, carrying a heavy load without moving it also not considered work.

5.2

WORK

Work is said to be done by a force if the point of application of force undergoes displacement either in the direction of the force or in the direction of the component of force.

m

F S Fig. 5.1 Work

Work done depends upon two factors as follows.

1. Force applied.

2. Distance travelled by the body in the direction of the force.

The work done by the force is measured by the product of the magnitude of the force and the displacement from the point of application in the direction of the force. i.e., W = FS •

The SI unit is joule (J) or kgm2 s−2 .

•

The dimensional formula of work is [ ML2 T−2 ] .

•

Work is a scalar quantity.

•

Some other important units of work are erg, eV, MeV, kWh, etc.

117

WORK AND ENERGY

Joule: One joule of work is said to be done whenever a force of one newton displaces a body through a distance of 1 m in its own direction.

5.2.1 Scientific conception of work To understand the scientific concept of work, let us examine some situations. When you push a pebble, and it moves, you have performed work. This is because you have applied a force to the pebble, causing it to be displaced. In the scientific definition, work is done when both force and displacement occur.

Similarly, when a girl pulls a trolley, and it moves, work is being done. This is because the girl is exerting a force on the trolley, resulting in its displacement. Again, both conditions of force and displacement are satisfied. Another example is lifting a book off the ground. In this case, work is done because you are applying a force to the book, causing it to move against gravity. This fulfils the criteria of both force and displacement. In the scientific conception of work, it is necessary for both force and displacement to exist for work to be considered as done.

5.2.2 Work done by a constant force When the displacement of the body is in the direction of constant applied force, the work done by  the force is measured as the product of the magnitude of force 'F ' and distance 'S' through which the body moves.

F

M

S W=FS Fig. 5.2 Work done by a constant force

When the displacement of the body makes an angle ' θ ' with the applied constant force, the work  done by a constant force F is given by W ( component of force along the displacement ) × ( displacement ) = (F cos θ)(S)   = F .S

F sinθ F θ

F cosθ S

Fig. 5.3 Components of force 118

IL Foundation Series Class 9

Work done by a constant force can be positive, negative, or zero.

•

If the displacement is in the direction of the force or the direction of the component of force, then the work done is positive.

•

If the displacement is in the direction opposite to that of force or component of force, then the work done is negative.

•

If the displacement and force are perpendicular to each other, then the work done is zero.

SOLVED EXAMPLES

Example 1: A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force. Find the work done.

Solution: If the force acts on the object all through the displacement, then work done is W = FS ⇒ 5 N × 2 m =10 Nm or 10 J. Example 2: A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.

Solution: Mass of luggage, m = 15 kg and displacement, s = 1.5 m.

W = F ´ s = mg ´ s = 15 kg ´10 ms-2 ´1.5 m = 225 kg ms-2 m = 225 N m = 225 J

Therefore, the work done is 225 J. Example 3: If the work done is 50 J by a force of 10 N, which displaces an object by 10 m, determine the angle between the force and the displacement.

Solution: Given, Work = 50J, Force = 10N, and s = 10m. = W F.s.cosθ We know, ⇒ 50 J = 10 × 10 × cos θ ⇒ cos θ = 1 / 2 ⇒ θ= 60° Therefore, the angle between force and displacement is 60°.

119

WORK AND ENERGY

5.3

ENERGY

Energy is the capacity to do work. It is present in various forms and is essential for everything we do. The unit of energy is the same as that of 'work'.

•

The SI unit of energy is joule (J).

•

The dimensional formula is [ ML2 T−2 ].

•

Energy is a scalar quantity.

5.3.1 Forms of energy There are different types of energy. Energy can be further classified into various well-defined forms, such as 1.

Heat or thermal energy

2.

Chemical energy

3.

Sound energy

4.

Electrical energy

5.

Nuclear energy

6.

Solar energy

7.

Mechanical energy

Heat or thermal energy

Heat is the energy that is transferred between a system and its environment because of a temperature difference that exists between them. Heat is an internal energy that consists of the kinetic and potential energies associated with the random motion of the atoms, molecules and other microscopic bodies within the object.

Chemical energy

Chemical energy is the energy that is released during chemical reactions.

Sound energy

Sound energy is the energy that is created when sound waves move outward from a vibrating object or a sound source.

Electrical energy

Electrical energy refers to the energy generated by the movement of electrons in a conductor that is connected to a battery.

Nuclear energy

The energy released when two nuclei of light elements combine with each other to form a heavy nucleus or when a heavy nucleus breaks into two light nuclei is known as nuclear energy.

120

IL Foundation Series Class 9

Solar energy

Solar energy refers to the energy emitted by the sun.

Mechanical energy

Mechanical energy is the energy that a body possesses due to its speed, position, or changes in shape. It is the combined total of the body's kinetic energy and potential energy.

5.3.2 Kinetic energy The energy possessed by a body by virtue of its motion is known as kinetic energy.

It is given by KE =

1 mv 2 .

2

 Consider a body of mass 'm', which is initially at rest, moving along a straight line with a velocity V   by the application of constant force F . Let the displacement be S.



The work done by the force F is

  W= F ⋅ S= FS ( θ= 0° )

F

F

S Fig. 5.4 Work done by a constant force

If 'a' is the acceleration produced, then, according to Newton's second law F = ma ∴ W = FS = maS

.........................................Eq.(1)

We know that v 2 − u2 = 2aS , put u = 0 v 2 − 02 = 2aS (∴ u = 0) v2 = 2aS v2 ..................................................Eq.(2) 2a Substitute Eq.(2) in Eq.(1) S=

W = maS  v2  = ma    2a  1 W = mv 2 2 But, the kinetic energy of a body is equivalent to the work done in giving the body this velocity.

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WORK AND ENERGY

1 Hence, KE = mv 2 2 Example: An object is moving with a velocity of 54 km/h. Calculate the kinetic energy of the object if its mass is 40 kg.

Solution: To calculate the kinetic energy of the object, we need to use the formula:

K .E =

1 mv 2 2

Given: Mass of the object (m) = 40 kg Velocity of the object (v) = 54 km/h First, we need to convert the velocity from km/h to m/s:

1km 1000m 5 m = = h 3600s 18 s Therefore, the velocity of the object in m/s is: v=

54km 5 m × = 15m / s h 18 s

Now, we can calculate the kinetic energy: 2

1  m 4500 joules K.E =× 40kg × 15  = 2  s 

Therefore, the kinetic energy of the object is 4500 joules.

5.3.3 Potential energy The energy possessed by a body by virtue of its position or configuration is known as potential energy. Potential energy is a form of energy that is stored within an object based on its position or condition. It is the energy that an object has because of its potential to do work.

Example: A block attached to a compressed or elongated spring possesses potential energy.

Types of potential energy: 1.

Gravitational potential energy

2.

Elastic potential energy

3.

Electric potential energy

Gravitational Potential Energy

This is the energy an object has due to its height or position in a gravitational field.

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IL Foundation Series Class 9

The higher an object is, the greater its gravitational potential energy. For example, if you lift a book off the ground, it gains potential energy because of its increased height.

The gravitational potential energy of two particles of masses 'm1' and 'm2' separated by a distance 'r' is given by U=

−Gm1m2 r

The negative sign indicates that the force is attractive in nature. If a body of mass 'm' is raised to a height 'h' from the surface of the Earth.

The change in PE of the system = mgh [when h << R (radius of earth)] Elastic potential energy

This is the energy stored in an elastic object, like a stretched rubber band or a compressed spring.

When an elastic object is stretched or compressed, it gains potential energy. When released, that potential energy is converted into kinetic energy, causing the object to move. When a spring is stretched or compressed by an amount 'x' from its unstretched position, the elastic potential energy stored in the spring is 1 U = kx 2 2

Where k is called the spring constant, its unit is Nm-1.

(Natural state) x (Elongated state) Fig. 5.5 Elastic potential energy

Electric potential energy

Electric potential energy refers to the energy stored in an object or system due to the configuration or arrangement of electric charges. It arises from the interaction between charged objects or within an electric field. The electric potential energy of two point charges q1 and q2 separated by a distance 'r' in a vacuum is given by U=

1 q1q2 4π0 r

0 is called the permittivity of free space.

1 = 9 ×109 Nm2 / C2 4π0

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WORK AND ENERGY

SOLVED EXAMPLES Example 1: A man lifts 20 kg of mass up to a height of 1.6 m. What is the work done by the man?

(Assume g = 10 m / s2 ) Solution: We know that work done on an object is equal to the change in its energy.

The energy possessed by an object due to its position relative to other objects is the potential energy.

When a man lifts 20 kg of mass up to a height of 1.6 m, the work done is stored as potential energy.

So, the potential energy will be counted here.

Work done = potential energy ⇒ W = m×g×h Given: Mass, m = 20 kg Height, h = 1.6 m g = 10 m/s2 Work done, W = 20 ×1.6 ×10 = 320 J The work done by the man is 320 J. Example 2: A bag of wheat weighs 200 kg. To what height should it be raised so that its potential energy becomes 9800 J? ( g = 9.8 ms−2 ) Solution: Given: Mass of the bag, m = 200 kg Potential energy, PE = 9800 J Acceleration due to gravity, g = 9.8 ms-2 We know, PE = mgh Now,

P.E ⇒h= mg 9800 ⇒h= 200 × 9.8 ⇒h= 5m Hence, the bag of wheat should be raised to a height of 5 m.

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IL Foundation Series Class 9

5.4

THE WORK-ENERGY THEOREM

The work done by all the forces acting on a particle is equal to the change in its kinetic energy. In other words, when work is done on an object, it transfers energy to or from the object, resulting in a change in its motion.

W = Kf − Ki 1 1 W mv 2 − mu2 = 2 2

Consider a particle of mass 'm' that is moving with an initial velocity 'u'. When it is under the action of a constant net force 'F', its uniform acceleration is 'a'.

Its velocity becomes 'v' after a displacement 'S'.

Work done by the net force W = FS = maS ( F = ma)  v 2 − u2   v 2 − u2  = m = ( S ) m      2S   2  1 1 =mv 2 − mu2 =− Kf Ki 2 2 = Kf − Ki W

Kf and Ki are the final and initial kinetic energies of the particle, respectively.

The work-energy theorem is applicable not only for a single particle but also for a system of particles.

Example: A bullet of mass 2.5 g moving with a velocity of 500 ms-1 enters a wooden block and comes out of it with a velocity of 100 ms-1. Find the work done by the bullet while passing through the wooden block.

Solution: Mass m = 2.5 g = 2.5 × 10-3 kg; Initial velocity u = 500 m/s; Final velocity v = 100 m/s 1 1 mv 2 − mu2 2 2 = 1/2 × 2.5 × 10-3 × [1002 - 5002] = W By work-energy theorem,

= - 300 J Therefore, the work done is - 300J

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5.5

CONSERVATION OF ENERGY

5.5.1 Interconvertible nature of various energy forms Energy can change its form, like from kinetic energy (energy of motion) to potential energy (stored energy), but the total amount of energy remains the same. For example, imagine a ball rolling down a slope. As the ball rolls down, it gains kinetic energy as its speed increases, but it loses potential energy since it is moving downhill. At the bottom of the slope, the ball has converted all of its potential energy into kinetic energy. This demonstrates the conservation of energy.

Energy has an interconvertible nature. This principle of energy conversion, also known as the principle of energy conservation, holds true across various natural processes. An example is the conversion of chemical energy into thermal energy during combustion, the transformation of electrical energy into light energy through a light bulb, or the interplay between potential and kinetic energy when an object falls. In all these instances, energy constantly changes form.

5.5.2 Law of conservation of energy According to the law of conservation of energy, energy can neither be created nor be destroyed, but it can be transformed from one form to another form without any loss or gain. The total energy of a closed system remains constant.

In the case of a freely falling body and a vertically projected body, the sum of potential and kinetic energies (mechanical energy) at any point in its path is constant.

•

Mechanical energy remains constant under the action of conservative forces.

•

Mechanical energy doesn't remain constant under the action of non-conservative forces.

According to the work-energy theorem, the work done by all the forces, i.e., conservative, non-conservative, and external forces, is equal to the change in the kinetic energy.

Wc + Wnc + Wext = Kf - Ki The negative of the work done by the conservative internal forces is equal to the change in the potential energy of the system.

Wc = - ( Uf − Ui ) Work done by all the forces except conservative forces is equal to change in the mechanical energy.

Wnc + Wext = Ef - Ei If the internal forces are conservative, the work done by the external forces is equal to the change in mechanical energy.

Wext = Ef - Ei 126

IL Foundation Series Class 9

SOLVED EXAMPLES Example: A body of mass 10 kg is kept at a height of 10 m from the ground. When it is released after some time, its kinetic energy becomes 450 J. Find the potential energy of the body at that instant?

Solution: Potential energy is the amount of energy stored in an object due to its position.

The potential energy of the object decreases as it approaches the ground while the kinetic energy increases.

We know, Potential energy = mgh

1 Kinetic energy = × m × v 2 2 Given: Mass, m = 10 kg Height, h = 10 m At a height of 10 m, the mechanical energy of the body, E = Kinetic energy + Potential energy

1 E = × m × v 2 + mgh 2 1 E = ×10 × (0)2 + 10 ×10 ×10 (initial velocity of the body is zero) 2

E = 10 ×10 ×10 = 1000 joule We are given that after some time, the kinetic energy is 450 J. Suppose at that height, the instant potential energy is u, then by the law of conservation of energy

= E 450 + u 1000 = 450 + u ⇒ u= 1000 − 450 ⇒u= 550 J Therefore, the potential energy of the body at that instant is 550 J.

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WORK AND ENERGY

5.5.3 Mass-energy equivalence Mass-energy equivalence is a fundamental concept in physics that describes the relationship between mass and energy. A material particle itself is a form of energy. 'm' kg of mass is equivalent to mc2 joule of energy. i.e., E = mc2 Where, c = 3 × 108 ms-1 is the speed of light in vacuum.

5.6

POWER

Power is a fundamental concept in physics and engineering, representing the rate at which work is done, or energy is transferred. It is defined as the amount of work done or energy transferred per unit of time. In simpler terms, power measures how quickly a task is performed.

The work done in unit time is called power. work done time taken W P= t

Power =

•

The SI unit of power is 'watt' or 'W' or kgm2 s-3.

•

The dimensional formula of power is [M1 L2 T-3].

•

The power is a scalar quantity.

1W =1J / s

•

Horsepower is a unit of power commonly used to measure the rate at which work is done.

1 horsepower = 1 hp = 746 W

  If a force F acts on the body which has velocity ' V ' at that instant, then the instantaneous power  due to F is given by   P = F ⋅ V = FV cos θ

Example: An automobile is moving at 100 kmph and its engine is exerting an attractive force of 3920 N. What horsepower must the engine develop if 20% of the power developed is wasted?

Solution: Velocity = 100 kmph = 100 × Wastage of power = 20%

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5 m / s ; Force = 3920 N; 18

IL Foundation Series Class 9

Used power = 80% Power =

80% P = FV; 80 5 P = 3920 ×100 × 100 18

W Fs = t t

100 5 × 3920 ×100 × = 13.61 ×104 W= 182.5 hp 80 18 Therefore, the horsepower developed by the engine is 182.5 hp.

∴ P=

5.7 COMMERCIAL UNIT OF ENERGY Kilowatt-hour is the commercial unit of energy. A kilowatt-hour is the amount of electrical energy consumed or produced when a power of one kilowatt (1 kW) is used continuously for one hour. In commercial settings, kilowatt-hours measure the amount of electricity consumed by customers. The cost of electricity is usually based on the number of kilowatt-hours used.

1= kWh 1000 W ×1 h We know,

1 W = 1 J s−1 1 h =60 × 60 s =3600 s So, it can be rewritten as

1 kilowatt hour = 1000 W ×1 h = 1000 ×1 J s−1 × 3600 s So, 1 kWh = 3, 600, 000= J 3.6 ×106 J Therefore, 1 kilowatt hour (1 kWh) is equal to 3.6 × 106 joules.

QUICK REVIEW •

Work and energy are integral to both living beings and machines. In science, work is defined as the displacement of an object caused by a force.

•

Everyday activities may be considered work, but in scientific terms, work involves both force and displacement.

W (there component of force along displacement ) × ( displacement ) Work is done by a force when is displacement in thethe direction of the force.

It depends on the force applied, and the distance in the direction of the force. = (F costravelled θ)(S)   The formula for work is W == F . S , measured in joules (J).

• •

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WORK AND ENERGY

•

Work can be positive, negative, or zero based on the direction of displacement relative to the force.

•

Energy, the capacity to do work, comes in various forms, such as heat, chemical, sound, electrical, nuclear, solar, and mechanical energy.

•

Kinetic energy is related to the motion of an object, while potential energy is associated with its position or configuration.

•

The work done on a particle is equal to the change in its kinetic energy.

•

Energy is interconvertible; it can transform from one form to another without loss or gain.

•

The law of conservation of energy states that the total energy of a closed system remains constant.

•

Power is the work done per unit time.

•

P = W/t, measured in watts (W). It depends on force, velocity, and the angle between them.

The instantaneous power is given by P = FVcos θ.

•

The kilowatt-hour (kWh) is the commercial unit of energy, equivalent to 3.6 × 106 joules.

WORKSHEET - 1 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER I.

Work

1. A man pushes a wall and fails to displace it. Based on this information, which of the following is true? a. Work done by him is positive

b. Work done by him is negative

c. No work is done by him

d. All of the above

2. The work done by a centripetal force a. Increases by decreasing the radius of the circle b. Decreases by increasing the radius of the circle c. Increases by increasing the mass of the body d. Is always zero 3. When the force applied and the displacement of the body are inclined at 90° with each other, the work done is a. Infinite

b. Maximum

c. Zero d. Unity

4. A particle is acted upon by a force of constant magnitude that is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that:

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a. Its velocity remains unchanged.

b. Its acceleration remains in the same direction.

c. Its kinetic energy decreases over time.

d. It moves in a circular path.

IL Foundation Series Class 9

5. No work is done by force on an object if: a. The force is always perpendicular to its velocity b. The force is always perpendicular to its acceleration c. The point of application of force moves on the object d. The object moves in such a way that the point of application

of force also changes 6. The work done by a force on a body does not depend upon a. Mass of the body. b. The displacement of the body. c. The initial velocity of the body. d. The angle between the force vector and the displacement vector.

7. A force of 7 N acts on an object. The displacement is 8 m in the direction of the force. How much is the work done in this case? a. 56 J

b. 64 J

c. 42 J

d. 32 J

8. A body moves 20 cm when 90 N of force is applied to it. If the work done in this process is 18 J, what angle should there be between the force and the displacement? a. 45° II.

b. 60°

c. 90°

d. 0°

Energy

1. Elastic potential energy of a spring is independent of a. Spring constant

b. Elongation of spring

c. Compression of spring

d. Both b and c

2. If two electrons are forced to come closer to each other, then the PE of the system of 2 electrons a. Becomes zero

b. Increases

c. Decreases

d. Becomes infinity

3. Potential energy is zero a. At a height equal to double the radius of the earth. b. At infinity. c. At a height equal to half the radius of the earth. d. At a depth equal to 1/3 radius of the earth.

4. A body at rest can have a. Kinetic energy

b. Momentum

c. Both momentum and kinetic energy

d. Potential energy

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WORK AND ENERGY

5. Identify which of the following forms of energies can be only positive. a. Kinetic energy

b. Potential energy

c. Mechanical energy

d. Both kinetic and mechanical energies

6. If the force acting on a body is inversely proportional to its speed, then the kinetic energy of the body is a. Constant

b. Directly proportional to time

c. Inversely proportional to time

d. Directly proportional to square of time

7. A certain weight is attached to a spring. It is pulled down and then released making it oscillate up and down. Its KE will be a. Maximum in the middle of the movement

b. Maximum at the bottom

c. Maximum just before it is released

d. Constant

8. The energy possessed by a body by virtue of its motion is called a. Kinetic energy

b. Potential energy

c. Mechanical energy

d. All the above

9. KE of a body is directly proportional to (at constant velocity).

1 m 10. A body of mass 5 kg is moving with a velocity of 18m/s. Find its kinetic energy.

a. m

b. m2

c.

a. 760 J

b. 810 J

c. 980 J

m

d.

d. 788 J

11. A metal ball of 4kg is dropped from a height of 2.5 m. What will the ball's kinetic energy be when it strikes the ground? a. 76 J

b. 84 J

c. 98 J

d. 88 J

12. An object of mass 8 kg is at a certain height above the ground. If the potential energy of the object is 480 J, find the height at which the object is with respect to the ground. [Given: g = 10 ms-2]. a. 6 m

b. 4 m

c. 2 m

d. 8 m

13. A body of mass 2 kg is thrown vertically upwards with an initial velocity of 20m/s. The potential energy at the end of 2.0 s is (use, g = 10 m/s2) a. 360 J

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b. 400 J

c. 420 J

d. 280 J

IL Foundation Series Class 9

III. The work-energy theorem

1. The total work done on a particle is equal to the change in its kinetic energy a. Only if the forces acting on it are conservative. b. Always c. Only if gravitational force alone acts on it. d. Only if an elastic force acts on it.

2. The work done by all the forces (external and internal) on a system equals the change in a. Total energy

b. Kinetic energy

c. Potential energy

d. Both b and c

c. Both a and b

d. Rigid body

3. Work-energy theorem is applicable for a. Single particle

b. System of particles

4. The expression for work-energy theorem a. W= Ki - Kf

b. W= Kf - Ki

1 d. both b and c ( mv2 − mu2 ) 2 5. A body of mass 2 kg is moving at a constant velocity of 4 ms-1. To bring it to rest in 4 meters, the work done is

c. W =

a. 16 J

b. 8 J

c. 12 J

d. 10 J

6. How

much work should be done on a bicycle of mass 20kg to increase its speed from 2ms-1 to -1 5ms ? a. 170 J

b. 210 J

c. 270 J

d. 290 J

IV. Conservation of energy

1. The

negative of the work done by the conservative internal forces on a system equals the change in a. Total energy

b. Kinetic energy

c. Potential energy

d. Both b and c

2. Identify the correct statement about work-energy theorem a. Work done by all the conservative forces is equal to the decrease in potential energy. b. Work done by all the forces except the conservative forces is equal to the change in mechanical energy. c. Work done by all the forces is equal to the change in kinetic energy. d. Work done by all the forces is equal to the change in potential energy.

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WORK AND ENERGY

3. A heavy stone is thrown from a cliff of height ' h' with a speed 'v'. The stone will hit the ground with maximum speed if it is thrown a. Vertically downward.

b. Vertically upward.

c. Horizontally.

d. Speed does not depend on initial direction.

4. The work done by all the forces (external and internal) on a system equals the change in a. Total energy

b. Kinetic energy

c. Potential energy

d. both b and c

5. If the kinetic energy of a body is increasing, then a. Work done by conservative forces maybe positive b. Work done by conservative forces maybe zero c. Both a and b d. Neither a nor b V.

Power

1. A train is moving with velocity 'V' and a force 'F' is acting on it. The power on it is a. FV

b. FV2

c. F/V

d. F2 V

2. A force of 200 N is required to move a body with a velocity of 20 ms-1; the power delivered is a. 100 W

b. 500 W

c. 1000 W

d. 4000 W

3. Two children, say A and B, weigh the same. Both start climbing up a rope separately and reach a height of 8 m in 15 s and 20 s, respectively. If the power of persons A and B are given

by PA and PB respectively, then a. PA > PB

b. PA < PB

c. PA = PB

d. PA ≥ PB

c. 1.341

d. 7460

4. One kW = __________ hp. a. 746

b. 3.602

5. A 12hp motor has to be operated 8 hours per day. How much will it cost (in Rupees) at the rate of 50 paise per kWh in 10 days? a. 3970

b. 358

c. 3750

d. 3500

VI. Commercial unit of energy

1. How many joules are equivalent to 1 kilowatt hour? a. 360 J

b. 3600 J

c. 3.6 × 105 J

d. 3.6 × 106 J

c. Kilowatt

d. Kilowatt-hour

2. What is the commercial unit of energy? a. Joule

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b. Watt

IL Foundation Series Class 9

3. A device consumes 2 kilowatt hours of energy in 1 hour. How many joules of energy does it consume? a. 7.2J

b. 7.2 × 103 J

c. 7.2 × 104 J

d. 7.2 × 106 J

4. How many kilowatt hours are equivalent to 1.5 × 107 joules? a. 4.17 kWh

b. 41.7 kWh

c. 417 kWh

d. 4170 kWh

5. A 100-watt light bulb is switched on for 5 hours. How many kilowatt hours of energy does it consume? a. 0.05 kWh

b. 0.5 kWh

c. 0.005 kWh

d. 0.0005 kWh

WORKSHEET - 2 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Assertion (A): Power is a scalar quantity.

Reason (R): Power is the vector product of force and velocity. a. Both Assertion (A) and Reason(R) are true, and Reason (R) is the correct explanation of Assertion (A). b. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). c. Assertion (A) is true, but Reason (R) is false. d. Assertion (A) is false, but Reason (R) is true.

2. Assertion (A): Power of a constant force is also constant.

Reason (R): Net constant force will always produce a constant acceleration. a. Both Assertion (A) and Reason(R) are true, and Reason (R) is the correct explanation of Assertion (A). b. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). c. Assertion (A) is true but Reason (R) is false. d. Assertion (A) is false but Reason (R) is true.

3. Statement (A): Power is inversely proportional to time taken by the body.

Statement (B): Power is directly proportional to work done. a. Both A and B are true.

b. Both A and B are false.

c. A is true and B is false.

d. A is false and B is true.

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4. If the kinetic energy possessed by a man of 50 kg is 625 J, then the speed of man is a. 5 ms-1

b. 50 ms-1

c. 0.5 ms-1

d. 500 ms-1

5. A car is moving with a velocity of 54 km/h. The kinetic energy of a boy of mass 40 kg sitting in the car is a. 4500 J

b. 450 J

c. 0.45 J

d. 45 J

6. Two bodies of equal masses move with uniform velocities V and 2 V, respectively. The ratio of their kinetic energies is a. 2:1

b. 4:1

c. 1:2

d. 1:4

7. A man has a box of mass 10 kg. The energy of the box when the man runs with a constant velocity of 2 m/s along with the box behind the bus is a. 10 J

b. 30 J

c. 20 J

d. 2 J

8. A car and a bus are moving with the same kinetic energy. They are brought to rest by applying brakes which provide equal retarding forces. The distances covered by them before coming to rest will be a. Inversely proportional to their masses b. Directly proportional to their masses c. Equal d. Inversely proportional to the square of their masses 9. If a lorry and a car moving with same KE are brought to rest by applying the same retarding force, then a. Lorry will come to rest in a shorter distance

b. Car will come to rest at a shorter distance

c. Both come to rest at the same distance

d. Cannot predict

10. Assertion (A): Whenever a force acts on a body, its KE always increases.

Reason (R): KE of a particle remains constant in uniform circular motion. a. Both Assertion (A) and Reason(R) are true, and Reason (R) is the correct explanation of Assertion (A). b. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). c. Assertion (A) is true, but Reason (R) is false. d. Assertion (A) is false, but Reason (R) is true.

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IL Foundation Series Class 9

11. Assertion (A): The change in KE of a particle is equal to the work done on it by the net force.

Reason (R): Change in KE of a particle is equal to the work done only in case of a system of one particle. a. Both Assertion (A) and Reason(R) are true, and Reason (R) is the correct explanation of Assertion (A). b. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). c. Assertion (A) is true, but Reason (R) is false. d. Assertion (A) is false, but Reason (R) is true.

12. A stone falls from a height on the sand. If it penetrates more a. Resistance to motion is more. b. Resistance to motion is less. c. It does not depend on the extent of penetration. d. A stone never penetrates into sand but only into mud.

13. Statement (A): A body at rest can have mechanical energy.

Statement (B): Mechanical energy of a freely falling body decreases gradually. a. Both A and B are true.

b. Both A and B are false.

c. A is true, B is false.

d. A is false, B is true.

14. Assertion (A): The total energy of a system is always conserved irrespective of whether external forces act on the system or not.

Reason (R) : If external forces act on a system, the total momentum and energy will increase. a. Both Assertion (A) and Reason(R) are true, and Reason (R) is the correct explanation of Assertion (A). b. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). c. Assertion (A) is true but Reason (R) is false. d. Assertion (A) is false but Reason (R) is true.

15. The incorrect statement(s) among the following a. Only conservative forces are responsible for change in potential energy, but for change in kinetic energy other forces are responsible. b. If kinetic energy is constant, it means work done by conservative force is zero. c. If conservative forces are doing negative work, then potential energy will increase and kinetic energy will decrease. d. All of the above are wrong.

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WORK AND ENERGY

16. A body of mass m = 1 kg is dropped from a height h = 40 cm on a horizontal platform fixed to one end of an elastic spring the other being fixed to a base as shown in fig. As a result, the spring is compressed by an amount =10 cm. The force constant of the spring is ( g = 10 ms−2 ) a. 600 N/m

b. 800 N/m

c. 1000 N/m

d. 1200 N/m

m h

Platform

Base

17. A child is sitting on a swing. Its minimum and maximum height from the ground are 0.75m and 2m, respectively. Its maximum speed will be a. 100 m/s

b. 5 m/s

c. 8 m/s

d. 15 m/s

18. A body of mass 10 kg at rest is acted upon simultaneously by two forces 4 N and 3 N at right angles to each other. The kinetic energy of the body at the end of 10 s is a. 100 J

b. 300 J

c. 50 J

d. 125 J

19. A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest, assuming it faces constant resistance to motion? a. 3.0 cm

b. 2.0 cm

c. 1.5 cm

d. 1.0 cm

20. A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m, and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is a. 40 m/s

b. 20 m/s

c. 10 m/s

d. 10 30 m / s

21. A body of mass 'm' is accelerated uniformly from rest to a speed 'v' in a time 'T'. The instantaneous power delivered to the body as a function of time is given by mv 2 t

mv 2 t2

1 mv 2 t2

1 mv 2 t2

a. 0.5 J

b. -0.5 J

c. -1.25 J

d. 1.25 J

a. T2 b. T2 c. 2 T2 d. 2 T2 22. A particle of mass 100 g is thrown vertically upwards with a speed of 5 m/s, the work done by the force of gravity during the time that the particle goes up is 23. A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg, and then they move together as a single body. The energy loss during the collision is a. 0.16 J

138

b. 1.00 J

c. 0.67 J

d. 0.34 J

IL Foundation Series Class 9

24. On a level road, a scooterist applies brakes to slow down from a speed of 36 km/h to 18 km/hr.

The mass of the scooter is 86 kg and the mass of the scooterist and petrol is 64 kg. The work done by the brakes is a. 5265 J

b. 5625 J

c. 5256 J

d. 5526 J

25. A body of mass 0.5 kg is moving at a constant velocity of 2 ms-1. To bring it to rest in 2 metres, the magnitude of work done is a. 0.5 J

b. 1.0 J

c. 2.0 J

d. 4.0 J

26. A force of magnitude 10 N acts on a body of 4 kg moving initially in the direction of the force with a speed of 5 m/s. The distance through which the force must act in order to increase the speed of the body to 10 m/s is a. 5 m

b. 10 m

c. 15 m

d. 25 m

27. An object of mass 5 kg falls from rest through a vertical distance of 20 m and attains a velocity of 10 m/s. The work done by the resistance of air on the object is ( g = 10 ms-2 ) a. 750 J

b. -750 J

c. 500 J

d. -500 J

28. A box is pushed through 4.0 m across a floor offering 100 N resistance, then the work done by the resisting force is a. 40 J

b. 400 J

c. 4000 J

d. zero

c. Both PE & KE

d. Chemical Energy

29. The energy possessed by a bent bow is a. KE

b. PE

30. ___________of a two-particle system depends only on the separation between the two particles a. Kinetic energy

b. Total mechanical energy

c. Potential energy

d. Total energy

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6

SOUND

In daily life, we encounter a variety of sounds from different sources, such as humans, birds, bells, machines, vehicles, televisions, radios, and more. Sound is a type of energy that makes us hear things.

6.1 PRODUCTION OF SOUND 6.1.1 Sound is produced by a vibrating body Sound is produced when objects vibrate, which involves a back-and-forth motion. For example: •

S triking a tuning fork or a stretched rubber band causes them to vibrate, resulting in the production of sound.

•

The human voice is a product of the vibration of vocal cords.

•

String instruments generate sound as their strings vibrate.

•

The movement of a bird’s wings creates a sound.

•

A flute produces sound by the vibration of the air column as air passes through it.

Now, when the objects vibrate, they push the air around them. This pushing of air creates something we call sound waves. These sound waves travel through the air and reach your ears.

6.1.2 Experiment to verify that vibrating bodies produce sound

Fig. 6.1 Experiment to verify that vibrating bodies produce sound

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Consider a scenario where a tuning fork is attached to a rigid support. A pith ball, suspended from support, is positioned close to one of the prongs of the tuning fork. Upon exiting the fork with a rubber hammer, and if the pith ball comes into contact with the fork, it is observed that the ball is propelled away by the fork. This indicates that the tuning fork, which generates sound, is in a state of vibration, and the vibrations of the fork cause the pith ball to be propelled away. This observation leads us to the conclusion that sound is created by vibrating objects.

In the context of human speech, when we talk, the vocal cords, located in a cavity in our throat known as the larynx, vibrate, producing sound. Similarly, the vibrating strings of musical instruments such as guitars and veenas contribute to the generation of sound in the resonating chambers, or 'sounding boxes,’ of these instruments.

6.2 PROPAGATION OF SOUND 6.2.1 Sound needs a medium for propagation Sound originates from the vibrations of objects. The substance or material through which sound travels is referred to as a medium, which can be a solid, liquid, or gas. As sound travels, it moves through the medium from its source to the listener.

When an object vibrates, it causes the particles of the surrounding medium to vibrate as well.

It’s important to note that these particles don’t physically travel from the vibrating object to the ear. Instead, a particle in direct contact with the vibrating object is initially displaced from its equilibrium position. This displaced particle then imparts a force on its adjacent particle, causing it to move from its resting position. After this displacement, the initial particle returns to its original position. This process continues throughout the medium until the sound reaches the listener’s ear.

It’s crucial to understand that the disturbance created by the sound source in the medium travels through the medium itself and not by the physical movement of individual particles within the medium.

6.2.2 Experiment to verify that sound requires a medium for propagation Consider an electric bell suspended within a glass jar equipped with an outlet. The bell is hung from the cork lid of the jar using strings, and two small holes in the lid allow for the connection of electric wires to the bell. Initially, the jar contains air. When the circuit is energised by turning on the switch, the bell rings, and the sound is audible to an observer standing in proximity to the jar.

Lid made of cork Wires Electric bell Glass jar

Outlet to vacuum pump

Fig. 6.2 Bell jar experiment 141

SOUND

Subsequently, the outlet of the jar is attached to a vacuum pump, and the air is evacuated from the jar. Consequently, there is no longer air or any other medium surrounding the bell within the jar.

Upon activating the circuit, the bell still visibly rings, but the sound becomes inaudible. This shows that sound is unable to propagate through a vacuum.

6.3 TYPES OF WAVES 6.3.1 Wave A wave is described as a disturbance that travels through a medium, initiated by the particles of the medium setting neighbouring particles into motion.

This sequential motion among particles generates similar movements in adjacent particles. Notably, the particles of the medium do not advance themselves; instead, the disturbance is transmitted forward. This phenomenon is evident during the propagation of sound in a medium, and hence, sound is often known as a wave.

Specifically, sound waves are identified by the motion of particles in the medium and are termed mechanical waves.

The most common medium through which sound travels is air. When a vibrating object moves forward, it pushes and compresses the air in front of it, creating a region of high pressure known as compression (C). As illustrated in Fig 6.3, this compression starts to move away from the vibrating object. Conversely, when the vibrating object moves backwards, it generates a region of low pressure called rarefaction (R), as depicted in Fig 6.3. The rapid back-and-forth movement of the object creates a sequence of compressions and rarefactions in the air, constituting the sound wave that propagates through the medium.

Compression B

ARarefaction

Compression Rarefaction

Fig. 6.3 Compressions (C) and rarefactions (R)

In this context, compression represents the region of high pressure, while rarefaction denotes the region of low pressure. The pressure is intricately linked to the number of particles in a given 142

IL Foundation Series Class 9

volume of the medium. A higher density of particles results in increased pressure, and conversely, lower density leads to decreased pressure.

6.3.2 Types of waves Waves in a medium can be classified into two main types: longitudinal waves and transverse waves.

Longitudinal waves

In a longitudinal wave, the particles of the medium move parallel to the direction of the wave.

Importantly, the particles do not physically relocate from one place to another; instead, they oscillate back and forth around their position of rest.

In sound, regions where the coils are closer together are called compressions (C), and where they are further apart are called rarefactions (R) or expansion. Individual particles of the medium oscillate back and forth about their position of rest, contributing to the propagation of sound. Therefore, sound waves are an example of longitudinal waves.

Example

Expansion Source

Music System Sound Waves

Direction

Wavelength

Compression

Fig. 6.4 Longitudinal waves Transverse waves

In a transverse wave, particles of the medium oscillate up and down about their mean position as the wave travels. An example of a transverse wave is the ripples on the surface of water when a pebble is dropped into a pond.

Source

Light is also considered a transverse wave, but in the case of light, the oscillations are not related to the particles, pressure, or density of the medium; it is not a mechanical wave. Further exploration of transverse waves will be covered in more advanced classes.

Example Crest

Television Visible light

Direction

Amplitude

Trough

Trough

Wavelength

Fig. 6.5 Transverse waves

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SOUND

Note: Sound relies on a medium for propagation and cannot travel through a vacuum where no air or particles are present. The absence of particles in a vacuum means that there is no medium for sound waves to traverse. Unlike sound, light does not depend on a material medium for its propagation and can travel through the vacuum of space.

6.4 CHARACTERISTICS OF A SOUND WAVE A graphic representation of a sound wave is depicted in Fig. 6.6(c), illustrating the variations in density and pressure as the sound wave transverses through the medium. At any given time, the density and pressure of the medium fluctuate above and below their average values.

Fig. 6.6(a) and Fig. 6.6(b) specifically show the variations in density and pressure as a sound wave propagates.

Compressions (C), where particles are closely packed, are represented by the upper portion of the curve in Fig. 6.6(c). The peak of the curve signifies the maximum compression, indicating regions of high density and pressure. In contrast, rarefactions (R), where particles are spread apart, are depicted by the valley or the lower portion of the curve in Fig. 6.6(c). The highest point of the wave is called the crest, while the lowest point is called the trough.

Density variations Speaker (Source of sound) C

C

R

R

(a)

C

R

C

Pressure variations Speaker (Source of sound) C

C

R

R

(b)

λ

Density or Pressure

A

Distance

(c)

C

R

C

Crest Average density or pressure Trough

Fig. 6.6 Sound propagates as density or pressure variations

6.4.1 Wavelength (λ) The distance between two consecutive compressions or two consecutive rarefactions is termed wavelength, as illustrated in Fig. 6.6(c).

It is typically denoted by λ, and its SI unit is the metre (m).

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6.4.2 Frequency Frequency, denoted by ν (Greek letter nu), represents how often an event occurs. In the context of sound, it refers to the number of compressions or rarefactions passing a fixed point or the number of oscillations it completes per second, and its SI unit is hertz (Hz).

For example, 10 Hz indicates 10 oscillations per second.

Low Frequency

High Frequency

Fig. 6.7 Frequency

6.4.3 Time period The time required for two consecutive compressions or rarefactions or one full oscillation of a sound wave is known as the time period of the sound wave, and its SI unit is the second (s).

Period T

t Time

Fig. 6.8 Time period

Note: The relationship between frequency and time period is f =

1 T

6.4.4 Pitch The pitch of a sound is influenced by the frequency of its sound wave. A higher frequency results in a higher pitch. Consider the comparison between the sound of a baby and that of an adult. Despite having equal loudness, these sounds exhibit differences. This discrepancy becomes apparent when we explore the role of frequency in determining the shrillness or pitch of a sound.

Higher frequency leads to a shrill and higher-pitched sound, while lower frequency produces a lower-pitched sound. For instance, a drum, vibrating with low frequency, emits a low-pitched sound, while a whistle, with a high frequency, produces a higher-pitched sound.

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SOUND

Low frequency - Low pitch

High frequency - High pitch

Fig. 6.9 Pitch

6.4.5 Amplitude The amplitude of the wave, represented by the letter A, and it is defined as the maximum displacement of particles from their average position due to vibrations.

Amplitude

Time Fig. 6.10 Amplitude

6.4.6 Loudness The loudness or softness of a sound is primarily influenced by its amplitude, with large amplitudes producing loud sounds, and small amplitudes resulting in feeble sounds.

When the amplitude is greater, it leads to a larger displacement of particles, resulting in a louder sound. The relationship between the loudness of a sound and its amplitude is direct and proportional to the square of the amplitude. The standard unit for measuring sound loudness is decibels (dB).

146

IL Foundation Series Class 9 Wave disturbance Amplitude Time

Soft sound Wave disturbance Amplitude

Time

Louder sound

Fig. 6.11 Loudness

6.4.7 Quality or timber The quality or timber of sound distinguishes one sound from another with the same pitch and loudness. A sound with a more pleasant quality is described as rich. A sound of a single frequency is termed a tone, while a note is produced by a mixture of several frequencies, creating a pleasant listening experience. In contrast, noise is considered unpleasant to the ear, while music, characterised by its rich quality, is pleasing to hear.

Pure Sound

Noise

Fig. 6.12 Quality

6.4.8 Speed of sound The speed of sound is a measure defined as the distance travelled by a point on a wave, such as a compression or a rarefaction, per unit time.

This can be expressed mathematically using the formula: speed (v = distance/time)

Therefore, v T

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SOUND

Here, λ represents the wavelength of the sound wave, indicating the distance travelled by the sound wave in one time period (T) of the wave.

Thus,

1 v 

T or v

That is, speed = wavelength × frequency Note: It is essential to note that the speed of sound remains nearly constant for all frequencies within a given medium under the same physical conditions. This principle underscores the fact that, regardless of the frequency, sound travels at a consistent speed within a specific medium and set of conditions.

SOLVED EXAMPLES Example 1: The frequency of a sound wave produced by a vibrating body is 30 Hz. Find the 1 wavelength of the sound wave produced. v air 330 ms .

Solution: v f

v 330

11 m f 30

Example 2: If the speed of sound in air is 340 ms −1, calculate the frequency when the wavelength is 1.33 m.

Solution: v f f

v 340

255.6 Hz 1.33

Example 3: A source of wave produces 20 crests and 20 troughs in 0.2 s. The distance between a crest and the next trough is 50 cm. Find the wavelength, frequency and time period of the wave.

Solution: A source of wave produces 20 crests and 20 troughs in 0.2 s. The distance between a crest and the next trough is 50 cm.

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IL Foundation Series Class 9

Since the distance between a crest and the next trough is

λ

, therefore, λ = 50 cm .

2 2

= λ 100 = cm 1 m

If there are 20 crests and 20 troughs in 0.2 seconds, it means that there are 20 complete waves in that time period.

To calculate the frequency, we can use the formula: v = Number of Waves / Time Taken 20 v= 0.2 v = 100 Hz Therefore, the frequency of the wave is 100 Hz, which means it completes 100 cycles or oscillations per second.

The time period ( T ) of the wave is the time it takes to complete one full cycle. It can be calculated as the reciprocal of the frequency: 1 T= v 1 T= 100 T = 0.01 seconds Hence, the wavelength of the wave is λ = 1 m, the frequency of the wave, v = 100 Hz and the

time period of the wave T = 0.01 s.

6.4.9 Speed of sound in different media Sound moves through a medium, like air, at a certain speed. Interestingly, sound doesn’t travel as fast as light. We notice this when we first see lightning (light) and hear the thunder (sound) a bit later. This tells us that sound is slower than light.

Few examples to understand movement of sound through various mediums

When a person hits one end of the table, the resulting sound is almost instantly heard by another person placing their ear on the opposite end of the table. In comparison, if the same sound travels through the air, it takes considerably longer (approximately 14 times more time) to be heard across the same distance, with a notable reduction in intensity. Consequently, it is evident that sound travels more rapidly in solids than in gases or air.

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SOUND

Fig. 6.13 Sound through solid

Now consider another example, when two divers submerged underwater, positioned at a significant separation. If one generates a sound, the other diver perceives it after a specific duration. In a parallel scenario above water, the same distance apart, if one of the individuals produces a sound, the other takes slightly longer to hear it than underwater. This suggests that sound moves more swiftly in liquids than in gases or air.

Now, if the two divers underwater are situated at the ends of a lengthy metal rod and one taps the rod at their end, the other diver notices that they can hear the sound sooner if they place their ear against the rod. This observation indicates that sound travels faster in solids than in liquids.

In conclusion, the speed of sound is highest in solids, less in liquids, and the least in gases or air, i.e., Speed of sound in solids > Speed of sound in liquids > Speed of sound in gases or air.

The speed of sound is also influenced by the various properties of the medium it travels through such as temperature. In general, as we raise the temperature of a medium, the speed of sound in that medium increases. o

For instance, the speed of sound in the air changes with temperature. At 0 C, it’s 331 metres per o second, and at 22 C, it’s a bit faster at 344 metres per second.

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State

Substance

Speed in m/s

Solids

Aluminium

6420

Nickel

6040

Steel

5960

Iron

5950

Brass

4700

Glass (Flint)

3980

IL Foundation Series Class 9

State

Substance

Speed in m/s

Liquids

Water (Sea)

1531

Water (distilled)

1498

Ethanol

1207

Methanol

1103

Hydrogen

1284

Helium

965

Air

346

Oxygen

316

Sulphur dioxide

213

Gases

Table 6.1 Speed of sound in different media at 25 C.

6.5 REFLECTION OF SOUND Sound is like a bouncing ball - when it hits a surface, it bounces back, just like a rubber ball bounces off a wall. Similar to light, sound follows certain rules when it bounces off surfaces.

6.5.1 Laws of reflection of sound •

The incident sound, the reflected sound, and the normal all lie in the same plane.

•

he angle of incidence of incident sound is equal to the angle of reflection formed by the T reflected sound, i.e., i = r.

av

Normal

In

e

ci

av

de

w

nt s

d un

so

ou nd

r

ed ct

w

i

le ef

R

e

Reflecting surface

Fig. 6.14 Laws of reflection of sound

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SOUND

6.5.2 Echo Have you ever shouted or clapped near a big thing like a tall building or a mountain and heard your sound coming back to you?

When a sound repeats itself in a place, we call it an echo. This echo happens because the sound bounces off surfaces and comes back to us. Our brain remembers the sound for a short time, about 0.1 seconds, and for us to hear a clear echo, the time between our shout and the echo must be at least 0.1 seconds.

Sometimes, if the sound bounces more than once, we might hear many echoes, one after the other.

If we assume the speed of sound is 344 metres per second, then the sound needs to travel back to our ears in 0.1 seconds to make the echo. That’s a total distance of at least (344 m/s. × 0.1 s = 34.4 metres). So, to hear a good echo, the object making the echo should be at least half of this distance away, which is 17.2 metres.

ECHO CLAP 17.2 m

Fig. 6.15 Echo

Note: Remember, this distance can change with the air temperature.

Example: A person shouts near a wall and hears an echo after 20 s. What is the distance of the wall from the person? (Speed of sound = 346 ms-1) Solution: Speed of sound, v 346 ms 1 Time taken, t = 20 s Distance travelled by the sound v t 346 20 = 6920 m In the 20 s, the sound had to travel from person to wall and back. Hence, sound travels twice the distance.

6920 Distance between person and wall = = 3460 m.

2 Therefore, the distance of the wall from the person is 3460 m .

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IL Foundation Series Class 9

6.5.3 Reverberation When a sound is made in a large hall, it keeps bouncing off the walls over and over until it becomes so quiet that we can’t hear it anymore. This prolonged bouncing and persistence of sound is known as reverberation.

Places like auditoriums and big halls often face this challenge. To manage the reverberation, the roofs are made from soundproof materials like fibreboard, and the chairs in the halls are covered with fabrics that can soak up sound. This way, the unwanted echoing is reduced, making it more pleasant for everyone in the hall.

ive Reflect

aths

Sound P

Direct Sound Path

Fig. 6.16 Reverberation

6.5.4 Uses of multiple reflections of sound 1. Megaphones and musical instruments like trumpets are designed to direct sound precisely.

Using a tube and conical opening, they reflect sound successively, channelling waves forward for targeted projection. This construction optimises sound projection in a specific direction, as depicted in Fig. 6.17.

Megaphone

Horn

Fig. 6.17 A megaphone and a horn 153

SOUND

2. Stethoscopes, a medical instrument employed for listening to internal body sounds, particularly in the heart or lungs, utilise multiple reflections of sound to transmit the patient’s heartbeat to the doctor’s ears, as demonstrated in Fig. 6.18. This method enhances the precision of sound transmission for medical diagnosis.

Fig. 6.18 Multiple reflections of sound in a stethoscope

3. Ceilings in venues like concert halls, conference halls, and cinema halls are often curved to facilitate the even distribution of sound throughout the space after reflection, as depicted in Fig. 6.19 (a). The curvature of the ceiling aids in ensuring that sound reaches all corners of the hall. Additionally, a curved soundboard may be strategically placed behind the stage in some instances. This soundboard reflects sound, promoting an even spread across the width of the hall, as shown in Fig. 6.19 (b). Such architectural considerations contribute to an enhanced and uniform auditory experience for the audience in large spaces.

Sound board

Source of sound

(a) Curved ceiling(a) of a conference hall

(b)in a big hall (b) Sound board used

Fig. 6.19 Multiple reflections of sound

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IL Foundation Series Class 9

6.6 RANGE OF HEARING In sound production, it is crucial to understand that a vibrating body is necessary. However, not all vibrations can be perceived by the human ear.

The audible range of sound for human beings spans approximately 20 Hz to 20,000 Hz (one Hz equals one cycle per second). Children under the age of five and certain animals, such as dogs, have the ability to hear up to 25 kHz (1 kHz equals 1000 Hz). However, as individuals age, their ears tend to become less sensitive to higher frequencies.

Sounds with frequencies below 20 Hz are classified as infrasonic sound or infrasound. If humans could perceive infrasound, they could hear the vibrations of a pendulum, similar to how they hear the vibrations of a bee’s wings. Rhinoceroses utilise infrasound frequencies as low as 5 Hz for communication. Whales and elephants also produce sounds within the infrasound range.

Interestingly, some animals show signs of disturbance before earthquakes, possibly due to the low frequency infrasound produced before the main shock waves.

Frequencies higher than 20 kHz are referred to as ultrasonic sound or ultrasound. Animals such as dolphins, bats, and porpoises produce ultrasound. Certain moths possess highly sensitive hearing, allowing them to detect the high-frequency squeaks of bats and evade capture. Rats engage in games involving the production of ultrasound.

Infra Sound

Audible Sounds

Ultra Sound

Over 20,000 Hz Below 20 Hz

20 Hz to 20,000 Hz

Fig. 6.20 Range of hearing

Note: For individuals with hearing loss, hearing aids become essential. A hearing aid is an electronic, battery-operated device that receives sound through a microphone. The microphone converts sound waves into electrical signals, which are then amplified by an amplifier. The amplified electrical signals are transmitted to a speaker in the hearing aid, converting them back into sound and delivering them to the ear, facilitating clear hearing for individuals with hearing impairment.

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6.7 APPLICATIONS OF ULTRASOUND Ultrasound waves, characterised by their high frequency, have the ability to travel long distances even in the presence of obstacles along their paths.

1. In industrial applications, ultrasound waves are utilised to clean intricate and hard-to-reach parts of objects, such as spiral tubes or electronic components. Objects are submerged in a solution, and ultrasonic waves are then passed through the solution. This process causes dust particles on the object to detach and fall off.

2. Ultrasound waves play a crucial role in identifying tiny cracks in metallic objects, commonly employed in the construction of large structures, buildings, and scientific equipment. These cracks can compromise the strength of structures and machinery. Ultrasound waves are passed through the metallic objects to detect cracks, and detectors are used to identify waves passing through the cracks. If a crack is present, the ultrasound waves reflect back.

Ultrasound

Detectors

Defect or flaw

Metal Block Fig. 6.21 Ultrasounds can be used to detect cracks

3. Medical applications of ultrasound include Echocardiography, where ultrasound waves are passed through different parts of the heart to create images of the organ.

4. Another medical procedure, Ultrasonography, involves passing ultrasound waves through internal organs to obtain images. This enables doctors to diagnose diseases or abnormalities in the organs. Ultrasound waves travel through body tissues, reflecting back when encountering changes in tissue density. The reflected waves are then converted into electrical signals, forming images of the internal organs.

5. Additionally, ultrasound waves are employed in medical procedures to break down kidney stones, providing a non-invasive method for dealing with this medical condition.

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6.7.1 SONAR – sound navigation and ranging What is SONAR?

The SONAR, an acronym for Sound Navigation and Ranging, serves the purpose of determining the distance, direction, and speed of objects submerged underwater, utilising ultrasonic waves.

It comprises two main components: the transmitter and the detector (or receiver). The sonar operates by producing and transmitting ultrasonic waves into the water. As these waves travel through the water, they encounter objects, and upon hitting an object, they reflect back to the detector. The detector then converts these reflected sound waves into electrical signals, which are subsequently interpreted.

Boat (or ship)

Water surface

Detector Transmitter

Sea bed

Fig. 6.22 SONAR

To calculate the distance of the object, the sonar relies on the speed of sound in water and the time taken for the waves to reach the detector. This technique is known as echo ranging.

Applications of SONAR

1. Determining Water Depth: Sonar is frequently used to find the depth of water bodies, such as seas and oceans.

2. Detecting Underwater Objects: Sonar is instrumental in detecting the presence of underwater objects, including submarines, hills, icebergs, and ships. This capability is particularly crucial for navigation, defence, and safety purposes.

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SOLVED EXAMPLES Example 1: The time taken by ultrasonic sound to reach a sonar receiver is 3 seconds. What is the depth of the sea in this region? (Speed of sound in water = 1500 m/s).

Solution: Given: Time, t = 3 s Speed of sound in water given in the problem, v w = 1500 m / s . It is known that the distance between the transmission and receiver point is 2 d.

2 d 1500 3 4500 Therefore, d = 2 d v t 2 2substituting d 1500 3 the values, we get: On d 2250 m 4500 d= 2 d 2250 m So, the depth of the sea is 2250 m.

Example 2: A man standing in front of a vertical cliff fires a gun. He hears the echo after 3 s. On moving closer to the cliff by 82.5 m , he fires again and hears the echo after 2.5 s. Find a. The distance of the cliff from the initial position of man, b. The speed of sound.

Solution: a. Let the distance of the cliff from the initial position of man be d m and the speed of sound be V ms −1 . t For the first echo,=

2d = 3 s (given)....(i) V

On moving closer to the cliff by 82.5 m , the distance of the cliff from the new position becomes

d - 82.5 m , then for a second echo: 2 ( d − 82.5 ) = 2.5 s (given) ………(ii) V Dividing eqn. (i) by eqn. (ii), we get:

= t

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IL Foundation Series Class 9

d 3 6 = = d − 82.5 2.5 5 6 d − 495 = 5d d = 495 m 2d t Substituting the value of d = 495 m , and t = 3 s ,

b. From eqn. (i), V =

= V

2 × 495 = 330 ms −1 3

6.8 THE HUMAN EAR Sounds are generated as waves in the air or any medium they travel through, and our ears play a crucial role in receiving audible frequencies from our surroundings and translating them into electrical signals, which are then transmitted through a specialised nerve called the auditory nerve to our brain. The brain interprets these signals and responds accordingly.

6.8.1 Structure of the human ear The ear is comprised of several key components: 1. Pinna: The outer part of the ear gathers sound from the environment.

2. Auditory Canal: Collected sound from the surroundings passes through the auditory Canal.

3. Eardrum or Tympanic Membrane: Located at the end of the auditory canal, the eardrum responds to compressions and rarefactions in sound waves. It moves inwards due to increased pressure (compression) and outwards due to decreased pressure (rarefaction), resulting in vibrations that correspond to the incoming sound wave.

4. Middle Ear: This section includes ossicles, the three bones known as the hammer, anvil, and stirrup. These bones serve to amplify the vibrations produced by the eardrum. The amplified vibrations are then transmitted to the inner ear.

5. Cochlea: Situated in the inner ear, the cochlea plays a crucial role in converting the amplified vibrations into electrical signals. These signals are then carried to the brain through the auditory nerve.

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Pinna Hammer Anvil Stirrup Oval window Auditory nerve Cochlea Tympanic membrane or eardrum Auditory canal Eustachian tube Outer ear

Middle Inner ear ear

Fig. 6.23 Structure of human ear

QUICK REVIEW

160

•

Sound is produced when objects vibrate, which involves a back-and-forth motion.

•

he substance or material through which sound travels is referred to as a medium and this T medium can be in the form of a solid, liquid, or gas.

•

wave is described as a disturbance that travels through a medium, initiated by the particles A of the medium setting neighbouring particles into motion.

•

hen the vibrating object moves forward, it pushes and compresses the air in front of it, W creating a region of high pressure known as compression (C).

•

hen the vibrating object moves backwards, it generates a region of low pressure called W rarefaction (R).

•

aves in a medium can be classified into two main types: longitudinal waves and transverse W waves.

•

In a longitudinal wave, the particles of the medium move parallel to the direction of the wave.

•

I n a transverse wave, particles of the medium oscillate up and down about their mean position as the wave travels.

•

he distance between two consecutive compressions or two consecutive rarefactions is termed T wavelength.

•

requency refers to the number of compressions or rarefactions passing a fixed point or the F number of oscillations it completes per second, and its SI unit is the hertz (Hz).

IL Foundation Series Class 9

•

he time required for two consecutive compressions or rarefactions or one full oscillation of a T sound wave is known as the time period of the sound wave, and its SI unit is the second (s).

•

he pitch of a sound is influenced by the frequency of its sound wave. A higher frequency T results in a higher pitch.

•

he amplitude of the wave, represented by the letter A, is defined as the maximum T displacement of particles from their average position due to vibrations.

•

he loudness or softness of a sound is primarily influenced by its amplitude, with large T amplitudes producing loud sounds and small amplitudes resulting in feeble sounds.

•

he quality or timber of sound distinguishes one sound from another with the same pitch and T loudness. A sound with a more pleasant quality is described as rich.

•

he speed of sound is a measure defined as the distance travelled by a point on a wave, such as T a compression or a rarefaction, per unit time. Speed of sound, v .

•

Speed of sound in solids > Speed of sound in liquids > Speed of sound in gases or air.

•

Sound is like a bouncing ball - when it hits a surface, it bounces back.

•

The incident sound, the reflected sound, and the normal, all lie in the same plane.

•

he angle of incidence of incident sound is equal to the angle of reflection formed by the T reflected sound, i.e., i = r.

•

When a sound repeats itself in a place, we call it an echo.

•

hen a sound is made in a large hall, it keeps bouncing off the walls over and over until it W becomes so quiet that we can’t hear it anymore. This prolonged bouncing and persistence of sound is known as reverberation.

•

ses of multiple reflections of sound - Megaphones, Stethoscopes, and Ceilings in venues like U concert halls.

•

The audible range of sound for human beings spans approximately 20 Hz to 20,000 Hz.

•

Sounds with frequencies below 20 Hz are classified as infrasonic sound or infrasound.

•

Frequencies higher than 20 kHz are referred to as ultrasonic sound or ultrasound.

•

he SONAR, an acronym for Sound Navigation and Ranging, serves the purpose of determining T the distance, direction, and speed of objects submerged underwater, utilizing ultrasonic waves.

•

Parts of the human ear: Pinna, Auditory Canal, Eardrum, Middle Ear, and Cochlea.

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WORKSHEET - 1 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER I.

Production and propagation of sound

1. The matter or substance through which sound is transmitted is called a __________. a. Object

b. Substance

c. Medium

d. Wood

2. The device used in the lab to generate sound with a fixed frequency is _____________. a. Simple pendulum

b. Tuning fork

c. Thermometer

d. Meter scale

3. Which one of the following processes is responsible for sound production? a. Heating

b. Vibration

c. Oscillation

d. Rotation

4. We can produce sound by _________ different objects. a. Plucking

b. Scratching

c. Shaking

d. All of the above

5. Sound cannot travel through ____________. a. Solid

b. Liquid

c. Vacuum

d. Air

6. In the bell jar experiment, the sound gradually grew fainter after the vacuum pump was turned on until we could only hear a feeble sound. Why? a. The Vacuum pump absorbs the sound b. The medium is sucked by the pump c. The bell jar behaves as an isolated medium d. The bell does not produce such a strong sound as it was produced earlier.

7. A sound wave consists of: a. A number of compression pulses one after the other. b. A number of rarefactions pulse one after the other. c. Compression and rarefaction pulses one after the other. d. Compression and rarefaction are usually separated by a distance equal to one wavelength.

II. Types and characteristics of a sound wave

1. Which of the following terms refers to the region of low pressure created when a vibrating object moves backwards? a. Phase

162

b. Compression

c. Rarefaction

d. Wavelength

IL Foundation Series Class 9

2. Which type of waves is produced when a stone is dropped on the surface of the water in a pond? a. Longitudinal waves

b. Transverse waves

c. Electromagnetic waves

d. Surface waves

3. Sound waves are called a. Electrical waves

b. Matter waves

c. Electromagnetic waves

d. Mechanical waves

4. The distance between two consecutive compressions or rarefactions is called the ___________. a. Displacement

b. Frequency

c. Time period

d. Wavelength

5. Which of the following statements accurately describes the relationship between frequency and pitch? a. Lower frequency corresponds to a higher pitch b. Higher frequency always corresponds to a higher pitch c. Frequency and pitch are unrelated concepts d. Higher frequency always corresponds to a lower pitch 6. Which of the following statements accurately describes the relationship between the loudness of sound waves and their amplitude? a. The loudness of sound waves is unrelated to their amplitude b. The loudness of sound waves is inversely proportional to their amplitude c. The loudness of sound waves is directly proportional to their amplitude d. The loudness of sound waves is determined by their frequency, not their amplitude 7. The loudness of sound is measured in units of _______________. a. hertz (Hz)

b. metre (m)

c. decibels (dB)

d. (m/s)

8. A pendulum oscillates 70 times in 5 seconds. Find its time period. a. 0.1 s

b. 0.051 s

c. 0.071 s

d. 0.035 s

9. If pendulum A makes 15 oscillations in 5 seconds, and pendulum B makes 12 oscillations in 3 seconds. Which pendulum has a higher frequency, A or B? a. Pendulum A

b. Pendulum B

c. Pendulums A and B have equal frequencies d. Insufficient information 10. Which of the following terms refers to the number of compressions or rarefactions per unit of time? a. Wavelength

b. Frequency

c. Velocity

d. Amplitude

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11. A wave on a string is moving with a speed of 570 m/s and a wavelength of 1.3 m. Find the frequency of the wave. a. 458.66 Hz

b. 468.56 Hz

c. 408.26 Hz

d. 438.46 Hz

12. In which of the following mediums sound travels the fastest? a. Vacuum

b. Air

c. Water

d. Steel

13. Sound travels: a. Slower in warm air than in cold air b. Faster in solids and liquids than in air c. Slower in water than air

d. Slower in solids and liquids than in air

14. Lightning can be seen at the moment when it occurs. Paheli observes lightning in her area.

She hears the sound 5s after the observed lightning. How far is she from the place where lightning occurs? (Speed of sound = 330 m/s) a. 16.5 km

b. 1.65 km

c. 165 km

d. 1.065 km

15. An object completes 50 vibrations in 5 sec when some waves pass through the surface of the water. If the wavelength of the wave is 20 cm, then find the velocity of the waves. a. 220 cm/s

b. 180 cm/s

c. 300 cm/s

d. 200 cm/s

c. i < r

d. i not equal to r

III. Reflection of sound

1. For the reflection of sound waves, a. i = r

b. i > r

2. An echo is returned in 6 seconds. What is the distance of the reflecting surface from the source? [Speed of sound is 342 m/s]. a. 1026 m

b. 1006 m

c. 1076 m

d. 1016 m

3. The speed of sound in water is 1500m/s. If the sound produced by an instrument in the sea is heard in the minimum time of echo, the depth of the sea is a. 150 m

b. 75 m

c. 55 m

d. 125 m

4. Which of the following options shown below does not follow the application of reflection of sound? a. Stethoscope

b. Tape recorder

c. Megaphone

d. Speaker

5. Which of the following statements about concert hall ceilings is true? a. The ceilings of concert halls are flat b. Curved concert hall ceilings ensure uniform sound distribution c. Sound reflection is not affected by the shape of the ceiling d. Curved concert hall ceilings prevent sound from reaching the audience

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IL Foundation Series Class 9

IV. Range of hearing

1. An object vibrates with a frequency of 10 Hz. Which of the following is true? a. It produces a sound that we can hear b. It does not produce sound c. It produces a sound that we cannot hear d. It produces sound that we can hear if we strain our ears 2. Which frequency corresponds to ultrasonic waves: a. 15000 Hz

b. 30000 Hz

c. 2500 Hz

d. 10 Hz

3. Which sound can be heard by humans? a. The sound produced by rhinoceroses to communicate b. Infrasound c. Ultrasounds d. Car Siren V. Applications of ultrasound and sonar

1. Ultrasound is a technique used to clean hard-to-reach places. Which of the following statements best describes how ultrasound achieves this? a. Ultrasound waves generate heat to melt and dissolve dirt. b. Ultrasound waves create powerful air currents to blow away dirt. c. Ultrasound waves generate high-frequency sound waves to dislodge dirt particles. d. Ultrasound waves emit ultraviolet light to kill bacteria and remove dirt 2. In SONAR, we use: a. Radio waves

b. Audible sound waves

c. Ultrasonic waves

d. Infrasonic waves

3. A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02s. If the speed of sound in salt water is 1531 m/s, then how far away is the cliff from the submarine? a. 740.81 m

b. 700.81 m

c. 780.81 m

d. 730.81 m

4. A sonar device on a submarine sends out a signal and receives an echo 5 seconds later.

Calculate the speed of sound in water if the distance of the object from the submarine is 3625m: a. 1450 ms −1

b. 1650 ms −1

c. 1350 ms −1

d. 967 ms −1

5. A submarine sends out an ultrasound that returns from the seabed and is detected after 4.52 s. If the speed of ultrasound through seawater is 1470 m/s, what is the distance of the seabed from the submarine? a. 2233 m

b. 2844 m

c. 3322 m

d. 3422 m

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SOUND

6. A ship sends an ultrasound that returns from the seabed and is detected after 2.42 s. If the speed of ultrasound through seawater is 1531 m/s, what is the approximate distance of the seabed from the ship (in km)? a. 1.85 km

b. 1.42 km

c. 2.21 km

d. 1.65 km

VI. The human ear

1. The visible ear is a flap of tissue. It is called the _______. a. Eardrum

b. Pinna

c. Cochlea

d. Ear canal

2. The cochlea is a part of the inner ear which converts sound vibrations into __________________. a. Light signals

b. Mechanical signals

c. Heat signals

d. Electrical signals

3. Which part of the ear amplifies sound waves in the middle ear? a. Cochlea

b. Vestibular system

c. Tympanic membrane

d. Ossicles

4. What is the primary function of the tympanic membrane? a. Transmit sound waves to the cochlea b. Protect the inner ear from foreign objects c. Amplify sound waves in the middle ear d. Convert sound vibrations into electrical signals 5. Middle ear consists of: a. Three small bones

b. Two small bones

c. Four small bones

c. No bones

6. In which part of the ear sound is magnified? a. Outer

b. Middle

c. Inner

d. Both the outer and middle

7. What is another name for the tympanic membrane? a. Pinna

d. Cochlea

c. Eardrum

d. Meatus

WORKSHEET - 2 MULTIPLE CHOICE QUESTIONS WITH SINGLE CORRECT ANSWER 1. Sound waves do not travel through _____. a. Solids

b. Liquids

c. Gases

d. Vacuum

2. Which of the following will remain unchanged when a sound wave travels in air or in water a. Amplitude 166

b. Wavelength

c. Frequency

d. Speed

IL Foundation Series Class 9

3. The outer ear is called: a. Pinna

b. Malleus

c. Incus

d. Stapes

4. The frequency which is not audible to the human ear is : a. 50 Hz

b. 500 Hz

c. 5000 Hz

d. 50, 000 Hz

5. On a slinky, we can produce: a. Transverse waves only

b. Longitudinal waves only

c. Both transverse and longitudinal

d. Neither transverse nor longitudinal waves.

6. The walls of a hall built for musical concerts should: a. Amplify sound

b. Reflect sound

c. Transmit sound

d. Absorb sound

7. In an orchestra, the musical sounds of different instruments are distinguished from one another by the characteristics of: a. Loudness

b. Pitch

c. Quality or timbre

d. All of these

8. The bells of a temple are made of large size. It is for: a. Producing sound of high pitch

b. Producing loud sounds

c. Producing sound of high quality

d. Enhancing the beauty

9. The loudness of a sound depends upon: a. amplitude

b. Pitch

c. Velocity

d. Wavelength

10. Note is a sound that is a: a. Mixture of several frequencies

b. Mixture of two frequencies only

c. Single frequency

d. Always unpleasant to listen

11. A key of a mechanical piano struck gently and then struck again, but much harder this time.

In the second case,: 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 the loudness and pitch will remain unaffected.

12. In SONAR, we use: a. > 20000 Hz

b. < 20 Hz

c. = 20000 Hz

d. 20-20000 Hz

13. Sound travels in the air if: a. Particles of medium travel from one place to another b. There is no moisture in the atmosphere c. Disturbance moves d. Both particles and disturbance travel from one place to another.

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SOUND

14. When we change feeble sound to loud sound, we increase its: a. Frequency

b. Amplitude

c. Velocity

d. Wavelength

15. In the curve, half the wavelength is :

A

a. AB

B

D

C

b. BD

E

c. DE

d. AE

16. Earthquake produces which kind of sound before the main shock wave begins: a. Ultrasound

b. Infrasound

c. Audible sound

d. None of the above

c. Rhinoceros

d. Human beings

c. Vacuum

d. Oxygen

c. Gases

d. Vacuum

17. Infrasound can be heard by: a. Cat

b. Bat

18. Sound cannot travel through a. Air

b. Wood

19. Speed of sound is more in a. Solids

b. Liquids

20. Sound waves are waves a. Transverse-mechanical waves

b. Transverse-non mechanical waves

c. Longitudinal-mechanical waves

d. Longitudinal-non mechanical waves

21. Stethoscope works on the principle of a. Multiple reflection

b. Diffraction

c. Both a and b

d. Resonance

c. N

d. None

λ 4

d. 2λ

22. The SI unit of loudness of sound a. Hz

b. dB

23. The distance between two adjacent crests is λ a. b. λ 2 24. Bats make use of

c.

a. Infrasonic sound

c. Subsonic

d. None

c. Liquids

d. Wood

b. Ultrasonic sound

25. Velocity of sound is minimum in a. Steel

b. Air

26. The audible range of hearing for average human beings is:

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IL Foundation Series Class 9

a. 20 Hz − 20000 Hz

b. 20kHz − 40kHz

c. 10kHz − 20kHz

d. 20kHz − 30kHz

27. If the time period is 20 s and the wavelength is 20 m , then velocity V = a. 2 ms −1

b. 3 ms −1

c. 1 ms −1

d. 5 ms −1

28. The speed of sound depends on: a. Temperature of medium

b. Nature of medium

c. Elasticity of medium

d. All the above

29. Sound waves can: a. Reflect

b. Diffract

c. Only reflect

30. Persistence of the human ear is: 1 1 1 s s a. b. s c.

100 10 50 31. Which of the following waves is a longitudinal wave: a. Radio waves

b. Sound waves

c. Waves on the surface of water

d. Light waves

d. Both a & b

d.

1 s 20

32. If the temperature of the air is increased, then the velocity of sound: a. Decreases

b. Increases

c. Remains the same

d. May decrease or increase

33. If the distance between a crest and the next trough is 0.5 m, what is the wavelength? a. 1 m

b. 0.25 m

c. 0.5 m

d. 2 m

34. Time period of a wave is 0.5 sec and its speed is 100 m / s. What is the wavelength of the wave? a. 200 m

b. 100 m

c. 50 m

d. 5 m

35. Audible range for human beings is: a. Above 20 kHz

b. Below 20 Hz

c. Between 20 Hz to 20 kHz

d. None

36. When we say 'sound travels in the medium’, we mean: a. The particles of medium travel

b. The source travels

c. The disturbance travels

d. The medium travels

37. A sound wave consists of: a. A number of compression pulses, one after the other b. A number of rarefaction pulse one after the other c. Compression and rarefaction pulses one after the other

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SOUND

d. A compression and a rarefaction are usually separated by a distance equal to one wavelength 38. The time period of a sound wave travelling in a medium is T. At a given instance t 0 a particular region in the medium has minimum density. The density of the region will be minimum again at: a. t = T

b. t = T / 2

c. t = T / 3

d. t = T / 4

b. second-1

c. metre

d. metre-1

39. Hertz stands for: a. second

40. The frequency of a source is 20kHz. The frequencies of the sound waves produced by it in water and air will: a. Be the same as that of the source b. Depends on the velocity of the waves in these media c. Depends on the wavelength of the waves in these media d. Depends on the density of the media 41. If the density of air at a point through which a sound wave is passing is maximum at an instant, the pressure at that point will be: a. Minimum

b. Same as the density of air

c. Equal to the atmospheric pressure

d. Maximum

42. The properties of ultrasound that make it useful are: a. High power and high speed b. Good directionality and high power c. High speed and frequency d. Good directionality and ability to move around objects 43. Ultrasonic waves are used for detecting objects underwater. What technique/device is used for this? a. Ultrasonography

b. Echocardiography

c. Phacoemulsification

d. Sonar

44. A tuning fork of frequency 384 Hz produces a lower note than one of frequency: a. 256 Hz

b. 512 Hz

c. 288 Hz

d. 320 Hz

45. The speed of sound in air at 0 C is approximately: a. 332 ms −1

b. 1450 ms −1

c. 5100 ms −1

d. 3 108 ms 1

46. If the frequency of a wave is 330 Hz, and the velocity of the wave in air is 330 ms −1, then the wavelength of the wave is:

170

IL Foundation Series Class 9

a. 0.5 m

b. 1 m

c. 2 m

d. Non-certain

47. The minimum distance required for producing echo is a. 1.7 m

b. 17 m

c. 10 m

d. 300 m

48. Shyamal calculated the velocity of wave using a slinky. He asked his teacher regarding the features of spring to be used. The teacher replied that the spring should be: a. Long, soft, and flexible

b. Short, soft, and flexible

c. Short, hard, and flexible

d. Long, soft, but not flexible

49. Factors affecting the velocity of sound are: a. Density of the medium

b. Amplitude of sound waves

c. Temperature of the medium

d. Both a and c

50. In an experiment on studying the laws of reflection of sound, the tube facing the clock is placed as shown. The position of the second tube, at which the ear will get the best-reflected sound, is obtained when q equals:

40°

Clock

a. 20

b. 30

θ

Ear

c. 40

d. 50

51. Assertion (A): Compression is a region of high pressure.

Reason (R): When the vibrating particles come closer to one another than they normally are, there is a momentary reduction in volume and a compression is formed. a. Both A and R are correct, and R is the correct explanation of A. b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, but R is incorrect. d. A is incorrect, but R is correct.

52. Assertion (A): The distance between the centre of compression and an adjacent rarefaction is equal to half the wavelength Reason (R): The minimum distance in which a sound wave repeats itself is called its wave length. a. Both A and R are correct, and R is the correct explanation of A.

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SOUND

b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, R is incorrect. d. A is incorrect, R is correct.

53. Assertion (A): The frequency of a wave is the reciprocal of its time - period.

Reason (R): 1 Hertz is equal to 1 vibration per second. a. Both A and R are correct, and R is the correct explanation of A. b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, R is incorrect. d. A is incorrect, R is correct.

54. Assertion (A): The sound of our voice is produced by the vibrations of two vocal cords in our throat.

Reason (R): The energy required to make an object vibrate and produce sound is provided by some outside source. a. Both A and R are correct and R is the correct explanation of A b. Both A and R are correct but R is not the correct explanation of A c. A is correct, R is incorrect d. A is incorrect, R is correct 55. Assertion (A): We cannot talk to one another directly on the moon as we do on Earth, though we may be very close.

Reason (R): Sound cannot travel through a vacuum. a. Both A and R are correct, and R is the correct explanation of A. b. Both A and R are correct, but R is not the correct explanation of A. c. A is correct, but R is incorrect. d. A is incorrect, but R is correct.

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ANSWER KEY 1: UNITS AND MEASUREMENT Worksheet 1 I. The international system of units 1. a 2. c 3. b 4. c 5. d 6. c 7. b 8. a 9. b 10. a 11. a 12. d 13. a 14. a 15. d 16. b II. Dimensions of physical quantities, formulae, equations and analysis 1. d 2. c 3. c 4. c 5. a 6. d 7. b 8. c 9. c 10. b 11. a 12. d 13. b Worksheet 2 1. b 2. c 3. c 6. d 7. b 8. d 11. b 12. b 13. c 16. c 17. a 18. a 21. d 22. c 23. c 26. c 27. a 28. a

4. d 9. c 14. b 19. c 24. d 29. a

5. d 10. b 15. a 20. c 25. a 30. b

2: MOTION Worksheet 1 I. Motion and rate of change of velocity 1. d 2. a 3. c 4. c 5. b 6. b 7. b 8. b 9. b 10. b 11. c 12. a II. Graphical representation of motion 1. a 2. a 3. c 4. b 5. c 6. d 7. c 8. a III. Equations of motion 1. c 2. c 3. a 4. a 5. a 6. d 7. a IV. Circular motion 1. d 2. b 3. c 4. d 5. c 6. c 7. c 8. c 9. a Worksheet 2 1. d 2. a 3. c 6. b 7. c 8. a

4. c 9.d

5. a 10. c

11.a 12. d 13. b 14. c 15. b 16. a 17. b 18. c 19. b 20. a 21. b 22. d 23. c 24. d 25. a 26. c 27. d 28. a 29. c 30. a 3: FORCE AND LAWS OF MOTION Worksheet 1 I. Types of forces and Newton's laws of motion 1. c 2. d 3. b 4. c 5. c 6. a 7. b 8. c 9. b 10. c 11. a 12. d 13. b 14. c 15. a 16. a 17. a II. Second law of motion 1. b 2. a 3. c 4.c 5. a 6. c 7. d 8. b 9. a 10. a 11. b 12. c 13. c 14. a 15. b 16. b 17. a 18. c III. Third law of motion 1. c 2. a 3. d 4. c 5. d 6. b 7. a 8. b 9. d Worksheet 2 1. b 2. c 3. c 4. a 6. d 7. b 8. b 9. c 11.b 12. c 13.d 14. c 16. b 17. c 18. a 19. b 21. c 22. d 23. a 24. a 26. c 27. a 28.a 29. d 31. b 32. d 33. a 34. c 36. d 37. c 38. c 39. c 41. c 42. c 43. b 44. c 4: GRAVITATION Worksheet 1 I. Gravitation 1. c 2. d 3. d 6. c 7. a 8. d 11. b 12. b 13. b 16. b 17. c 18. b

5.a 10. b 15. c 20. d 25. a 30. c 35. d 40. c 45. c

4. a 5. d 9. a 10. c 14. b 15. b 19. c

173

ANSWER KEY II. Acceleration due to gravity 1. c 2. c 3. c 4. d 5.d 6. a 7. d 8. a 9. d 10. d 11. c 12. c 13. d 14. a III. Motion of objects 1. a 2. a 3. b 4. b 5. a 6. a 7. b 8. a 9. b 10. c 11. d 12. d 13. c 14. c 15. b 16. c 17. b 18. a 19. c 20. c 21. a 22. d 23. b IV. Mass and weight 1. c 2. b 3. b V. Pressure, thrust, and pressure exerted by liquids and gases 1. c 2. d 3. a 4. a 5. c 6. d 7. d 8. b 9. a 10. b 11. c VI. Buoyancy 1. b 2. c 3. b 4. d 5. a 6. b 7. b 8. a 9. a 10. a 11. b 12. c 13. c 14. d 15. a 16. a 17. c 18. c 19. d 20. a 21. c 22. d 23. c 24. d 25. c 26. c 27. b 28. d 29. c 30. c 31. b 32. d Worksheet 2 1. d 2. b 3. b 4. c 5. b 6. a 7. c 8. a 9. c 10. a 11. b 12. a 13. d 14. a 15. b 16. c 17.b 18. b 19. c 20. b 21. b 22. a 23. b 24. a 25. a 26. a 27. b 28. d 29. d 30. a 31. a 32. a 33. b 34. b 35. c 36. a 37. a 38. c 39. b 40. b 41. a 42. d 43. c 44. b 45. a 46. a 47. d 48. b 49. b 50. d 51. d 52. d 53. a 54. b 55. d

174

5. WORK AND ENERGY Worksheet 1 I. Work 1. c 2. d 3. c 4. d 5. a 6. c 7. a 8. d II. Energy 1. a 2. b 3. b 4. d 5. a 6. b 7. a 8. a 9. a 10. b 11. c 12. a 13. b III. The work-energy theorem 1. b 2. a 3. c 4. d 5. a 6. b IV. Conservation of energy 1. c 2. c 3. d 4. a 5. c V. Power 1. a 2. d 3. a 4. c 5. b VI. Commercial unit of energy 1. d 2. d 3. d 4. a 5. b Worksheet 2 1. c 2. d 3. a 6. d 7. c 8. c 11. a 12. b 13. c 16. c 17. b 18. d 21. a 22. c 23. c 26. c 27. b 28. b

4. a 9. c 14. c 19. d 24. b 29. b

5. a 10. d 15. c 20. a 25. b 30. c

6: SOUND Worksheet 1 I. Production and propagation of sound 1. c 2. b 3. b 4. d 5. c 6. b 7. c II. Types and characteristics of a sound wave 1. c 2. b 3. d 4. d 5. b 6. c 7. c 8. c 9. b 10. b 11. d 12. d 13. b 14. b 15. d III. Reflection of sound 1. a 2. a 3. b 4. b 5. b IV. Range of hearing 1.c 2. b 3. d V. Applications of ultrasound and sonar

ANSWER KEY 1. c 2. c 3. c 6. a VI. The human ear 1. b 2. d 3. d 6. b 7. c Worksheet 2 1. d 2. c 3. a 6. d 7. c 8. b 11. a 12. a 13. c 16. b 17. c 18. c 21. a 22. b 23. b 26. a 27. c 28. d 31. b 32. b 33. a 36. c 37. c 38. a 41. d 42. b 43. d 46. b 47. b 48. a 51. a 52. a 53. b

4. a

5. c

4. b

5. a

4. d 9. a 14. b 19. a 24. b 29. d 34. c 39. b 44. b 49. d 54. a

5. c 10. a 15. b 20. c 25. d 30. b 35. c 40. a 45. a 50. c 55. a

175