Volume 25 Managing Editor Mahabir Singh Editor Anil Ahlawat

No. 12

December 2017

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CONTENTS

Class 11 NEET | JEE Essentials

8

Exam Prep

23

Ace Your Way CBSE : Series 6

29

MPP-8

38

Brain Map

42

Class 12 Brain Map

43

NEET | JEE Essentials

44

Exam Prep

60

Ace Your Way CBSE : Series 7

67

MPP-8

75

Competition Edge Olympiad Problems

79

Physics Musing Problem Set 53

82

JEE Work Outs

84

Physics Musing Solution Set 52

88

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PHYSICS FOR YOU | DECEMBER ‘17

7

Unit

6

PROPERTIES OF BULK MATTER i.e., Stress v Strain Stress = K Strain

=5381>93C?6C?<94C Elasticity The property of the material to recover its original shape or size on the removal of deforming force is known as elasticity. x If a body completely regains its original form after removal of external force, it is called a perfectly elastic body. x If the body remains deformed and shows no tendency to recover its original condition on removal of external forces, is said to be perfectly plastic. x Strain : The ratio of change in length, volume or shape of body to its original configuration is called strain. Strain may be longitudinal, volume or shearing. As it is the ratio of two similar quantities, it is a pure number and has no unit. x Stress : The restoring force developed per unit area in the body when subjected to deforming force is Restoring force = F/A called stress, i.e., Stress = area –2 SI unit is N m The dimensional formula for stress is [ML–1 T –2]. Hooke’s Law and Stress-Strain Curve x Hooke’s Law : For small deformations the stress developed in the body is directly proportional to the strain of the body. 8

PHYSICS FOR YOU | DECEMBER ‘17

K=

x

Stress where, K = modulus of elasticity Strain

Modulus of Elasticity (K) : The ratio of stress developed to the strain produced in the body is called modulus of elasticity. Stress K= Strain Its SI unit is N m–2 and dimension is [ML–1T –2] Stress-strain curve The given graph is a tensile simple case of metallic wire which is subjected to increasing tension. O ¾ From O to A, the curve is linear and body can regain its original shape after removing stress i.e., Hooke’s Law is obeyed. ¾ From A to B, stress is not proportional to strain but the body still can regain its original shape. ¾ Point B, is the yield point (or elastic limit) on further increasing stress beyond the point B, the body will have a permanent set after removing the stress.

PHYSICS FOR YOU | DECEMBER â€˜17

9

¾ ¾

From B to C and further upto D strain changes rapidly for smaller stress. E is the fracture point where the body breaks on applying further strain.

Elastic Potential Energy 1 x Energy density = × stress × strain 2 1 = × Y × (strain)2. 2 1 x Workdone = × stress × strain × volume 2 (stress)2 x Workdone per unit volume = . 2Y x For elastic constants : inter relations Y = Young’s modulus, K = Rigidity modulus B = Bulk modulus, V = Poisson’s ratio ¾ Y = 2K(1 + V) ¾ Y = 3B(1 – 2V) 9B η 9 3 1 . = + or Y = ¾ η + 3B Y η B 3B − 2η σ= ¾ 6B + 2η x Applications ¾ Designing a beam for construction of roofs l and bridges. b = breadth of slab Y = Young’s modulus of slab d = thickness of slab l = length Wl 3 The depression in rectangular beam, δ = 4Y bd 3 or G v(l)3 ?To minimize the depression in the beam or slab short length slabs are used. For a beam with circular cross-section of radius r depression is, δ =

¾

12πr 4Y A hollow shaft is stronger than a solid shaft of same mass, length and material. Torque required to produce a unit twist in solid πηr 4 shaft (bar) τ solid = 2L and torque for hollow shaft, τ hollow =

10

Wl 3

(

πη r2 4 − r14 2L

)

PHYSICS FOR YOU | DECEMBER ‘17

∴ or

τ solid r4 = 4 τ hollow r − r 4 2 1

(

)(

r22 + r12 r22 − r12 τ hollow r2 4 − r14 = = τ solid r4 r4

τ hollow r22 + r12 ≅ >1 τ solid r2

{'

(

)

r 2 ≅ r22 − r12

)}

Illustration 1 : A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm. Then what is the elastic energy stored in the wire ? Soln.: Elastic potential energy 1 1 F ΔL = × stress × strain × volume = × × × (AL) 2 2 A L 1 1 = F ΔL = × 200 × 10 −3 = 0.1 J 2 2

8I4B?CD1D93C*6<E94C1DB5CD Pressure Due to Liquid x The thrust exerted by a liquid at rest per unit area of the substance in contact with the liquid is called pressure. If F is the thrust exerted by a liquid on a surface of small area 'A, then pressure is given by P = lim (F / ΔA) ΔA→0

x

x

x

x x x

The unit of pressure is dyne cm–2 in CGS system and N m–2 in SI. A pressure of one N m–2 is also called pascal. Pascal’s law : Pressure applied to enclosed liquid is transmitted equally in all directions, to every position of liquid and wall of container. Hydrostatic pressure of liquid column : Pressure = Ugh where U = density of liquid, h = height of liquid column. Brahma’s hydraulic press is based upon Pascal’s law of liquid pressure. Unit of pressure is pascal. Its symbol is Pa. Bar = 105 Pa, torr = 1 mm of Hg column. Density and relative density : Density = mass/volume. For water, density = 103 kg m–3. 3 –2 3 ¾ One litre = 1000 cc = 1000 cm = 1000 × (10 m) –3 3 = 10 m . Density of substance ¾ Relative density = Density of water at 4 °C

PHYSICS FOR YOU | DECEMBER â€˜17

11

¾

Weight of substance in air = Loss of weight in water m + m2 (m1 + m2 )ρ1ρ2 = Density of mixture = 1 V1 + V2 (m1ρ2 + m2ρ1) 2ρ ρ 2 1 1 If m1 = m2 = m, ρ = 1 2 or = + ρ1 + ρ2 ρ ρ1 ρ2

? Density of mixture of two liquids is harmonic mean of the two densities. If V1 = V2 = V, m + m2 ρ V + ρ2V ρ1 + ρ2 ρ= 1 = = 1 . 2 V1 + V2 V +V

x

? Density of mixture of two liquids is arithmetic mean of two densities. ¾ Density of liquid varies with pressure. ⎡ ΔP ⎤ ρ = ρ0 ⎢1 + , where 'P = change in K ⎥⎦ ⎣ pressure, K = bulk modulus of elasticity of liquid. ¾ Relative density, also known as specific gravity, has no unit, no dimension. Laws of floatation : Weight of floating body = weight of liquid displaced. ¾ Volume of body immersed = volume of liquid displaced. ¾ When a solid body is immersed, partly or wholly, in a liquid at rest, it loses a weight which is equal to weight of the liquid displaced by the immersed portion of solid body. ¾ Observed weight = true weight – weight of liquid displaced or T = Mg – mg = aUgh – aVgh = ahg(U – V)

Surface Tension x The property due to which the free surface Imaginary line of liquid tends to have minimum surface area and behaves like a stretched membrane is called surface tension. Surface tension is a force per unit length acting in the plane of interface between the liquid and the bounding surface. As per the definition, F S = where F = force acting an either side of L imaginary line on surface L = length of imaginary line 12

PHYSICS FOR YOU | DECEMBER ‘17

x x

x

x

SI unit = N m–1 Dimensions [S] = [MT –2] Factors affecting surface tension ¾ Temperature : Surface tension decreases with rise in temperature. ¾ Impurities : Highly soluble impurities increases surface tension and sparingly soluble impurities decreases surface tension. Surface Energy : As the molecules on the surface experiences a net downward force, so to bring a molecule from interior to the free surface of the molecule, the amount of work done against the internal force gets stored as the extra potential energy of surface molecule is called surface energy. SI unit = J, Dimensions [WS] = [ML2T –2]

x

Excess Pressure Inside the Drop and Bubbles ¾ Excess pressure inside a liquid drop 2S 2S ΔP = or, P − P0 = R R ¾ Excess pressure inside a soap bubble 4S 4S ΔP = or, P − P0 = R R ¾ Excess pressure inside an air bubble in the liquid 2S Excess pressure ΔP = R Capillarity The rise and fall of a liquid in a tube of very fine bore is called capillarity. x

Angle of contact (T) : The angle between tangent to the liquid surface at the point of contact and solid surface inside the liquid is termed as angle of contact.

x

Rise of liquid in a capillary tube (Ascent formula) Here the angle of contact is acute and surface of water is concave. There is a pressure difference between two sides of the top surface. i.e., (Pi − P0 ) = 2S = 2S = 2S cos θ ...(i) R a sec θ a Thus pressure of water inside the tube at meniscus is less than atmospheric pressure. Points A and B are at same level so the pressure must be same. P0 + hUg = PA = Pi, ...(ii) hUg = Pi – P0

If T < 90° i.e., concave meniscus, h = positive rise of liquid. If T > 90° i.e., convex meniscus, h = negative fall of liquid.

From eqns (i) and (ii) 2S 2S cos θ hρg = cos θ or h = a ρga h = rise of capillary height x

Various properties of different liquids

Substances Angle of contact Meniscus shape Capillary action Sticking to solid Shape of liquid surface

water and glass almost zero/acute angle concave liquid rises sticks/wets almost round

Illustration 2 : Two soap bubbles A and B are kept in a closed chamber where the air is maintained at pressure 8 N m–2. The radii of bubbles A and B are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N m–1. Find the ratio nB/nA, where nA and nB are the number of moles of air in bubbles A and B, respectively. [Neglect the effect of gravity.] Soln.: The excess of pressure above atmospheric 4S pressure, due to surface tension in a bubble = . R The surrounding pressure P0 = 8 N m–2. 4 × 0.04 4S ? PA for 1st bubble = P0 + =8+ 0.02 RA PA = 16 N m–2 4 × 0.04 4S PB for 2nd bubble = P0 + =8+ = 12 N m − 2 0.04 RB PV = nRT 4 (16) π (0.02)3 = nART 3 n ⎛4 ⎞ (12) ⎝ π (0.04)3 ⎠ = nB RT ; B = 6 nA 3

8I4B?4I>1=93C*6<E94C9>=?D9?> Flow of Fluid x Streamline flow : The flow in which path taken by a fluid particle under a steady flow is a streamline in direction of the fluid velocity at that point. x Laminar flow : The liquid is flowing with a steady flow and moves in the form of layers of different velocities and do not mix with each other, is called laminar flow. x Turbulent flow : The flow in which velocity is greater than its critical velocity and the motion of particles becomes irregular is called turbulent flow.

water and silver right angle = 90° plane no effect does not wet spreads on surface x

x

mercury and glass obtuse angle = 135° convex liquid falls does not wet flat

Critical velocity : The velocity of liquid flow upto which the flow is streamlined and above which it becomes turbulent is called critical velocity. Reynold’s number : The number which determines the nature of flow of liquid through a pipe is called Reynold’s number. Inertial force per unit area Reynold’s number = Viscous force per unit area vρr or N R = η Where v = velocity of liquid, U = density of liquid, r = radius of tube, K = coefficient of viscosity of liquid. On the basis of Reynold’s number, we have, 0 < NR < 2000 o streamline flow. 2000 < NR < 3000 o streamline to turbulent flow. 3000 < NR o purely turbulent flow.

Equation of Continuity x If an incompressible v1 and non-viscous liquid flowing through a tube v2 A1 A2 of non-uniform crosssection with steady flow, the product of the area of cross-section and the velocity of flow remains same at every point in the tube. A1v1 = A2v2 Av = constant. This equation is known as continuity equation. Bernoulli’s Theorem x It states that for the streamline flow of an ideal liquid through a tube, the total energy (the sum of pressure energy, the potential energy and kinetic energy) per unit volume remains constant at every cross-section throughout the tube. PHYSICS FOR YOU | DECEMBER ‘17

13

1 P + ρgh + ρv 2 = constant 2 P 1 v2 or +h+ = another constant ρg 2 g x

dR = 4(H − h) + 4h(−1) = 4H – 4h – 4h dh 0 = 4H – 8h h = H/2. Range is maximum when liquid flows out from a point at mid-height of tank.

2R ⋅

If the liquid is flowing through a horizontal tube, then h is constant, then according to Bernoulli’s

Rmax = 2

P 1 v2 theorem, + = constant ρg 2 g x

Bernoulli’s theorem is based on law of conservation of energy.

x

Applications of Bernoulli’s Theorem x

Speed of Efflux : v = 2gh where v = velocity of efflux. ¾ Velocity of efflux is the velocity acquired by a freely falling body in falling through a vertical distance h which is equal to depth of a hole, below free surface of liquid, from which liquid flows out. range ¾ Horizontal h¢ When water flows h B out of a hole at depth H A C h below surface of h¢ water in a tank filled F E D with water upto a height H. Water flows out from point A. It covers a vertical distance = (H – h) before striking the ground. Time for this vertical 2(H − h) . g Horizontal velocity of water is velocity of efflux = 2gh . 2(H − h) Horizontal range FD = 2 gh × g Range R = FD = 2 h(H − h) Special case journey =

¾

For point B, range = 2 h′(H − h′) = FE

¾

For point C, range = 2 (H − h′)h′ = FE Range is same whether the point is at distance hc below top or hc above the bottom of a tank. Maximum range Range R = 2 h(H − h)

dR For maximum horizontal range, = 0. dh 2 R = 4h(H – h)

14

PHYSICS FOR YOU | DECEMBER ‘17

H 2

H⎞ ⎛ ⎜⎝ H − 2 ⎟⎠ = H same as height of

tank itself. Venturi-meter : It is a device used to measure the speed of an incompressible fluid using a U shaped manometer attached to it with one arm at narrow end and other at broader end. Applying Bernoulli’s principle A −1/2

x

x

1

A

2 2 ⎤ ⎛ 2ρm gh ⎞ ⎡ ⎛ A1 ⎞ v 1 ⎢ ⎥ v −1 v= ⎜ 2 ⎝ ρ ⎟⎠ ⎢⎣ ⎜⎝ A2 ⎟⎠ ⎥⎦ h Where U = density of fluid Um = density of fluid in manometer A1, A2 = area of cross sections Dynamic lift of ball due to Magnus effect When a spinning ball is thrown, it deviates from its usual path in air. Due to difference in velocities of air, a pressure difference arises in lower and upper faces and the ball lifts upward by a net upward force called the magnus effect. Lift on aircraft wings – Basic principle of an aeroplane.

Viscosity x It is the property of a fluid (liquid or gas) by virtue of which an internal resistance comes into play when the fluid is in motion and opposes the relative motion of its different layers. x According to Newton, viscous force (F) of a liquid between two layers is given by dv F = − ηA dx where K = coefficient of viscosity of liquid A = area of each layer, dv/dx = velocity gradient Here negative sign shows viscous force is acting in a direction opposite to the flow of a liquid. x

Coefficient of viscosity : It is defined as the ratio of shearing stress to the strain rate. Shearing stress F / A Fl η= = = Strain rate v / l vA

x

The dimensional formula of K is [ML–1T –1].

x

The SI unit of K is poiseuille (Pl) or Pa s or N m s.

x

The CGS unit of K is dyne cm–2 s called poise.

–2

1 Pl = 10 poise x

The value of viscosity for ideal fluid is zero.

x

Viscosity is due to transport of momentum.

x

Relative viscosity of liquid =

Viscosity of liquid η = Viscosity of water ηwater

x

Stokes’ law : According to Stokes’ law, if a sphere of radius r moves with a velocity v through a fluid of viscosity K the viscous drag opposing the motion is F = 6S Krv

x

Terminal velocity : The constant velocity attained by a sphere of radius r while going down in a fluid is called the terminal velocity. 2 r 2(ρ − σ)g 9 η where U = density of sphere, V = density of fluid medium, r = radius of sphere, K = coefficient of viscocity. v=

Poiseuille’s Formula x

The rate of volume of fluid coming out of the tube is V πPr 4 where P = pressure difference, l = length = t 8 ηl

Illustration 3 : A large open tank has two holes in the wall. One is a square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then find the value of R. Soln.: Velocity of efflux v = 2 gh where h denotes the depth of the hole. The quantities of water flowing out per second from both holes are given to be the same. ?

a1v1 = a2v2 or (L)2 × 2 gy = (πR 2 ) 2 g (4 y) L or L2 = 2SR2 or R = 2π

D85B=1<@B?@5BD95C?6=1DD5B Heat and Thermometry x Heat : Heat is the form of energy that flows between a body and its surrounding by virtue of temperature difference between them. x The SI unit of heat is joule. x The practical unit of heat is calorie. x 1 calorie is the amount of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C. x Temperature : Temperature is basically a measure of degree of hotness or coldness of a body. x

Name of Symbol Lower the scale for each fixed degree point (LFP)

of tube, r = radius of cross section of the tube ¾

Case I : Series combination of tubes (V1 = V2)

V P ; = t ⎡ 8ηl1 8ηl2 ⎤ Here, P = P1 + P2 ⎢ 4 + 4⎥ πr2 ⎦ ⎣ πr1 P1 and P2 are the pressure difference across the first and second tube. ¾

Case II : Parallel combination of tubes (P1 = P2)

⎡ πr 4 πr 4 ⎤ V = P ⎢ 1 + 2 ⎥ , Here V = V1 + V2 t ⎣ 8ηl1 8ηl2 ⎦

Different temperature scales

x

Upper Number of fixed point divisions (UFP) on the scale

Celsius

°C

0 °C

100 °C

100

Fahrenheit

°F

32 °F

212 °F

180

Reaumur

°R

0 °R

80 °R

80

Rankine

°Ra

460 Ra

672 Ra

212

Kelvin

K

273.15 K 373.15 K

100

Relationship between different temperature scales TC − 0 TF − 32 TR − 0 TRa − 460 TK − 273.15 = = = = 100 180 80 212 100 ¾ Temperature on one scale can be converted into another scale by using the following identity. PHYSICS FOR YOU | DECEMBER ‘17

15

It is principle of calorimetry which is based on the law of conservation of energy.

Reading on any scale − lower fixed point (LFP) Upper fixed point (UFP) − lower fixed point (LFP) ¾

= constant for all scales Thermometer is an instrument used to measure temperature of a body.

Thermal Expansions x Increase in configuration of a solid when its temperature increases is known as thermal expansion. There are three kinds of thermal expansions. Linear expansion : The Lf = L0[1 + D 'T] increase in length of a ΔL –1 body due to increase α = L ΔT ; SI unit { °C 0 in temperature of the body is called linear ΔL L0 expansion. D = coefficient of linear expansion. Superficial expansion : The increase in area of the body due to increase in its temperature is known as superficial expansion.

A = A0 (1 + E 'T) ΔA E= ; A0ΔT

x

x

x

During the change of phase temperature remains constant but heat is being supplied to the body. Latent heat of fusion (Lf) : When phase changes from solid to liquid the heat required for the process is called latent heat of fusion (Lf).

¾

Latent heat of vapourisation (Lv) : When the phase change from liquid to gas or vapour the heat required is called latent heat of vapourisation (Lv)

¾

Latent heat of sublimation (Ls) : When state changes directly from solid to gaseous state, the heat required is called the latent heat of sublimation (Ls).

'A

A0

SI unit { °C E = coefficient of superficial expansion.

0

α β γ = = 1 2 3

Calorimetry x When two thermodynamic systems at different temperatures are mixed the transfer of heat takes place from higher temperature body to the lower temperature one. The body at higher temperature releases heat while body at lower temperature absorbs it, so that Heat Lost = Heat gained 16

¾

–1

Volume expansion : The V = V0 (1 + J 'T) V V increase in volume of a ΔV body due to increase in γ = V ΔT ; 0 DV its temperature is called –1 SI unit { °C volume expansion. J = coefficient of volume expansion. Relation between D, E and J

Heat capacity : The amount of heat required to raise the temperature of an object by one degree. ΔQ s= ΔT Molar specific heat : The amount of heat required to raise the temperature of one mole of a substance by one degree. ΔQ C= nΔT Latent heat : The heat required to change the phase or state of a substance is called the Latent heat.

PHYSICS FOR YOU | DECEMBER ‘17

For water and ice Lf = 333 KJ kg–1 or 79.5 cal g–1 or 6.01 KJ mol–1 Lv = 2256 KJ kg–1 or 539 cal g–1 or 40.7 KJ mol–1 x

It can be graphically illustrated as shown in figure. If we keep ice on a burner, as time passes more heat will be transferred to it. So on the x-axis, we have plotted Q (heat transferred) or t(time). On the y-axis, we have T the temperature. T (°C) Boiling point of water Melting point of water

100 0

Q = m.Lf

Q = m.Lv Q = m.svap'T Q = m.swater'T

Q = m.sice'T Q (cal)

Heat Transfer

Conduction x

x

x

The process by which heat is transferred from one point to other point of a substance in the direction of fall of temperature, without any actual motion of particles of the substance. Coefficient of thermal conductivity. dQ ⎛ dT ⎞ = −KA ⎜ ⎟ dt ⎝ dx ⎠ ¾

x

Convection

Radiation

The process of heat transfer in which heat is transmitted from one point to another due to actual motion of heated particles in liquids and gases.

x

x

x

Application of convection : Formation of trade winds causes monsoons, land and sea breezes etc.

K = coefficient of thermal conductivity A = area of the hotter face

Units and dimension of K are J m–1 s–1 K–1 and [MLT–3 K –1].

Thermal resistance x Temperature difference T1 − T2 = ¾ R= = Rate of flow of heat dQ / dt KA Greater the K, smaller will be the thermal resistance. kelvin kelvin × sec = Unit of R = watt joule

Dimensions of R = [M–1L–2T3K] where K = dimensions of kelvin temperature. Illustration 4 : Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle ABC, right-angled at B. The points A and B are maintained at temperatures T and ( 2 ) T respectively. In the steady state, the temperature of the point C is TC. Assuming that only heat conduction takes place,What is ratio of TC/T ?

⎛ TC − 2 T ⎞ ⎛ T − TC ⎞ ⎟⎠ ⎟⎠ = KA ⎜⎝ a 2a

∴ KA ⎜ ⎝

or T − TC = 2 TC − 2T or 3T = TC ( 2 + 1) or x

¾

Soln.: For heat conduction,

( )

ΔQ ΔT = KA l Δt

x

ΔQ ⎛ T − TC ⎞ = KA ⎜ ⎝ 2a ⎟⎠ Δt

⎛T − 2 T ⎞ ΔQ ⎟⎠ For CB, = KA ⎜⎝ C Δt a

Exam √2a

90°

Equate the two equations for B (√2 T) steady state.

a

C (TC)

(

3

2 + 1)

.

5H1=3?B>5B

A(T)

a

TC = T

Perfectly Black Body : A perfectly black body is that which absorbs completely the radiations of all wavelengths incident on it. ¾ As a perfectly black body neither reflect nor transmit any radiation, therefore the absorptance of a perfectly black body is unity, i.e., t = 0 and r = 0 a = 1. Kirchhoff ’s Law : According to this law the ratio of emissive power to absorptive power is same for all surface at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.

Since B is at higher temperature than A, heat flows from B to A, A to C and then C to B, for steady state. For AC,

The process of heat transfer from one point to another point without heating the intervening medium is called radiation. When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted. Q = Qa + Qr + Qt Qa Qr Qt + + = a +r +t =1 Q Q Q a = Absorptance, r = Reflectance t = Transmittance

Date

JEE Main (Offline)

8th April, 2018

WB JEE

22nd April, 2018

JEE Advanced

20th May, 2018

AIIMS

27th May, 2018

PHYSICS FOR YOU | DECEMBER ‘17

17

Hence

E1 = A1

E2 ⎛E⎞ = ... ⎝ ⎠ A2 A Perfectly black body

E = Eb A If emissive and absorptive powers are considered for a particular wavelength O, ⎛ Eλ ⎞ ⎜⎝ A ⎟⎠ = (E λ )black λ

But for perfectly black body A = 1, i.e., ¾

x

Stefan’s Law : According to this law the radiant energy emitted per unit area per second by a perfectly black body (i.e., emissive power of black body) is directly proportional to the fourth power of its absolute temperature, i.e., E vT 4 E = VT 4 where V is a constant called Stefan’s constant having dimensions [MT –3K–4] and value 5.67 u 10–8 W m–2K–4. ¾

For ordinary body : E = eVT 4

¾

Radiant energy : If Q is the total energy radiated by the ordinary body then Q = eσT 4 ⇒ Q = eAσT 4t A×t

If an ordinary body at temperature T is surrounded by a body at temperature T0, then Stefan’s law can be put as E = eAV (T4 – T 40)t Wien’s Displacement T3 > T2 > T1 Law : According to EO Om < Om < Om 3 2 1 this law the product T3 of wavelength T2 corresponding to T1 Om 1 maximum intensity Om Om O 3 2 of radiation at a given temperature of body (in kelvin) is constant, i.e., OmT = b = constant where b is Wien’s constant and has value 2.89 u10–3 mK. ¾ As the temperature of the body increases, the wavelength at which the spectral intensity (EO) is maximum shifts towards left. Therefore it is also called Wien’s displacement law. ¾ This law is of great importance in astrophysics as through the analysis of radiations coming from a distance star, by finding Om the temperature of the star T = b/Om is determined. ¾

x

18

PHYSICS FOR YOU | DECEMBER ‘17

Newton’s Law of Cooling dQ According to this law, − = K (T − T0 ) where dt dQ = rate of loss of heat of a liquid dt (T – T0) = difference in temperatures of liquid and the surrounding. K = constant of proportionality x As Q = msT where m = mass, s = specific heat of liquid dT dQ dT K = ms = −K (T − T0 ) − = − (T − T0 ) dt dt dt ms Rate of fall of temperature = dT/dt dT Rate of cooling = ms dt dT Rate of loss of heat = rate of cooling = ms dt Illustration 5 : Two metallic spheres S1 and S2 are made of the same material and have got identical surface area. The mass of S1 is thrice that of S2. Both spheres are heated to the same high temperature and placed in the same room having lower temperature but are thermally insulated from each other. Find the ratio of the initial rate of cooling of S1 to that of S2. Sol. : According to Stefan’s law 'Q = eVAT 4 't x

Also, 'Q = ms 'T ? ms 'T = eVAT 4't ⎛ eσT 4 ⎞ A eσAT 4 ΔT ΔT ⎟ or = or = ⎜⎝ s ⎠m Δt ms Δt 4 3 ⎛ 3m ⎞ πr ρ or r = ⎜ ⎝ 4πρ ⎠⎟ 3

For a sphere, m =

1/3

? Area of sphere, A = 4Sr2 or A = 4S ⎛⎜ 3m ⎟⎞ ⎝ 4πρ ⎠ ∴ Rate of cooling

( )

2/3

⎛ eσT 4 ⎞ ΔT ⎛ 3 ⎞ ⎟⎠ 4π × ⎜ = ⎜⎝ ⎝ 4πρ ⎟⎠ s Δt

2/3

m m

2/3

= (Constant) × m–1/3 ∴

(ΔT / Δt )S1 (ΔT / Δt )S2

⎛ m1 ⎞ ⎝ m2 ⎟⎠

−1/3

=⎜

477*3(::?00 1. (d) 2. 6. (b) 7. 11. (b) 12. 16. (b) 17. 21. (a, b,c) 24. (7) 25. 29. (b) 30.

(a) (a) (d) (c) (4) (c)

3. 8. 13. 18. 22. 26.

⎛ 3m ⎞ ⎝ m ⎟⎠

−1/3

=⎜

⎛ 1⎞ ⎝ 3 ⎟⎠

1/3

=⎜

ANSWER KEY

(a) 4. (b) 9. (b) 14. (d) 19. (a, b, d) (6) 27.

(b) (a) (a) (a) (c)

5. 10. 15. 20. 23. 28.

(c) (c) (c) (a, c, d) (a, b, c) (a)

1. The length of a rubber cord is l1 m when the tension is 4 N and l2 m when the tension is 6 N. The length when the tension is 9 N, is (a) (2.5l2 – 1.5l1) m (b) (6l2 – 1.5l1) m (c) (3l1 – 2l2) m (d) (3.5l2 – 2.5l1) m 2. The lower end of a glass capillary tube is dipped in water. Water rises to a height of 8 cm. The tube is then broken at a height of 6 cm. The height of water column and angle of contact will be ⎛ 3⎞ ⎛ 3⎞ (a) 6 cm, sin −1 ⎜ ⎟ (b) 6 cm, cos −1 ⎜⎝ ⎟⎠ ⎝ 4⎠ 4 −1 ⎛ 1 ⎞ (c) 4 cm, sin ⎜⎝ ⎟⎠ 2

⎛ 1⎞ (d) 4 cm, cos −1 ⎜⎝ ⎟⎠ 2

3. A vessel contains oil (density = 0.8 g cm3) over mercury (density = 13.6 g cm–3). A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in g cm–3 is (a) 3.3 (b) 6.4 (c) 7.2 (d) 12.8 4. Two wires are made of the same material and have the same volume. However wire 1 has cross-sectional area A and wire 2 has cross-sectional area 3A. If the length of wire 1 increases by Δx on applying force F, how much force is needed to stretch wire 2 by the same amount? (a) F (b) 4F (c) 6F (d) 9F 5. A water barrel stands on a table of height h. If a small hole is punched in the side of the barrel at its base, it is found that the resultant stream of water strikes the ground at a horizontal distance R from the table. What is the depth of water in the barrel? R2 R2 R2 4 R2 (a) (b) (c) (d) 4h h 2h h 6. Hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is 20 °C. How long would the food take to cool from 71 °C to 69 °C? (a) 12 s (b) 25 s (c) 16 s (d) 42 s 7. Two identical rods AC and CB made of two different metals having thermal conductivities in the ratio 2 : 3 are kept in contact with each other at the end C as shown in the figure. A is at 100 °C and B is at 25 °C. Then the junction C is at

(a) (b) (c) (d)

55 °C 60 °C 75 °C 50 °C

8. A triangular lamina of area A and height h is immersed in a liquid of density U in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is 1 1 (a) Aρgh (b) Aρgh 2 3 2 1 (c) Aρgh (d) Aρgh 3 6 9. A wire of initial length L and radius r is stretched by a length l. Another wire of same material but with initial length 2L and radius 2r is stretched by a length 2l. The ratio of the stored elastic energy per unit volume in the first and second wire is (a) 1 : 4 (b) 1 : 2 (c) 2 : 1 (d) 1 : 1 10. Water is flowing in streamline motion through a horizontal tube. The pressure at a point in the tube is P where the velocity of flow is v. At another point, where the pressure is P/2, the velocity of flow is (density of water = U) P 2 P (a) v 2 + (b) v − ρ ρ (c)

v2 +

2P ρ

(d)

v2 −

2P ρ

11. Water leaves a faucet with a downward velocity of 3.0 m s–1. As the water falls below the faucet, it accelerates with acceleration g. The cross-sectional area of the water stream leaving the faucet is 1.0 cm2. What is the cross-sectional area of the stream 0.50 m below the faucet? (a) 0.56 cm2 (b) 0.69 cm2 2 (c) 0.29 cm (d) 0.66 cm2 12. A thin copper wire of length L increases its length by 1% when heated from temperature T1 to T2. What is the percentage change in area when a thin copper plate having dimensions 2L × L is heated from T1 to T2? (a) 0.5% (b) 1% (c) 2% (d) 4% PHYSICS FOR YOU | DECEMBER ‘17

19

13. Two rods A and B of different materials are welded together as shown in K1 A figure. Their thermal K T2 conductivities are K1 and T1 K B B K 2 K 2. The thermal conductivity of the d composite rod will be 3(K1 + K 2 ) (a) (b) K1 + K2 2 K + K2 (c) 2(K1 + K2) (d) 1 2 [NEET 2017] 14. A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in W would be (a) 450 (b) 1000 (c) 1800 (d) 225 [NEET 2017] 15. The bulk modulus of a spherical object is ‘B’. If it is subjected to uniform pressure ‘P’, the fractional decrease in radius is P 3P P B (a) (b) (c) (d) 3B B B 3P [NEET 2017] 16. A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagram). The density of the oil is Pa

F

A

E Oil

Pa 10 mm Final water level

65 mm D

Initial water level

65 mm B

C Water

(a) 425 kg m–3 (c) 928 kg m–3

(b) 800 kg m–3 (d) 650 kg m–3 [NEET 2017] 17. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of 1 1 (c) 81 (d) (a) 9 (b) 9 81 [JEE Main Offline 2017] 20

PHYSICS FOR YOU | DECEMBER ‘17

18. A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, the temperature of the system is found to be 75 °C. T is given by (Given : room temperature = 30 °C, specific heat of copper = 0.1 cal g–1 °C–1) (a) 800 °C (b) 885 °C (c) 1250 °C (d) 825 °C [JEE Main Offline 2017] 19. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and D is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by P P 3α (b) (c) (d) 3PKD (a) αK PK 3αK [JEE Main Offline 2017] 20. Two tubes of radii r1 and r2, and lengths l1 and l2, respectively, are connected in series and a liquid flows through each of them in streamline conditions. P1 and P2 are pressure differences across the two l

tubes. If P2 is 4P1 and l2 is 1 , then the radius r2 4 will be equal to r1 (a) 2r1 (b) 2 (c) 4r1 (d) r1 [JEE Main Online 2017] SOLUTIONS 1. (a) : Let the original unstretched length be l. T l ∴ Y= A Δl 4 l 6 l 9 l Now Y = = = A (l1 − l ) A (l2 − l ) A (l3 − l ) ? 4(l3 – l) = 9(l1 – l) or 4l3 + 5l = 9l1 ...(i) Again, 6(l3 – l) = 9(l2 – l) or 2l3 + l = 3l2 ... (ii) From equation (i) and (ii) ? l3 = (2.5l2 – 1.5l1) m 2. (b) : When a capillary tube is broken at a height of 6 cm, the height of water column will be 6 cm. 2 S cos θ h As h = = constant or ρrg cos θ 8 6 6 cos 0° 3 ∴ = or cos θ = = cos 0° cos θ 8 4 3 or θ = cos −1 ⎛⎜ ⎞⎟ ⎝4⎠

3. (c) : Upthrust force due to oil and mercury V V FB FB = ρm g + ρ0 g 2 2 Oil ' Weight of sphere is equal to the upthrust Mercury force W = FB V V W = V Vg VVg = ρm g + ρ0 g 2 2 ρ + ρ0 13.6 + 0.8 14.4 V= m = = 7.2 g cm–3 = 2 2 2 4. (d) : For the same material, Young’s modulus is the same and it is given that the volume is the same and the area of cross-section for the wire 1 is A and that of 2 is 3A. V = V1 = V2, V = A × l1 = 3A × l2 or l2 = l1/3 Δl Δl F/A ∴ F1 = YA 1 and F2 = Y 3 A 2 As Y = l1 l2 Δl / l Given 'l1 = 'l2 = 'x (for the same extension) Δx ⎛ YAΔx ⎞ ∴ F2 = Y 3 A = 9⎜ = 9F1 = 9F l1 / 3 ⎝ l1 ⎟⎠ (' F1 = F2) 5. (c) : From Toricelli’s theorem ... (i) v = 2 gd where v is horizontal velocity and d is the depth of water in barrel. Time t to hit the ground is given by 1 2h h = gt 2 or t = g 2 2h ? R = vt = (2 gd ) (Using (i)) = 2 dh g R2 ? R2 = 4dh or d = 4h 6. (d) : According to Newton’s law of cooling (T1 − T2 )

⎞ ⎛ T +T = K ⎜ 1 2 − Ts ⎟ ⎝ ⎠ t 2 where Ts is the surrounding temperature.

For the Ist case (94 ° C − 86 °C) 2 min

or

8 °C 2 min

2 °C

or

t

...(ii)

= K (50 °C )

Divide (i) by (ii), we get or

8 °C / 2 min

t = 0.7 min = 42 s

2 °C/ t

=

K (70 °C ) K (50 °C)

7. (a) : Rate of flow of heat (i.e. heat current),

H=

KA(T1 − T2 )

L Let the temperature of junction be T K1 A(100 − T ) K 2 A(T − 25) = L L

∴

or or

K1(100 – T) = K2(T – 25) 3T – 75 = 200 – 2Tor 5T = 275 275 = 55 °C T= 5

or

x h = b′ b b b′ = x h

8. (d) :

x

bc b dA = b′ dx = x dx h P = Ugx b b dF = PdA = ρgx x dx h h ρgb h3 1 ρgb 2 F= x dx = = ρgbh2 h 3 3 h ∫0

dx

h

2 ⎛ bh ⎞ 2 = ρgh ⎜ ⎟ = ρgAh ⎝ 2⎠ 3 3 9. (d) : By definition, F /A FL FL = = 2 ΔL /L AΔL πr ΔL As both the wires are of the same material, therefore their Young moduli are same. F1L1 FL i.e., Y1 = Y2 or = 22 2 2 πr1 ΔL1 πr2 ΔL2 Young’s modulus, Y =

2

F ⎛ L ⎞ ⎛ r ⎞ ⎛ ΔL ⎞ or 1 = ⎜ 2 ⎟ ⎜ 1 ⎟ ⎜ 1 ⎟ F2 ⎝ L1 ⎠ ⎝ r2 ⎠ ⎝ ΔL2 ⎠ Substituting the given values, we get

⎛ (94 °C + 86 ° C ) ⎞ − 20 °C ⎟ = K⎜ ⎝ ⎠ 2

= K (70 ° C )

For the IInd case (71°C − 69 °C) ⎛ 71 °C + 69 °C ⎞ = K⎜ − 20 °C ⎟ t 2 ⎝ ⎠

2

...(i)

F1 ⎛ 2L ⎞ ⎛ r ⎞ ⎛ l ⎞ 1 = = F2 ⎜⎝ L ⎟⎠ ⎜⎝ 2r ⎟⎠ ⎜⎝ 2l ⎟⎠ 4

Elastic energy stored per unit volume in the wire is 1 1 F L u = × stress × strain = × × 2 2 A ΔL PHYSICS FOR YOU | DECEMBER ‘17

21

∴

u1 ⎛ F1 ⎞ ⎛ A2 ⎞ ⎛ L1 ⎞ ⎛ ΔL2 ⎞ = u2 ⎜⎝ F2 ⎟⎠ ⎜⎝ A1 ⎠⎟ ⎝⎜ L2 ⎟⎠ ⎜⎝ ΔL1 ⎟⎠ ⎛ 1 ⎞ ⎛ π(2r )2 ⎞ ⎛ L ⎞ ⎛ 2l ⎞ =⎜ ⎟⎜ =1 ⎝ 4 ⎠ ⎝ πr 2 ⎟⎠ ⎜⎝ 2L ⎟⎠ ⎜⎝ l ⎟⎠

10. (a) : According to Bernoulli’s theorem, P 1 1 1 P + ρv 2 = constant ∴ P + ρv 2 = + ρv ′ 2 2 2 2 2 or

P 1 2 1 P 1 + ρv = ρv ′2 or = ρ(v ′2 − v 2 ) 2 2 2 2 2

P P + v 2 or v ′ = v 2 + ρ ρ 11. (b) : Let the initial speed = v0 and the initial crosssectional area = A0. After the freely falling stream has or v ′2 =

2 descended a distance h, its speed v1 = v0 + 2 gh . ...(i) Under steady-flow conditions, the mass fluxes at the at the locations are equal, ? U0v0A0 = U1vA1. Since the water is effectively incompressible, U0 = U1 therefore A1 = (v0/v1)A0. Put the values in eqn, (i) –1

2

v1 = (3.0) + 2(9.8)(0.50) = 4.34 m s . A1 = (3.0/4.34)(1.0 cm2) = 0.69 cm2. 12. (c) : Length of the wire at temperature T2 is 2 1 ⎞ ⎛ 1 ⎞ 2 2⎛ Lt = L ⎜ 1 + ⎟ ∴ L = L + 2 2 1 ⎜⎝ ⎟ t ⎝ 100 ⎠ 100 ⎠

Now 2Lt2 = area of the plate at temperature T2 and 2L2 = area of the plate at temperature T1. 2

1 ⎞ 2 ⎞ 102 A ⎛ ⎛ ∴ At = A ⎜ 1 + = A ⎜1 + = ⎝ 100 ⎟⎠ ⎝ 100 ⎟⎠ 100 Thus, the area increases by 2%. 13. (d) : Equivalent thermal conductivity of the composite rod in parallel combination will be, K=

K1 A1 + K 2 A2 A1 + A2

=

K1 + K 2 2

14. (c) : According to Stefan-Boltzman law, rate of energy radiated by a black body is given as E = VAT 4 = V4SR2T 4 Given E1 = 450 W, T1 = 500 K, R1 = 12 cm R2 =

R1 , T2 = 2T1, E2 = ? 2 2

4

⎛ R ⎞ ⎛T ⎞ 1 = ⎜ 2 ⎟ ⎜ 2 ⎟ = × 16 = 4 4 E1 ⎝ R1 ⎠ ⎝ T1 ⎠ E2 = E1 × 4 = 450 × 4 = 1800 W E2

22

PHYSICS FOR YOU | DECEMBER ‘17

15. (c) : Bulk modulus B is given as − PV ...(i) B= ΔV The volume of a spherical object of radius r is given as − PB V r 4 , 'V = π (3r 2 )Δr ; ? =− V= 3Δr ΔV ΔV 3 Pr Put this value in eqn. (i), we get, B = − 3Δr p Δr Fractional decrease in radius is − = 3B r 16. (c) : Pressure at point C, PC = Pa + Uwater ghwater, where hwater = CE = (65 + 65) mm = 130 mm Pressure at point B, PB = Pa + Uoil ghoil where hoil = AB = (65 + 65 + 10) mm = 140 mm In liquid, pressure is same at same liquid level, PB = PC Uoil ghoil = Uwater ghwater 130 × 103 Uoil = = 928.57 kg m–3 140 17. (a) : We know stress is given by Force mg ρVg m⎞ ⎛ Stress = = = ⎜⎝' ρ = ⎟⎠ Area V A A i.e., Stress ∝

L3

(L is the linear dimension.) L2 Stress v L Since linear dimension increases by a factor of 9, stress also increases by a factor of 9. 18. (b) 19. (a) : Bulk modulus of the gas is given by P ΔV P … (i) K= or = ⎛ ΔV ⎞ V0 K ⎜⎝ V ⎟⎠ 0 ΔV Also, V = V0(1 + J'T) ; … (ii) = γΔT V0 Comparing eqn. (i) and (ii), we get P P P ( J = 3D) = = γΔT ⇒ ΔT = 3αK K Kγ 20. (b) : Rate of flow of liquid through narrow tube, dv πPr 4 = 8 ηl dt

As both the given tubes are connected in series so rate of flow of liquid is same. ⎛ P ⎞⎛l ⎞ πP1r14 πP2r24 = ⇒ r24 = ⎜ 1 ⎟ ⎜ 2 ⎟ r14 8 ηl1 8 ηl2 ⎝ P2 ⎠ ⎝ l1 ⎠ Here, P2 = 4P1, l2 = l1/4

∴

⎛ r1 ⎞ ⎛ P ⎞⎛ l ⎞ So, r24 = ⎜ 1 ⎟ ⎜ 1 ⎟ r14 ; = ⎜ ⎟ ⎝ 2⎠ ⎝ 4 P1 ⎠ ⎝ 4l1 ⎠

4

∴ r2 =

r1 2

EXAM

SS CLA XI

PREP 2018 Useful for Medical/Engg. Entrance Exams

CHAPTERWISE MCQs FOR PRACTICE OSCILLATION

1. A particle executes simple harmonic oscillation with an amplitude a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is (a) T/4 (b) T/8 (c) T/12 (d) T/2 2. A particle of mass 200 g is making SHM under the influence of a spring of force constant k = 90 N m–1 and a damping constant b = 40 g s–1. Calculate the time elapsed for the amplitude to drop to half its initial value. (Given ln(1/2) = – 0.693) (a) 7 s (b) 9 s (c) 4 s (d) 11 s 3. A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance X. Now the combined mass will oscillate on the spring with period (a) T = 2π (c) T =

π 2

( M + m) X mg (b) T = 2π mg X ( M + m) (M + m) mg (d) T = 2π (mg ) X X ( M + m)

4. Two identical springs are connected to mass m as shown, of spring constant k each. If the period of the configuration in (i) is 2 s, the period of the configuration in (ii) is (a)

2s

(b) 1 s

(c)

1 2

s

(d) 2 2 s

5. What is the spring constant for the combination of springs shown in figure? (a) k (b) 2k (c) 4k 5k (d) 2 6. A spring balance has a scale that reads from 0 to 60 kg. The length of the scale is 30 cm. A body suspended from this balance and when displaced and released, oscillates with a period of 0.8 s, what is the weight of the body when oscillating? (a) 350.67 N (b) 540.11 N (c) 311.64 N (d) 300.5 N 7. Figure depicts a circular motion of the particle P. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated on the figure. Obtain the SHM motion of the x-projection of the radial vector of the rotating particle P. π⎞ ⎛π (a) R sin ⎜ t + ⎟ ⎝2 4⎠

π⎞ ⎛π (b) R sin ⎜ t − ⎟ ⎝2 4⎠

π⎞ ⎛π (c) R cos ⎜ t + ⎟ ⎝2 4⎠

π⎞ ⎛π (d) R cos ⎜ t − ⎟ ⎝2 4⎠

8. A uniform spring of normal length l has a force constant k. It is cut into two pieces of lengths l1 and l2 such that l1 = nl2 where n is an integer. Then the value of k1 (force constant of spring of length l1) is PHYSICS FOR YOU | DECEMBER ‘17

23

(a)

kn n +1

(c)

k(n − 1) n

k(n + 1) n kn (d) n −1 (b)

9. A simple pendulum of length l1 has a time period of 4 s and another simple pendulum of length l2 has a time period 3 s. Then the time period of another pendulum of length (l1 – l2) is 3 s (d) 7 s 4 10. Let T1 and T2 be the time periods of springs A and B when mass M is suspended from one end of each spring. If both springs are taken in series and the same mass M is suspended from the series combination, the time period is T. Then 1 1 1 = + (a) T = T1 + T2 (b) T T1 T2 (a)

3s

(b) 1 s

(c) T 2 = T12 + T22

(c)

(d)

1 T

2

=

1 T12

+

1 T22

11. A man measures time period of a simple pendulum inside a stationary lift and find it to be T. If the lift starts accelerating upwards with an acceleration g/4, then the time period of pendulum will be 2 2T 5 5T (a) (b) (c) (d) 5T 2T 2 5 12. The amplitude of a damped oscillator becomes rd

⎛1⎞ 1 ⎜⎝ ⎟⎠ in 2 s. If its amplitude after 6 s is times 3 n the original amplitude, the value of n is (a) 32

(b) 3 3 (c) 23 (d) 33 13. The displacement of a particle is represented by the equation by y = sin3 Zt. The motion will be (a) non-periodic (b) periodic but not simple harmonic 2π (c) simple harmonic with periodic ω π (d) simple harmonic with periodic ω 14. What fraction of the total energy is kinetic when the displacement is one-half of the amplitude? 2 3 7 1 (a) (b) (c) (d) 4 4 4 4 24

PHYSICS FOR YOU | DECEMBER ‘17

15. A disc of radius R = 10 cm oscillates as a physical pendulum about an axis perpendicular to the plane of the disc at a distance r from its centre. R If r = , the approximate period of oscillation is 4 (Take g = 10 m s–2) (a) 0.84 s (b) 0.94 s (c) 1.26 s (d) 1.42 s WAVES

16. The apparent frequency observed by a moving observer away from a stationary source is 20% less than the actual frequency. If the velocity of sound in air is 330 m s–1, then the velocity of the observer is (a) 660 m s–1 (b) 330 m s–1 –1 (c) 66 m s (d) 33 m s–1 17. A vibrating string of certain length l under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency X. Now when the tension of the string is slightly increased the number of beats reduces to 2 per second. Assuming the velocity of sound in air to be 340 m s–1, the frequency X of the tuning fork (in Hz) is (a) 344 (b) 336 (c) 117.3 (d) 109.3 18. An organ pipe is closed at one end and open at the other. What is the ratio of frequencies of the 3rd and 4th fundamental modes of vibration? 3 5 9 3 (b) (c) (d) (a) 4 7 11 5 19. Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 s and the velocity of the wave is 300 m s–1. What is the phase difference between the oscillations of two points ? (a) S (b) S/6 (c) S/3 (d) 2S/3 20. A longitudinal wave is represented by x⎞ ⎛ x = x0 sin 2 π ⎜ υt − ⎟ . The maximum particle ⎝ λ⎠ velocity will be four times the wave velocity if πx (a) λ = 0 (b) O = 2Sx0 4 πx (c) λ = 0 (d) O = 4Sx0 2 21. An organ pipe A, with both ends open, has fundamental frequency 300 Hz. The third harmonic of another organ pipe B, with one end open, has the

same frequency as the second harmonic of pipe A. The lengths of pipe A and B are (a) 57.2 cm and 42.9 cm (b) 57.2 cm and 45.8 cm (c) 42.9 cm and 32.2 cm (d) 42.9 cm and 34.0 cm (Speed of sound in air = 343 m s–1) 22. A travelling wave at a rigid boundary or a closed end, is reflected with a phase change of I1 and the reflection at an open boundary takes place with a phase change of I2, then (a) I1 = 0, I2 = 0 (b) I1 = S, I2 = 0 (c) I1 = 0, I2 = S (d) I1 = S, I2 = S 23. A string of mass 3 kg is under tension of 400 N. The length of the stretched string is 25 cm. If the transverse jerk is stuck at one end of the string how long does the disturbance take to reach the other end? (a) 0.043 s (b) 0.055 s (c) 0.034 s (d) 0.065 s 24. The equation of a stationary wave is ⎛ πx ⎞ y = 4 sin ⎜ ⎟ cos (96 St). ⎝ 15 ⎠ The distance between a node and its next antinode is (a) 7.5 units (b) 1.5 units (c) 22.5 units (d) 30 units 25. Pick out the correct statement in the following with reference to stationary wave pattern. (a) In a tube closed at one end, all the harmonics are present. (b) In a tube open at one end, only even harmonics are present. (c) The distance between successive nodes is equal to the wavelength. (d) In a stretched string, the first overtone is the same as the second harmonic. 26. If two waves of the same frequency and amplitude respectively on superposition produce a resultant disturbance of the same amplitude the waves differ in phase by (a) S (b) zero (c) S/3 (d) 2S/3 27. Two uniform strings A and B made of steel are made to vibrate under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the lengths of the strings is (a) 1 : 2 (b) 1 : 3 (c) 1 : 4 (d) 1 : 6 28. If the temperature is raised by 1 K from 300 K the percentage change in the speed of sound in a gaseous mixture is (R = 8.31 J mole–1 K–1) (a) 0.167% (b) 2% (c) 1% (d) 0.334%

29. A wave is represented by the equation : y = 0.sin (100St – kx). If wave velocity is 100 m s–1, its wave number is equal to (a) 1 m–1 (b) 2 m–1 (c) Sm–1 (d) 2S m–1 30. A transverse wave is travelling along y C a string from left B D to right. The given figure represents the A x E shape of the string H F at a given instant. At G this instant, among the following, choose the wrong statement. (a) Points D, E and F have upwards positive velocity. (b) Points A, B and H have downwards negative velocity. (c) Point C and G have zero velocity. (d) Points A and E have minimum velocity. SOLUTIONS 1. (c) : In simple harmonic motion, the displacement x(t) of a particle from equilibrium position at any time t is given by x(t) = a sinZt where a is the amplitude 1 a a = sin ωt At x (t ) = , = a sin ωt or 2 2 2 π 2π ⎛π⎞ or sin ⎜ ⎟ = sin ωt or = t ⎝6⎠ 6 T 2π ('ω = , where T is the time period of oscillation) T T or t = 12 2. (a) : The amplitude of the damped oscillator at any instant t is given by A(t) = A0e–bt/2m ...(i) where A0 is the initial amplitude and b is the damping constant. At t = T1/2, the amplitude drop to half of its initial value. From (i), we get A0 −bT / 2m −bT / 2m 1 = A0e 1/2 , = e 1/2 2 2 Taking natural logarithm on both sides, we get bT ln(1 / 2) ⎛1⎞ ...(ii) ln ⎜ ⎟ = − 1/ 2 or T1/ 2 = − ⎝2⎠ b / 2m 2m Substituting values in eqn. (ii) we get ⎛ 0.693 ⎞ T1/2 = ⎜ × 2 × 200 ⎟ s = 7 s ⎝ 40 ⎠ PHYSICS FOR YOU | DECEMBER ‘17

25

3. (b) : Time period of spring block system is given mass of block by, T = 2π spring constant mg Here, mass of block = (M + m); spring constant, k = X At equilibrium, (M + m)g = k(Xo + X) or, mg = kX (Initially, Mg = kXo)] ∴ T = 2π

( M + m) ( M + m) X = 2π mg mg X

4. (b) 5. (c) : Since all the forces are acting upwards, ? F = –(2k + k + k)x = –4kx keq = 4k. 6. (c) : The 30 cm length of the scale reads upto 60 kg. ? Maximum force, F = 60 kg wt = 60 u 9.8 N = 588 N and maximum extension, x = 30 – 0 = 30 cm = 30 u 10–2 m Spring constant of the spring balance is 588 N F = 1960 N m–1 k= = x 30 × 10−2 m Let a body of mass m is suspended from this balance. Then, time period of oscillation, T = 2π ∴ m=

T 2k

m k

(0.8)2 × (1960)

= 31.8 kg 4 π2 4 × (3.14)2 Weight of the body = mg = (31.8 kg) (9.8 m s–2) = 311.64 N π 7. (c) : At t = 0, OP makes an angle 45° or with 4 2πt the x-axis. After time t, it covers an angle of T in the anticlockwise direction and makes an angle 2 πt π + with the x-axis. of T 4 The projection of OP on the x-axis at time t is given by. ⎛ 2 πt π ⎞ + ⎟ x = R cos ⎜ ⎝ T 4⎠ 2 πt π ⎞ ⎛ πt π ⎞ For T = 4 s, x = R cos ⎛⎜ + ⎟ = R cos ⎜ + ⎟ . ⎝ T ⎝ 2 4⎠ 4⎠ 8. (b) : l1 = nl2 The entire length of the spring is (n + 1)l2. 26

=

2 or T 2 = 4 π m k

PHYSICS FOR YOU | DECEMBER ‘17

Each l2 spring has spring constant kc. 1 1 1 + + ...(n + 1)times = k′ k′ k n +1 1 or kc = (n + 1)k = k′ k 1 n = Now, l1 consists of n such springs of kc, or k ′ k1 k ′ (n + 1)k k1 = = n n 9. (d) : T1 = 4 = 2 π

l1 4g or l1 = 2 g π

l 9g T2 = 3 = 2 π 2 or l2 = 2 g 4π 9 ⎞ 7g g ⎛ (l1 − l2 ) = ⎜ 4 − 4 ⎟⎠ = 2 π2 ⎝ 4π T ′ = 2π

(l1 − l2 ) 7g = 2π = 7s g 4 π2 g

10. (c) : T1 = 2 π and T2 = 2 π

M k1

or k1 =

4 π2 M T12

4 π2 M M or k 2 = k2 T22

In series combination, keff =

k1k2 4π2 M = k1 + k2 T12 + T22

M = T12 + T22 or T 2 = T12 + T22 keff 11. (a) : In a stationary lift, time period of simple l pendulum is T = 2π , g where l is the length of the simple pendulum. When the lift accelerates upwards with an acceleration g/4, the effective acceleration on the bob of pendulum is g 5g g′ = g + = 4 4 ∴ T = 2π

l 4⎛l ⎞ l = 2π = 2 π ⎜ ⎟ = 2T 5⎝g⎠ g′ ⎛ 5g ⎞ 5 ⎜⎝ 4 ⎟⎠ 12. (d) : Amplitude of a damped oscillator at any instant t is A = A0e–bt/2m, where A0 is the original amplitude. A When t = 2 s, A = 0 3 A0 −2b / 2m or 1 = e −b /m ...(i) ∴ = A0e 3 3 ∴ T ′ = 2π

A0 A0 ∴ = A0e −6b / 2m n n 3 1 −3b /m = (e–b/m)3 = ⎛ 1 ⎞ (Using (i)) or ⎜⎝ ⎟⎠ =e 3 n ? n = 33 13. (b) : Given equation of motion is y = sin3 Zt = (3sin Zt – sin 3Zt)/4 dy ⇒ 4 = 3ω cos ωt − 3ω cos 3ωt dt When t = 6 s, A =

⇒ 4×

d2 y dt 2

= −3ω2 sin ωt + 9ω2 sin 3ωt

−3ω2 sin ωt + 9ω2 sin 3ωt = 4 dt 2 2 d y is not proportional to y. ⇒ dt 2 Hence, motion is not SHM. As the expression is involving sine function, it will be periodic. 14. (c) 15. (b) : Time period of a physical pendulum is I T = 2π , where I is the moment of inertia mgh of the pendulum about an axis through the pivot, m is the mass of the pendulum and h is the distance from the pivot to the centre of mass. In this case, a solid disc of R oscillates as a physical pendulum r about an axis perpendicular to the R plane of the disc at a distance r from its centre. ⇒

d2 y

2

∴ I=

2 mR2 mR2 ⎛ R ⎞ 9mR + mr 2= + m⎜ ⎟ = ⎝4⎠ 2 2 16

9mR2 9R R and h = r = ∴ T = 2π 16 = 2π mgR 4g 4 4 Here, R = 10 cm = 0.1 m 9 × 0. 1 3 1 = 2π × × = 0.94 s ∴ T = 2π 4 × 10 2 10 16. (c) : Let the velocity of the observer be vo. If X is the actual frequency, then the apparent frequency observed by the moving observer away from the stationary source is ⎛ v − vo ⎞ υ′ = υ ⎜ ⎝ v ⎟⎠

where v is the velocity of sound in air. 20 80 4 υ= υ = υ (given) 100 100 5 v v − ⎛ ⎞ 4 4 1 o υ = υ⎜ or vo = v − v = v 5 ⎝ v ⎟⎠ 5 5

But υ ′ = υ − ∴

Here, v = 330 m s–1 1 ∴ vo = (330 m s−1 ) = 66 m s−1 5 17. (a) : As the string and the tube are in resonance X 1 = X 2. also |X1 – X| = 4 Hz. When T increases, X1 also increases. It is given that beat frequency decreases to 2 Hz. X – X1 = 4 or X = 4 + X1 As X1 = X2 , X = 4 + X2 3v 3 × 340 = = 340 Hz 4l 4 × (3 / 4) ? X = 344 Hz. Now υ2 =

18. (b) : Let L be length of a closed organ pipe (i.e., a organ pipe closed at one end and open at the other). The frequency of 3rd fundamental mode of vibration in the closed pipe is 5v υ3 = , where v is the speed of sound in air. 4L The frequency of 4th fundamental mode of vibration 7v in the closed pipe is υ4 = 4L υ 5 Their ratio is 3 = υ4 7 19. (d) : y = A sink(x – vt) 1 1 = 20 Hz T = 0.05 s, ∴ υ = = T 0.05 s Velocity, v = 300 m s–1 −1 v 300 m s = = 15 m υ 20 Hz The point at 10 m from the source has a phase 2π 2π 4π ×x= × 10 = rad λ 15 3

∴ λ=

The point at 15 m from the source has phase 2S rad ? The phase difference between the points = 2π −

4 π 2π = rad 3 3 PHYSICS FOR YOU | DECEMBER ‘17

27

20. (c) : The given wave equation is x⎞ ⎛ x = x0 sin 2 π ⎜ υt − ⎟ ⎝ λ⎠ 2 πx ⎞ ⎛ x = x0 sin ⎜ 2 πυt − ⎟ ⎝ λ ⎠ Compare it with the standard wave equation x = Asin(Zt – kx) we get, Z = 2SX; k = 2S/O 2 πυ ω Wave velocity, v = = = υλ k (2 π / λ) Particle velocity,

The distance between successive nodes is equal to half the wavelength. Except (d), all the other statements are wrong. 26. (d) 27. (b) : For first overtone (2nd harmonic) n = 2 and for second overtone (3rd harmonic) n = 3

...(i)

dx 2 πx ⎞ ⎛ vp = = 2 πυx0 cos ⎜ 2 πυt − ⎟ ⎝ dt λ ⎠ Maximum particle velocity, (vp)max = 2SXx0 ...(ii) According to given problem, (vp)max = 4v (Using (i) and (ii)) ? 2SXx0 = 4XO 2 πυx0 πx0 λ= = 4υ 2 21. (a) 22. (b) : In case of a travelling wave, the reflection at a rigid boundary will take place with a phase reversal or with a phase difference of S or 180°. On the other hand, in case of a travelling wave for a reflection at an open boundary such as the open end of an organ pipe, the reflection takes place without any phase change. 23. (a) : Mass per unit length of the string, 3 kg m μ= = = 12 kg m–1 L 25 × 10−2 m Speed of the wave on the string is 400 N T v= = = 5.77 m s–1 μ 12 kg m−1 Time taken by disturbance to reach the other end

t=

−2

L 25 × 10 m = = 0.043 s v 5.77 m s−1

⎛ πx ⎞ 24. (a) : y = 4 sin ⎜ ⎟ cos (96 St) ⎝ 15 ⎠ π 2π π or = ⇒ λ = 30 units λ 15 15 Distance between node and neighbouring antinode λ 30 is = = 7.5 units. 4 4 25. (d) : In a tube open at both ends, all the harmonics are present. In a tube open at one end, only odd harmonics are present. ⇒ k=

28

PHYSICS FOR YOU | DECEMBER ‘17

2 ∴ υA = lA DA

T πρ

T 3 = πρ l D A B 2 As XA = XB (Given) υB =

∴

3 lB DB

T = πρ

2 lADA

lA 2 1 = = lB 6 3

3 ⎛D ⎞ lB ⎜ A ⎟ ⎝ 2 ⎠

T πp

(' DA = 2DB)

T πρ

γRT ∴ Δv = 1 ΔT , v 2 T M Δv 1 1 = × × 100 = 0.167% 100 × v 2 300 29. (c) : y = 0.1 sin(100St – kx) ω 100 π Z = 100S ∴ k = = = π m−1 v 100 dy ⎛ dy ⎞ 30. (d) : Particle velocity vPa = = −v ⎜ ⎟ ⎝ dx ⎠ dt dy = − wave velocity × slope of the wave or dt 28. (a) : As v =

y C B A

D H

E

x

F G

(a) For upward velocity, vPa = positive, so slope must be negative which is at the points D, E and F. (b) For downward velocity, vPa = negative, so slope must be negative which is at the points A, B and H. (c) For zero velocity, slope must be zero which is at C and G. (d) For maximum magnitude of velocity, |slope| should be maximum, which is at A and E. Hence, option (d) is wrong.

Series 6 CHAPTERWISE PRACTICE PAPER THERMODYNAMICS | KINETIC THEORY Time Allowed : 3 hours

Maximum Marks : 70 GENERAL INSTRUCTIONS

(i) (ii) (iii) (iv) (v) (vi) (vii)

All questions are compulsory. Q. no. 1 to 5 are very short answer questions and carry 1 mark each. Q. no. 6 to 10 are short answer questions and carry 2 marks each. Q. no. 11 to 22 are also short answer questions and carry 3 marks each. Q. no. 23 is a value based question and carries 4 marks. Q. no. 24 to 26 are long answer questions and carry 5 marks each. Use log tables if necessary, use of calculators is not allowed.

SECTION - A

1. If an inflated tyre bursts, the air escaping out is cooled. Why? 2. What is the nature of P-V diagram for isobaric and isochoric processes? 3. The graph shows the variation of PV with P of given masses of three gases A, B and C. If the temperature is kept constant, which of these is ideal graph? 4. When a gas filled in a closed vessel is heated through 1Â°C, its pressure increases by 0.4%. What is the initial temperature of the gas? 5. Discuss the results when a thermos flask containing hot tea is vigorously shaken. SECTION - B

6. Show that the pressure exerted by the gas is two-third of the average kinetic energy per unit volume of the gas molecules.

7. Write at least four fundamental assumptions regarding the model of a gas on which the kinetic theory is based. 8. There are n molecules of a gas in a box. If the number of molecules is increased to 2n, what will be the effect on the pressure and the total energy of the gas? 9. By applying the first law of thermodynamics to isobaric process, obtain a relation between two specific heats of a gas. OR State two conditions of a reversible process. What is the quasi-static process? 10. No real engine can have an efficiency greater than that of a Carnot engine working between the same two temperatures. Give reason. SECTION - C 11. A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom? PHYSICS FOR YOU | DECEMBER â€˜17

29

12. A gaseous mixture contain 16 g of helium and 16 g of oxygen, then calculate the ratio of CP/CV of the mixture. 13. Two thermally insulated vessels 1 and 2 are filled with air at temperatures (T1, T2), volumes (V1, V2) and pressures (P1, P2) respectively. If the valve joining the two vessels is opened, what will be the temperature inside the vessel at equilibrium ? 14. Temperatures of the hot and cold reservoirs of a Carnot engine is raised by equal amounts. How the efficiency of the Carnot engine affected? 15. Derive the P-V relation for a monatomic gas undergoing an adiabatic process. 16. How much energy in watt hour may be required to convert 2 kg of water into ice at 0°C, assuming that the refrigerator is ideal? Given temperature of freezer is –15°C, room temperature is 25°C and initial temperature of water is 25°C, specific heat of water = 4.2 × 103 J kg–1 K–1, latent heat of fusion = 3.36 × 105 J kg–1. 17. At what temperature the root mean square velocity is equal to escape velocity from the surface of earth for hydrogen molecule and for oxygen molecule? (Given radius of earth = 6.4 × 106m, g = 9.8 m s–2). 18. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 °C. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J mol–1 K–1, molecular mass of O2 = 32 u). 19. A gaseous mixture enclosed in a vessel consists of 5⎞ ⎛ 1 g mole of a gas A ⎜ γ = ⎟ and some amount of ⎝ 3⎠ 7⎞ ⎛ gas B ⎜ γ = ⎟ at a temperature T. The gases A and B ⎝ 5⎠ do not react with each other and are assumed to be ideal. Find number of gram moles of the gas B, if J for the gaseous mixture is ⎜⎛ 19 ⎞⎟ . ⎝ 13 ⎠ 20. The pressure of a gas changes linearly with volume from 10 kPa, 200 cm3 to 50 kPa, 50 cm3 as shown in figure. (a) Calculate the work done by the gas. (b) What is the effect on the internal energy of the gas? 30

PHYSICS FOR YOU | DECEMBER ‘17

21. Two glass bulbs of equal volumes are connected by a narrow tube and are filled with a gas at 0°C and a pressure of 76 cm of mercury. One of the bulbs is then placed in melting ice and the other in a waterbath maintained at 62°C. What is the new value of the pressure inside the bulbs? The volume of the connecting tube is negligible. OR N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel which is maintained at a temperature T. If the mean square velocity of the molecules of B type is denoted by v2 and the mean square of the x component of the velocity of A is denoted by Z2, then what is the value of

ω2

? v2 22. Two cylinders A and B of equal capacity are connected to each other via a stopcock. The cylinder A contains a gas at standard temperature and pressure, while the cylinder B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following : (a) What is the final pressure of the gas in A and B? (b) What is the change in internal energy of the gas? (c) What is the change in the temperature of the gas? (d) Do the intermediate states of the system (before setting to the final equilibrium state) lie on its P-V-T surface?

SECTION - D 23. Manvi and her friends were enjoying the birthday party of one of their friends in her home. They all were enjoying themselves by playing various songs and dancing. One of her friends became sweaty and was feeling very hot. As there was no air conditioner, she got an idea. She opened the door of the refrigerator thinking that this might relieve her from heat. On seeing this, Manvi immediately rushed towards her and made her understand that

opening the door of refrigerator would increases the temperature of the room. She told her to rest for sometime, if she was feeling very hot. Her friend understood this and closed the door of refrigerator at once. (a) What qualities of Manvi do you appreciate? (b) Temperature inside an ideal refrigerator is 275 K. How much heat is delivered to room for every one joule of work done on working substance when room temperature is 315 K? (c) A room cannot be cooled by opening the door of refrigerator in a closed room. Why? SECTION - E

24. Two Carnot engines A and B are operated in series. The first engine A receives heat at 800 K and rejects to a reservoir at temperature T. The second engine B receives the heat rejected by the first engine and in turn rejects to a heat reservoir at 300 K. Calculate the temperature T for the following cases : (a) When the outputs of the two engines are equal. (b) When the efficiencies of the two engines are equal. OR (a) Define an adiabatic process. Derive an expression for work done during an adiabatic process for one mole of gas. (b) A cylinder containing one gram molecule of the gas was compressed adiabatically until its temperature increases from 27°C to 97°C. Calculate the work done on the gas [Take J as 1.5.] 25. Using the law of equipartition of energy, predict the specific heat of water. How does the specific heat of water vary with temperature? Define one calorie. OR Using kinetic theory of gases, derive : (a) Boyle’s law (b) Charles’ law (c) Ideal gas equation (d) Avogadro’s law (e) Dalton’s law of partial pressure. 26. Consider one mole of perfect gas in a cylinder of unit cross section with a piston attached as shown in figure. A spring (spring constant k) is attached (unstretched length L) to the piston and to the bottom of the cylinder. Initially the spring is unstretched and the gas is in equilibrium. A certain amount of heat Q is supplied to the gas causing an increase of volume from Vo to V

(a) What is the initial pressure of the system? (b) What is the final pressure of the system? (c) Using the first law of thermodynamics, write down a relation between Q, Pa, V, Vo and k. OR (a) Derive an expression for work done during isothermal process. (b) Three moles of an ideal gas kept at constant temperature of 300 K are compressed from a volume of 4 L to 1 L. Calculate the work done in the process. Take R as 8.31 J mol–1K–1. SOLUTIONS

1. When the tyre bursts, there is adiabatic expansion of air because the pressure of the air inside the tyre is sufficiently greater than the atmospheric pressure. During the expansion, the air does some work against the surroundings, therefore, the internal energy decreases and as such the temperature falls. 2. (i) For an isobaric process, the P-V diagram is a straight line parallel to the volume-axis. (ii) For an isochroic process, the P-V diagram is a straight line parallel to the pressure-axis. 3. Gas C is ideal because PV is constant for the gas. It means C obeys Boyle’s law at all pressures. 4. Consider that initially gas has pressure P at temperature T. According to question, 0.4 P′ = P + P, T ′ = T + 1 100 0.4 ⎞ ⎛ P⎟ ⎜⎝ P + P 100 ⎠ By Gay Lussac’s law, = T T +1 On solving, T = 250 K 5. Work done in vigorously shaking a thermos flask is converted into internal energy of the tea inside it which raises its temperature. Further, along with the hot tea, a lot of vapours are also present. Due to an increase in temperature, the pressure of these vapours may further increase and in the extreme cases, the cork of the bottle may be blown out. 6. According to kinetic theory of gases, the pressure P exerted by a gas of density U and r.m.s. velocity v is 1 given by P = ρv 2 3 Average kinetic energy of translation per unit volume of the gas is given by, 1 (' m = UV) E = ρv 2 2 PHYSICS FOR YOU | DECEMBER ‘17

31

1 2 ρv 2 2 P 3 ? = = ;P= E 3 E 1 2 3 ρv 2 2 P = × Average kinetic energy per unit volume 3 7. The fundamental assumptions are (i) A gas comprises of large number of small particles called molecules. (ii) The molecules of a gas are in random motion and obey Newton’s law of motion. (iii) The collisions are elastic and of negligible duration. (iv) The volume occupied by the molecules is negligibly a small fraction of the volume occupied by the gas. 1 mnv 2 , i.e., P v n, therefore, pressure 3 V will be double when number of molecules n becomes 2n. 1 As kinetic energy of the gas = mnv 2 , therefore, 2 kinetic energy of the gas also becomes double when n changes to 2n.

8. Since P =

9. In an isobaric process, pressure remains constant. If an amount of heat dQ is supplied to one mole of a gas at constant pressure and its temperature increases by dT, then dQ = CPdT Here CP is molar specific heat of the gas at constant pressure. Therefore, for an isobaric process, the first law of thermodynamics becomes CP dT = dU + PdV ...(i) From perfect gas equation it follows that PdV = RdT and dU = CV dT In the eqn. (i), substituting PdV and dU, we have CPdT = CVdT + RdT C P = CV + R OR (i) The process should take place very slowly, so that it satisfies the following requirements: x The system should be in thermal equilibrium. x The system should be in mechanical equilibrium. x The system should be in chemical equilibrium. 32

PHYSICS FOR YOU | DECEMBER ‘17

(ii) There should be no frictional losses. A process in which the system passes through a continuous sequence of equilibrium states is said to be quasi-static, i.e., apparently static. 10. A Carnot engine is an ideal engine satisfying the following conditions : (i) There is no friction between the walls of the cylinder and the piston. (ii) The working substance is an ideal gas i.e., the gas molecules have point sizes and have no attractive forces between them. Real engines cannot fulfil these conditions. Hence no heat engine working between the same two temperatures can have efficiency greater than that of a Carnot engine. 11. When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air 15 cm. A length of 9 cm of the tube will be left at the open end, figure (a). The pressure of air enclosed in tube will be atmospheric pressure. Let area of crosssection of the tube be 1 cm2.

? P1 = 76 cm of mercury and V1 = 15 cm3 When the tube is held vertically, 15 cm air gets another 9 cm of air filled in the right hand side (in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure, figure (b). Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively. The pressure of air = 76 – (76 – h) = h cm of mercury ? V2 = (24 + h) cm3 and P2 = h cm of mercury. If we assume that temperature remains constant, then P1V1 = P2V2 or 76 × 15 = h × (24 + h) or h2 + 24h – 1140 = 0 or h =

−24 ± ( 24)2 + 4 × 1140 2

= 23.8 cm or – 47.8 cm Since h cannot be negative (because more mercury cannot flow into the tube), therefore h = 23.8 cm. Thus, in the vertical position of the tube, 23.8 cm of mercury flows out.

16 =4 4 16 1 Number of moles of oxygen (nO ) = = 2 32 2 As helium is monatomic, so degrees of freedom of helium f = 3, so f 3 CV = R = R He 2 2 As oxygen, is diatomic, so degrees of freedom of oxygen f = 5, so 5 f CV = R = R O2 2 2

12. Number of moles of helium (nHe ) =

∴ CV

mixture

=

nHe CV

He

+ nO CV 2

3 1 5 4 × R + × R 29 2 2 2 = = R 1 18 4+ 2 R R γ mixture = 1 + = 1+ = 1.62 29 CV R mixture 18 ∴ n=

...(ii) Dividing eqn. (ii) by (i), we have ⎛ T1 − T2 ⎞ ⎜ T1 η′ ⎝ T1 + T ⎟⎠ = <1 = η ⎛ T1 − T2 ⎞ T1 + T ⎜⎝ T ⎟⎠ 1 As Kc < K i.e., the efficiency of Carnot engine decreases.

O2

nHe + nO 2

13. Using gas equation, PV = nRT PV For first vessel, n1 = 1 1 RT1

As given the temperature of both the reservoirs is raised by equal amount T, so T1c = T1 + T and T2c = T2 + T. The final efficiency of the Carnot engine will be T ′ − T2 ′ (T1 + T ) − (T2 + T ) T1 − T2 = = η′ = 1 (T1 + T ) (T1 + T ) T1 ′

PV RT

P2V2 For second vessel, n2 = RT2

P(V1 + V2 ) For the combined vessel, n = RT But n = n1 + n2 P (V1 + V2 ) P1V1 P2V2 ∴ = + RT RT1 RT2 T T P(V1 + V2 ) or T = 1 2 P1V1T2 + P2V2T1 Using energy conservation, P(V1 + V2) = P1V1 + P2V2 T T (P V + P V ) Hence T = 1 2 1 1 2 2 P1V1T2 + P2V2T1

14. Let the initial temperatures of hot and cold reservoirs were T1 and T2. The efficiency of the Carnot engine is given by T −T ...(i) η= 1 2 T1

15. From first law of thermodynamics, dQ = dU + dW, For an adiabatic process, dQ = 0, ? dU = – dW ...(i) 3 For a monoatomic gas, U = RT and as such 2 3 ...(ii) dU = RdT 2 RT Also dW = PdV = ...(iii) dV V From eqns. (i), (ii) and (iii), RT 3 dV RdT = − V 2 dT 2 dV 2 + = 0 or ln T + ln V = constant T 3 V 3 2/3 or TV = constant PV As PV = RT, T = R ⎛ PV ⎞ 2 /3 Thus, ⎜ = constant or PV5/3 = constant V ⎝ R ⎟⎠ 16. Here, T1 = 25 + 273 = 298 K, T2 = –15 + 273 = 258 K Specific heat of water, s = 4.2 × 103 J kg–1 K–1 Latent heat of ice, L = 3.36 × 105 J kg–1 Amount of heat required to be removed from 2 kg of water at 25°C to change it into ice at 0°C, Q2 = ms(Ti – Tf) + mL = 2 × 4.2 × 103 (298 – 273) + 2 × 3.36 × 105 = 2.1 × 105 + 6.72 × 105 = 8.82 × 105 J Heat rejected to the surroundings, T 298 Q1 = 1 × Q2 = × 8.82 × 105 = 10.19 × 105 J T2 258 or

PHYSICS FOR YOU | DECEMBER ‘17

33

Energy supplied to convert water into ice, W = Q1 – Q2 = (10.19 – 8.82) × 105 = 1.37 × 105 J =

1.37 × 105 ≈ 38 W h 3600

[' 1 W h = 3600 J]

17. Escape velocity, ve = 2 gR From kinetic interpretation of temperature 1 2 3 mvrms = kBT 2 2 2mgR 3 1 m(2 gR) = kBT or T = or 3kB 2 2

=

2mH gR 3kB

2 ⎛ ⎞ × 10−3 ⎟ × 9.8 × 6.4 × 106 2×⎜ ⎝ 6.02 × 1023 ⎠ 3 × 1.38 × 10−23

= 1.01 × 104 K For oxygen, TO =

=

32 ⎞ ⎛ 2×⎜ × 10−3 ⎟ × 9.8 × 6.4 × 106 ⎠ ⎝ 6.02 × 1023 −23

4

= 16.1 × 10 K 18. Here, initially in the oxygen cylinder Volume V1 = 30 litres = 30 × 10–3 m3 Pressure P1 = 15 atm = 15 × 1.013 × 105 Pa Temperature T1 = 27 + 273 = 300 K If the cylinder contains n1 moles of the oxygen gas then P1V1 = n1RT1, n1 =

P1V1 15 × 1.013 × 10 5 × 30 × 10 −3 = = 18.285 RT1 8.31 × 300

Initial mass of oxygen in the cylinder m1 = n1 × molecular weight of O2 = 18.285 × 32 = 585.1 g Finally in the oxygen cylinder, let n2 moles of oxygen be left Final volume, V2 = 30 × 10–3 m3 Final pressure, P2 = 11 atm = 11 × 1.013 × 105 Pa Final temperature, T2 = 17 + 273 = 290 K Now, P2V2 = n2RT2 n2 =

34

P2V2 (11 × 1.013 × 10 5 ) × ( 30 × 10 −3 ) = 13.871 = 8.31 × 290 RT2 PHYSICS FOR YOU | DECEMBER ‘17

5 3 R = R , (CV )B = R 5 2 2 −1 3 13 R and (CV )mix = = R 19 6 −1 13

? (CV )A =

Now, from conservation of energy Umix = UA + UB or 'Umix = 'UA + 'UB (nA + nB) (CV)mix 'T = nA(CV)A'T + nB(CV)B'T

2mO gR 3kB

3 × 1.38 × 10

CP = γ and CP − CV = R CV So, combining above equations we get R CV = γ −1

19. For an ideal gas,

Given, vrms = ve = 2 gR

For hydrogen, TH =

Therefore, final mass of the oxygen gas in the cylinder is m2 = n2 × molecular weight of O2 = 13.871 × 32 = 443.9 g ? Mass of oxygen gas taken out = m1 – m2 = (585.1 – 443.9) | 141 g

n (C ) + nB (CV )B or (CV )mix = A V A n A + nB 3 5 1 × R + nB × R 13 2 2 R= 1 + nB 6 or 13 + 13nB = 9 + 15nB or nB = 2 g mol 20. (a) From figure, 'W = area of the shaded portion 1 = (50 + 10)(200 − 50) = 4500 kPa cm3 = 4.5 J 2 (' 1 kPa cm3 = 10–3 J) (b) As gas is compressed its pressure increases but volume decreases. Consequently temperature of gas is also increased. Hence internal energy of the gas increases. 21. According to the gas equation, PV = nRT, since the mass of a system remains constant, the sum of the terms PV/T for different parts of the system remains unchanged. Let V be the volume of the each bulb. Initially for the two bulbs, we have P1V1 P2V2 76 V 76 V 2 × 76 × V ...(i) + = + = T1 T2 273 273 273 When one bulb is placed at melting ice and another maintained at 62°C we have

P1′V1′

P ′V ′ PV PV + 2 2 = + 273 355 T1′ T2′

...(ii)

From (i) and (ii), we get 2 × 76 × V PV PV = + 273 273 335 or P =

2 × 76 273 × 335 × = 83.75 cm of Hg . 273 608

OR The mean square velocity of the gas molecules, v2 = 3kT/m For gas A : vA2 = 3kT/m and ...(i) For gas B : vB2 = 3kT/(2 m) = v2 The molecule A has equal probability of motion in all directions, therefore vx2 = vy2 = vz2 = Z2 (given) ? vA2 = vx2 + vy2 + vz2 = 3vx2 = 3Z2 v 2 1 ⎛ 3kT ⎞ kT or ω2 = A = ⎜ ⎟= 3 3⎝ m ⎠ m Dividing (ii) by (i), we get :

ω2 2

...(ii) =

2 kT / m = 3kT / 2m 3

v 22. (a) When the stopcock is suddenly opened, the volume available for the gas at 1 atm pressure will become two times. Therefore, pressure will decrease to one-half, i.e., 0.5 atm. (b) There will be no change in the internal energy of the gas as no work is done on/by the gas. (c) Also, there will be no change in temperature of the gas as gas does no work in expansion. (d) No, because the process called free expansion is rapid and cannot be controlled. The intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas returns to an equilibrium state. 23. (a) Manvi is sensible girl and she has scientific knowledge. (b) T2 = 275 K, T1 = 315 K, W = 1 J Q T2 275 β= 2 = = = 6. 9 W T1 − T2 315 − 275 Q2 = 6.9 W = 6.9 J ; Q1 = Q2 + W = 6.9 + 1 = 7.9 J (c) When a refrigerator is working in a closed room with its door opened, it will reject heat from inside to the air in the room. So, temperature of room increases gradually.

24. For engine, A, T1 = 800 K, T2 = T T T Efficiency, ηA = 1 − 2 = 1 − T1 800 Work output, WA = Q1 – Q2 = KA × Q1 Q2 ⎞ ⎛ ⎜⎝' ηA = 1 − Q ⎟⎠ 1 T ⎞ ⎛ or WA = ⎜1 − Q ⎝ 800 ⎟⎠ 1 For engine B, T1c = T, T2c = 300 K T ′ 300 Efficiency, ηB = 1 − 2 = 1 − T1 ′ T Work output, WB = Q1c – Q2c = KB × Q1c ⎛ 300 ⎞ or WB = ⎜⎝1 − ⎟Q ′ T ⎠ 1 Since, the engine B absorbs the heat rejected by the engine A, so ⎛ 300 ⎞ Q1c = Q2 ∴ WB = ⎜1 − ⎟Q ⎝ T ⎠ 2 (a) When outputs of the two engines are equal. WA = WB T ⎞ ⎛ ⎛ 300 ⎞ or ⎜1 − Q = 1− ⎟Q ⎝ 800 ⎟⎠ 1 ⎜⎝ T ⎠ 2 T ⎞ ⎛ 300 ⎞ Q2 ⎛ 300 ⎞ T ⎛ or ⎜1 − = 1− = ⎜1 − ⎟ ⎟ ⎝ 800 ⎟⎠ ⎜⎝ T ⎠ 800 T ⎠ Q1 ⎝ On solving, we get T = 550 K (b) When the efficiency are equal, KA = KB 300 T ⎞ ⎛ or ⎜1 − or T 2 = 24 × 104 = 1− ⎝ 800 ⎟⎠ T ? T = 489.9 K OR (a) When no exchange of heat energy is possible between system and surroundings, the process is known as adiabatic. Workdone in adiabatic process : Consider 1 mole of gas contained in a perfectly nonfrictionless non-conducting cylinder fitted with a piston (non conducting). Let P = pressure of gas, V = volume of gas and T = temperature of gas Compress the gas adiabatically by moving the piston inwards through a distance dx. Force acting on piston = P × A Work done, dW = Force × distance = PA dx = PdV where dV is the change in volume of the gas. PHYSICS FOR YOU | DECEMBER ‘17

35

During the adiabatic expansion, total work done by the gas, W=

V2

One calorie is defined as the amount of heat required to raise the temperature of 1 g of water from 14.5°C to 15.5°C. 1 calorie = 4.186 J

∫ P dV

V1

Adiabatic change is represented by PV J = K, where K is a constant. K ∴ P= Vγ Hence, W =

V2

K

∫ V γ dV

V1

=

K (V21− γ − V11− γ ) 1− γ

1 (P2V2γV21− γ − P1V1γV11− γ ) 1− γ (' P1V1J = P2V2J = K) R 1 (T − T ) (RT2 − RT1 ) = = γ 2 1 1 − 1− γ This is the work done during adiabatic process. (b) Given : Ti = 27°C = 27 + 273 = 300 K Tf = 97°C = 97 + 273 = 370 K, J = 1.5 =

Work done in adiabatic compression is given by R 8.31 W= (370 – 300) = –1163 J (Tf – Ti) = 1− γ 1 − 1.5 25. We may treat water like a solid. By the law of equipartition of energy, the average vibrational energy per atom is 3 kBT. Now a water molecule has three atoms : two hydrogen and one oxygen. ? Average vibrational energy per water molecule = 3 × 3kBT = 9 kBT The total vibrational or the internal energy of one mole of water molecules, U = NA × 9kBT = 9RT [' R = kBNA] Neglecting 'V, like for a solid, we get 'Q = 'U + P'V = 'U ΔQ ΔU ∴ CV = = = 9R ΔT ΔT This predicted value is found to be in good agreement with the observed value. The specific heat of water is nearly 75 J mol–1 K–1 | 9R. Variation of specific heat of water with temperature : Figure shows the variation of specific heat of water with temperature in the temperature range 0°C to 100°C. Water shows peculiar behaviour, its specific heat first decreases and then increases with temperature. For this reason, we have to specify the unit temperature interval for defining calorie. 36

PHYSICS FOR YOU | DECEMBER ‘17

OR 2 NK 3 where K is the average kinetic energy of translation per gas molecule. (a) Boyle’s law : At constant temperature, K is constant and for a given mass of the gas, N is constant. Thus, PV = constant for given mass of gas at constant temperature, which is the Boyle’s law. (b) Charles’ law : For a given mass of gas, N is constant 3 Since K = kBT , K ∝ T and as such PV v T. 2 If P is constant, V v T, which is the Charles’ law. (c) Ideal gas equation : As 2 3 PV = N K and K = kBT , 3 2 2 ⎛3 ⎞ PV = N ⎜ kBT ⎟ or PV = NkBT ⎠ 3 ⎝2 From kinetic theory of gases, PV =

which is the ideal gas equation (d) Avogadro’s law : Consider two gases 1 and 2. We can write 2 2 P1V1 = N1 K 1 , P2V2 = N 2 K 2 3 3 If their pressure, volumes and temperatures are the same, then P1 = P2 , V1 = V2 , K1 = K 2 Clearly, N1 = N2. Thus, equal volumes of all ideal gases existing under the same conditions of temperature and pressure contain equal number of molecules which is Avogadro’s law. (e) Dalton’s law of partial pressures : The kinetic theory of gases attributes the gas pressure to the bombardment on the walls of the containing vessel by molecules. In a mixture of ideal gases, we might therefore expect the total pressure (P) to be the sum of the partial pressures (P1, P2, ...) due to each gas, i.e.,

P = P1 + P2 + ... =

⎡ V PV ⎤ ∴ ΔU = CV ⎢(Pa + k(V − V0 )) − a 0 ⎥ R R ⎦ ⎣ ...(iii)

2 N1 2 N2 K 2 + ... K1 + 3 V1 3 V2

In equilibrium, the average kinetic energy of the molecules of different gases will be equal, i.e., 3 K 1 = K 2 = ... = K = kBT 2 2 ⎛3 ⎞ Thus, P = (n1 + n2 + ...) ⎜ kBT ⎟ ⎝2 ⎠ 3 = (n1 + n2 + ...) kBT, where n1 =

It represents Dalton’s law of partial pressures which states that the resultant pressure exerted by a mixture of gases or vapours which do not interact in any way is equal to the sum of their individual (i.e., partial) pressures. 26. Given, cross-sectional area of the piston A = 1 m2 Heat supplied = Q, atmospheric pressure = Pa Initial volume = V0, final volume = V (a) Since system is in equilibrium so, initial pressure of the system, Pi = Pa. (b) When the gas expands from V0 to V after heat Q is supplied. Change in volume of the gas = V – V0 (V − V0 ) = V − V0 A

(' A = 1 m2) Force applied by the spring on the piston F = kx = k(V – V0) So, pressure exerted by the extended spring on the piston of unit cross-section P=

1 [(P + k(V − V0 ))V − PaV0 ] + Pa (V − V0 ) (γ − 1) a 1 + k(V − V0 )2 (' x = V − V0 ) 2 OR (a) According to the first law of thermodynamics, dQ = dU + dW ...(i) In an isothermal expansion, temperature remains constant. As such there is no change in the internal energy of the gas, i.e., dU = 0. From eqn. (i), dQ = dW Thus, in this case, the entire heat is converted into work. As dW = PdV and for one mole of a gas i.e., n = 1, RT RT PV = RT, P = , dW = dV V V If the volume of the gas increases from V1 to V2, then total work done during expansion, =

N N1 , n2 = 2 ,.... V2 V1

Extension in the spring, x =

From (i), (ii) and (iii) 1 [(P + k(V − V0 ))V − PaV0 ] Q= (γ − 1) a ⎛ C 1 1 ⎞ + Pa (V − V0 ) + kx 2 ⎜' V = R (γ − 1) ⎟⎠ 2 ⎝

F k = (V − V0 ) = k(V − V0 ) A A

∫

V1

RT V dV = RT [lnV ]V2 1 V

or W = RT(ln V2 – lnV1) ⎛V ⎞ ⎛V ⎞ or W = RT ln 2 = 2.303 RT log 2 ⎜⎝ V ⎟⎠ ⎜⎝ V ⎟⎠ 1 1 For n moles of a gas,

Hence the final pressure of the system, Pf = Pa + P = Pa + k(V – V0). (c) Using the first law of thermodynamics, Q = 'Q = 'U + 'W Here, 'U = CV(T – T0) 1 ΔW = Pa (V − V0 ) + kx 2 2 PaV0 Also, T0 = R P f V [Pa + k(V − V0 )]V T= = R R

V2

W=

⎛V ⎞ ⎛V ⎞ W = nRT ln ⎜ 2 ⎟ = 2.303 nRT log 2 ⎜⎝ V ⎟⎠ ⎝ V1 ⎠ 1 ...(i) ...(ii)

(b) n = 3, T = 300 K, Vi = 4 L, Vf = 1 L, R = 8.31 J mol–1 K–1, W = ? Work done in isothermal process is given by Vf W = nRT ln Vi 1 = 3 × 8.31 × 300 ln = –1.037 × 104 J 4 PHYSICS FOR YOU | DECEMBER ‘17

37

MPP-8

Class XI

T

his specially designed column enables students to self analyse their extent of understanding of speciﬁed chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.

Mechanical Properties of Solids and Fluids Total Marks : 120

Time Taken : 60 min NEET / AIIMS

Only One Option Correct Type

1. If pressure at half the depth of a lake is equal to 2/3 pressure at the bottom of the lake then what is the depth of the lake? (Given : 1 atm = 105 Pa, g = 10 m s–2 and density of water = 103 kg m–3) (a) 10 m (b) 20 m (c) 60 m (d) 30 m 2. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30°. The angle of shear is (a) 12° (b) 1.2° (c) 0.12° (d) 0.012° 3. Two capillaries of length L and 2 L and of radii R and 2 R are connected in series. The rate of the flow of fluid through single capillary tube of length L, radius R is Q. The net rate of flow of fluid through the two capillaries in series is (a) (5/6) Q (b) (6/7) Q (c) (7/8) Q (d) (8/9) Q 4. A block of aluminium of mass 1 kg and volume 3.6 × 10–4 m3 is suspended from a string and then completely immersed in a container of water. The decrease in tension in the string after immersion is (Given : g = 10 m s–2 and density of water = 103 kg m–3) (a) 9.8 N (b) 6.2 N (c) 3.6 N (d) 1.0 N 5. Two very wide parallel glass plates are held vertically at a small separation r, and dipped in water of surface tension S. Some water climbs up in the gap between the plates. If P0 is the atmospheric pressure, then the pressure of water just below the water surface in the region between two plates is 38

PHYSICS FOR YOU | DECEMBER ‘17

2S 2S (b) P0 + r r 4S 4S (c) P0 − (d) P0 + r r 6. A cube floats in water with 1/3rd part is outside the surface of water and it floats in another liquid with 3/4th part is outside the liquid then the density of liquid is (a) 8/3 (b) 2/3 (c) 4/3 (d) 5/3 (a) P0 −

7. A closed vessel is half filled with water. There is a hole near the top of the vessel and air is pumped out from this hole. (i) The water level will rise up in the vessel. (ii) The pressure at the surface of the water will decrease. (iii) The force by the water on the bottom of the vessel will decrease. (iv) The density of the liquid will decrease. Which of the following statements are correct? (a) (i), (ii) only (b) (i), (iii) only (c) (ii), (iii) only (d) (i), (iv) only 8. A vertical capillary is brought in contact with the water surface (surface tension = T). The radius of the capillary is r and the contact angle T = 0°. The increase in potential energy of the water (density = U) is (a) independent of U (b) independent of r (c) independent of T (d) zero 9. The normal density of gold is U and its bulk modulus is K. The increase in density of a piece of gold when a pressure P is applied uniformly from all sides is

Kρ Pρ Kρ Pρ (b) (c) (d) K −P 2P 2K K −P 10. A cylindrical drum, open at the top, contains 30 litres of water. It drains out through a small opening at the bottom. 10 litres of water comes out in time t1, the next 10 litres in a further time t2 and the last 10 litres in a further time t3. Then, (a) t1 = t2 = t3 (b) t1 > t2 >t3 (c) t1 < t2 < t3 (d) t2 > t1 = t3 (a)

11. A uniform rod of length L has a mass per unit length O and area of cross-section A. The elongation in the rod is l due to its own weight if it is suspended from the ceiling of a room. The Young’s modulus of the rod is 2λgL λgl 2 λgL2 2λgL2 (b) (c) (d) Al AL Al 2 Al 12. A liquid of density U comes out with a velocity v from a horizontal tube of area of cross-section A. The reaction force exerted by the liquid on the tube is F. Choose the incorrect option. (a) F v v (b) F v v2 (c) F v A (d) F v U (a)

Assertion & Reason Type

'LUHFWLRQV In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false. 13. Assertion : The shape of an automobile is so designed that its front resembles the stream line pattern of the fluid through which it moves. Reason : The resistance offered by the fluid in case of streamline flow is maximum.

Reason : As the oil is poured in the situation of assertion, pressure inside the water will increase everywhere resulting in an increase in upward force on the object. JEE MAIN / JEE ADVANCED Only One Option Correct Type

16. There is a small hole in the bottom of a fixed container containing a liquid up to height h. The top of the liquid as well as the hole at the bottom are exposed to atmosphere. As the liquid comes out of the hole, the top surface of the liquid (Area of the hole is a and that of the top surface is A) (a) accelerates with acceleration = g a (b) accelerates with acceleration = g A a (c) retards with retardation = g A ga2 (d) retards with retardation = 2 A 17. A ball is thrown vertically upwards at time t = 0. Air resistance is not negligible and the acceleration of free fall is g. The ball reaches a maximum height at time t = T and the descends, reaching terminal speed. Which graph best shows the variation with time t of the acceleration a of the ball? a

a

+g

(a)

+g

O –g

T

t

(b) O

t

a

a +g

+g

(c) O

T

T

t

(d) O

T

t

14. Assertion : Viscosity of a liquid is the property of the liquid by virtue of which it opposes the relative motion amongst its different layers. Reason : The viscosity of liquid increases with increase in temperature.

18. A thin metallic rod of length 0.5 m and radius 0.01 m is rotated with an angular velocity 400 rad s–1 in a horizontal plane about a vertical axis passing through one of its ends. If the density of the material of the rod is 104 kg m–3 and Y = 2 × 1011 N m–2, the elongation in the rod is (a) 1/6 mm (b) 1/3 mm (c) 1/2 mm (d) 1 mm

15. Assertion : Consider an object that floats in water but sinks in oil. When the object floats in water half of it is submerged. If we slowly pour oil on top of water till it completely covers the object, the object moves up.

19. A liquid is flowing through horizontal pipes as shown in figure. Lengths of different pipes has the ratio L L LAB = LGH = CD = EF 2 2 PHYSICS FOR YOU | DECEMBER ‘17

39

C

A

section. The cross-sectional areas at point C is A and at E is A/2. Find the correct options.

D

B

G

E

F

Radii of different pipes has the ratio R RAB = RCD = RGH = EF 2 Pressure at A is 2 P0 and pressure at D is P0. The volume flow rate through the pipe AB is Q. Then ratio of the volume flow rate through the pipes CD and EF is (a) 1/17 (b) 1/16 (c) 16/17 (d) 17/16 More than One Options Correct Type

20. Two solid spheres A and B of equal volumes but of different densities UA and UB are connected by a string. They are fully immersed in a fluid of density UF. They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if (a) UA < UF (b) UB > UF (c) UA > UF (d) UA + UB = 2UF 21. A rod of length l and A B negligible mass is suspended at its two Al O Steel ends by two wires of steel (wire A) and aluminium m (wire B) of equal lengths (see figure). The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2, respectively. A block of mass m is suspended from any point O on the rod. (YAl = 70 × 109 N m–2 and Ysteel = 200 × 109 N m–2) (a) If point O is close to wire A then both the wires have equal stresses. (b) If point O is close to B then both the wires have equal stress. (c) If point O is at the middle of the wires then both the wires have equal stress. (d) If point O is close to wire A then both the wires have equal strain. 22. As shown in the figure a liquid of density U is standing in a sealed container to a height h. The container contains compressed air at a gauge pressure of p. The horizontal outlet pipe has non-uniform cross40

PHYSICS FOR YOU | DECEMBER ‘17

p

H

h

A

B

C

D

E

(a) The velocity of liquid at C will be 1/2

⎡ ( p + ρgh) ⎤ ⎢ 4ρ ⎥ . ⎣ ⎦ (b) The velocity of liquid at C will be 1/2

⎡ 2( p + ρgh) ⎤ ⎢ ⎥ . ρ ⎣ ⎦ A (c) The discharge rate is given by ( p + ρgh)1/2 . 2ρ A ( p + ρgh)1/2 . (d) The discharge rate is given by 2 ρ 23. A block of density 2000 kg m–3 and mass 10 kg is suspended by a spring of force constant 100 N m–1. The other end of the spring is attached to a fixed support. The block is completely submerged in a liquid of density 1000 kg m–3. If the block is in equilibrium position, (a) the elongation of the spring is 1 cm (b) the magnitude of buoyant force acting on the block is 50 N (c) the spring potential energy is 12.5 J (d) magnitude of spring force on the block is greater than the weight of the block. Integer Answer Type

24. Water is in streamline flow along a horizontal pipe with non uniform cross-section. At a point in the pipe where the area of cross-section is 10 cm2, the velocity of water is 1 m s–1 and the pressure is 2000 Pa. The pressure at another point where the cross-sectional area is 5 cm2 is 5 × 10x. Find x. 25. Two solid spheres of same metal but of mass M and 8M fall simultaneously on a viscous liquid and their terminal velocities are v and nv, then value of n is 26. A wire of initial length L and radius r is stretched by a length l. Another wire of same material but with initial length 2L and radius 2r is stretched by a length 2l. The ratio of the stored elastic energy per unit volume in the first and second wire is

Column I (A) Lift is accelerating vertically up. (B) Lift is accelerating vertically down with an acceleration less than the gravitational acceleration. (C) Lift is moving vertically up with constant speed. (D) Lift is falling freely.

Comprehension Type The figure shows an atomizer. Swift air flow is maintained by compressing the bulb A. The high velocity of air flow reduces pressure at the positions of vertical tube. Due to pressure difference liquid rises in the vertical tube and is sprayed out by high velocity air. If the gauge pressure in the bulb is P and v is the speed of the air in BC, take density of air V and that of liquid U. B

C

A h

(a) (b) (c) (d)

Liquid (density = U)

27. If atmospheric pressure is Pa, find the approximate pressure in BC. 1 1 (a) Pa + P + σv 2 (b) Pa + P − σv 2 2 2 1 2 1 2 (c) P − σv (d) P + σv 2 2 28. What is the velocity of liquid in BC ? (a) (c)

{ {

(P − ρgh) σ/2 (P + ρgh) σ/2

} }

1/2

(b)

1/2

(d)

{ {

(P + ρgh) σ (P − ρgh) σ

} }

1/2

1/2

Matrix Match Type 29. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift’s motion is given in column I and the distance where the water jet hits the floor of the lift is given in column II. Match the statements from column I with those in column II and select the correct answer using the given code.

A Q Q P Q

B R R P R

C Q P P P

Column II (P) d = 1.2 m (Q) d > 1.2 m

(R) d < 1.2 m (S)

No water leaks out of the jar

D S S S P

30. Column II is related to physical quantity and physical law given in column I. Match the following columns and select the correct option from the given codes. Column I Column II (A) Stoke’s law (P) Radius (B) Terminal velocity (Q) Density of material of the body (C) Excess pressure (R) Coefficient of inside mercury drop viscosity (D) Viscous force on (S) Surface tension a plate moving horizontally on a liquid surface (T) Velocity gradient A B C D (a) P, R P, Q, R P, S R, T (b) P, Q, R R, T P, R P, S (c) P, S P, R R, T P, Q, R, S (d) R P, R P P, S Keys are published in this issue. Search now! -

Check your score! If your score is > 90%

EXCELLENT WORK !

You are well prepared to take the challenge of ﬁnal exam.

No. of questions attempted

……

90-75%

GOOD WORK !

You can score good in the ﬁnal exam.

No. of questions correct

……

74-60%

SATISFACTORY !

You need to score more next time.

Marks scored in percentage

……

< 60%

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

PHYSICS FOR YOU | DECEMBER ‘17

41

Unit

6

MODERN PHYSICS

4E1<>1DEB5?6=1DD5B1>4B1491D9?> Electron Emission x x

x

44

The phenomenon of emission of electrons from the surface of a metal. The electron emission can be obtained from the following physical processes : ¾ Thermionic emission : It is the phenomenon of emission of electrons from the metal surface when heated suitably. ¾ Photoelectric emission : It is the phenomenon of emission of electrons from the surface of metal when light radiations of suitable frequency fall on it. ¾ Field emission or cold cathode emission : It is the phenomenon of emission of electrons from the surface of a metal under the application of a strong electric field. ¾ Secondary emission : It is the phenomenon of emission of electrons from the surface of metal in large number when fast moving electrons (called primary electrons) strike the metal surface. Work function : The minimum energy needed by an electron to come out from a metal surface is known as work function of the metal. It is denoted by I0 and measured in electron volt (eV). The work PHYSICS FOR YOU | DECEMBER ‘17

function depends on the properties of the metal and the nature of its surface. Photoelectric Effect x It is the phenomenon of emission of electrons from the surface of metal, when light radiations of suitable frequency fall on them. x The emitted electrons are known as photoelectrons and the current so produced is known as photoelectric current. x Einstein’s photoelectric equation : If a light of frequency X is incident on a photosensitive material having work function (I ), then maximum kinetic energy of the emitted electron is given as K = hX – I x For X > X or K = eV = hX – I = hX – hX (' X0 = Threshold frequency) ⎛1 1 ⎞ or eV0 = hc ⎜ − ⎟. ⎝ λ λ0 ⎠ where (V = the stopping potential) O = threshold wavelength O = incident wavelength x Einstein’s photoelectric equation is in accordance with the law of conservation of energy. x The photoelectric equation explains all the features of the photoelectric effect.

x

Experimental study of photoelectric effect

Variation of photoelectric current with intensity of incident light keeping accelerating potential and frequency constant

Graph

Inferences x

For a given metal and frequency of incident radiation, the number of photoelectrons emitted per second is directly proportional to the intensity of the incident light.

x

For a particular frequency of incident radiation, the minimum negative potential V0 given to the plate for which the photocurrent stops or become zero is called the cut-off or stopping potential. 1 2 Kmax = eV0 = mvmax 2 For a given frequency of incident light, the stopping potential is independent of the intensity.

Photoelectric current

Observation

Intensity of light

Photoelectric current

Variation of photoelectric current with collector plate potential for different intensities and same frequency of incident light.

I3 > I2 > I1 I3 I2 I1

Stopping potential â€“V0 Retarding potential

O Collector plate potential

x

x

Variation of photoelectric current with collector plate potential for different frequencies and same intensity of incident light.

x

x Collector plate potential

x

Variation of stopping potential V0 with frequency X of incident radiation for a given photosensitive material.

x B

The value of saturation current is independent of incident frequency. The stopping potential is more negative for higher frequencies of incident radiation. The energy of the emitted electrons depends on the frequency of incident light. For a given photosensitive material, stopping potential increases linearly with frequency of incident radiation. For a given metal, there exists a certain minimum frequency of the incident radiation below which no emission of photoelectrons takes place even if the intensity is large. This frequency is known as threshold frequency. PHYSICS FOR YOU | DECEMBER â€˜17

45

Illustration 1 : The photoelectric threshold wavelength for a metal is 10,000 Å. When light of wavelength 5461 Å is incident on it, the retarding potential in Millikan’s experiment is 1.02 V. Calculate the value of Planck’s constant. Soln.: If V0 is the stopping potential, then Einstein’s photoelectric equation can be written as 1 2 mvmax = eV0 = h(υ − υ0 ) 2 ⎡λ − λ ⎤ ⎡c c ⎤ or eV0 = h ⎢ − ⎥ = hc ⎢ 0 ⎥ λ λ 0⎦ ⎣ ⎣ λ0 λ ⎦ eλ 0 λ V0 h= (λ 0 − λ ) c =

1.6 × 10 −19 × 10000 × 10 −10 × 5461 × 10 −10 × 1.02

?

λp λα

=

=

x x

x

de Broglie wavelength is independent of the charge and nature of the material particle. In terms of kinetic energy K, de Broglie wavelength h . is given by λ = 2mK If a particle of charge q is accelerated through a potential difference V, its de Broglie wavelength is h . given by λ = 2mqV 1/2 h ⎛ 150 ⎞ ¾ For an electron, λ = = 2meeV ⎝ V ⎠

46

PHYSICS FOR YOU | DECEMBER ‘17

mα qα mpq p

=

h

(4m p )(2q p ) mpq p

=2 2

Davisson and Germer Experiment In 1927, Davisson and Germer designed an experiment to verify the wave nature of electron as shown. x A beam of electron emitted by electron gun is made to fall on nickel crystal cut along cubical axis at a particular angle. –

H.T.B.

+

Incident electron beam

Anode L.T.B.

Filament Vaccum chamber

Ni crystal

T I T Diffracted electron beam

To galvanometer Movable collector

x

x

x

1.227

nm V Ilustration 2 : An D-particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their de Broglie wavelengths are OD and Op respectively. λp . Find the ratio λα h Soln.: de Broglie wavelength, λ = 2mqV h h and λ α = λp = 2m p q pV 2mα qαV

2m p q pV

2mα qαV

×

x

(10000 − 5461) × 10 −10 × 3 × 108 = 6.545 × 10–34 J s Wave Nature of Matter x The waves associated with the moving material particles are called matter waves or de Broglie waves. x de Broglie Wavelength : The wavelength associated with moving particle is called de Broglie wavelength h h and it is given by λ = = p mv

h

The diffracted beam of electrons is received by the movable detector for different angles I about the point of incidence. The measured intensity of scattered electron beam was maximum at an angle I = 50q(or T = 65q) for an accelerating voltage of 54 V. If the de Broglie waves exist for electrons then these should be diffracted as X-rays. Using the Bragg’s formula 2d sin T = nO, we can determine the wavelength of these waves. For d = 0.91Å and T = 65°. This gives for n = 1, O = 2 × 0.91 × 10–10 sin 65° = 1.65Å. de Broglie wavelength associated with accelerating 12.27 12.27 electron is λ e = = Å = 1.67 Å. Thus the V 54 de Broglie hypothesis is verified.

1D?= Thomson’s Model of Atom An atom consists of a sphere of the radius of the order of 10–10 m. x The negatively charged particles, called electrons, are embedded in the positively charged sphere of the atom. x

x x

In an atom, total positive charge is equal to total negative charge. This model fails to explain: ¾ the emission of spectral lines from the atoms. ¾ the large angle of scattering of D-particles by thin metal foils.

Rutherford’s Model of Atom x The entire positive charge of the atom and most of the mass is concentrated in a small region called nucleus and electrons revolve around it in circular orbits. The most of an atom is empty space. x The atom as whole is electrically neutral, the total negative charge of the electrons is equal to the positive charge on the nucleus. x Rutherford inferred his model of atom from alpha particle scattering experiment and estimated the size of nucleus using the classical approach as follows: ¾ Number of D-particles per unit area reach on the screen at scattering angle of T, 1 N(θ) ∝ 4 sin (θ / 2) ¾ ¾

Ze 2 cot(θ / 2) Impact parameter, b = 4 πε0 K

The distance of closest approach of a particle is 2

x

2 Ze , given by d = 4 π ε0 K where K = kinetic energy of D-particles Z = atomic number of target atom. ¾ This distance d is the sum of the radii of the target nucleus and the D-particle. Drawbacks of Rutherford’s atomic model : According to electromagnetic theory, an accelerated charge particle always radiates energy. Electrons loses its energy and its radius of orbit decreases continuously and finally it would spiral into the nucleus. But in practical atoms do not collapse.

Bohr’s Model of Atom x Bohr’s postulates : ¾ An electron can revolve around the nucleus only in certain allowed circular orbits of definite energy and in these orbits it does not radiate. These orbits are known as stationary orbits. ¾ Angular momentum of an electron in a stationary orbit is an integral multiple of h/2S.

i.e., L = ¾

nh 2π

or, mvr =

nh 2π

The emission of radiation takes place when an electron makes a transition from a higher to a lower orbit. The frequency of the radiation is given by υ =

Ei − E f h

where Ei and Ef are the energies of the electron in the higher and lower orbits respectively. x

Bohr’s formulae for an electron orbit

Radius of nth orbit rn =

4 πε0n2h2 4 π2mZe 2

Velocity of electron in vn = the nth orbit

1 2πZe 2 4 πε0 nh

The kinetic energy of 2 ⎛ 1 ⎞ 2π2me 4 Z 2 the electron in the nth K n = ⎜⎝ 4 πε ⎟⎠ n2 h2 orbit 0 13.6 Z 2

=

eV

n2

The potential energy of 2 ⎛ 1 ⎞ 4 π2 me 4 Z 2 electron in nth orbit Un = − ⎜ 2 2 ⎝ 4 πε ⎟⎠ nh

0

=

−27.2 Z n2

2

eV

Total energy of electron 2 ⎛ 1 ⎞ 2π2me 4 Z 2 in nth orbit En = − ⎜ ⎝ 4 πε0 ⎟⎠ n2h2 =−

13.6 Z2 n2

eV

Frequency of revolution 2 ⎛ 1 ⎞ 4 π2 Z 2e 4 m of electron in nth orbit υn = ⎜ ⎝ 4 πε0 ⎟⎠ n3h3 Wavelength of radiation 1 1⎤ 2⎡ 1 in the transition from λ = RZ ⎢ 2 − 2 ⎥ , where ⎣⎢ n1 n2 ⎦⎥ n2 o n R is Rydberg’s constant Illustration 3 : An energy of 24.6 eV is required to remove one of the electrons from a neutral helium atom. What is the energy in (eV) required to remove both the electrons from a neutral helium atom? Soln.: When one of the electrons is removed from a neutral helium atom, energy is given by PHYSICS FOR YOU | DECEMBER ‘17

51

En = −

13.6 Z 2 n

2

eV per atom

For helium ion, Z = 2, when doubly ionised. For first orbit, n = 1 x

13.6

?

E1 = −

? ?

Energy required to remove it = 54.4 eV Total energy required = 54.4 + 24.6 = 79 eV.

2

(1)

× (2)2 = − 54.4 eV

Spectral series of Hydrogen Atom :

X-Rays

52

PHYSICS FOR YOU | DECEMBER ‘17

Characteristic X-Rays Few of the fast moving electrons having high velocity penetrate the surface O n=5 n=4 atoms of the target N MD ME material and knock M n=3 LD LE LJ M-series out the tightly n=2 bound electrons even L KD KE KJ L-series from the inner most K n=1 K-series shells of the atom. Now when the electron is knocked out, a vacancy is created at that place. To fill this vacancy, electrons from higher shell jumps in vacancy. This transition of electron lead to emission of radiation in the form of X-rays of very small but KD definite wavelength which KE L depends upon the target LJ LE D material. K-series Intensity

Intensity

Continuous X-Rays As an electron passes close to the positive nucleus of atom of the target, the electron is deflected from it’s path as shown in figure. This results in deceleration of the – X-rays photon electron. The loss in energy e + of the electron during deceleration is emitted in the form of X-rays. Minimum wavelength : When the electron looses whole of it’s energy in a single collision 30 kV 20 kV with the atom, an 10 kV X-ray photon of Wavelength maximum energy Intensity-wavelength graph hXmax is emitted. Minimum wavelength = cut off wavelength of X-ray hc 12375 = Å Omin = eV V (in volt)

L-series Omin Wavelength Intensity-wavelength graph

Moseley’s Law x According to Moseley’s observations, frequency of X-rays spectrum is given by υ = a ( Z − b) ; where a and b are constants. x x

a and b does not depend on the nature of target. Different values of b are follows: b=1 for K-series b = 7.4 for L-series b = 19.2 for M-series kE

X

kD

Z

Wave

x

⎛ 1 1 1 ⎞ = R(Z − b)2 ⎜ 2 − 2 ⎟ ⎜n ⎟ λ ⎝ 1 n2 ⎠ Energy of X-ray radiations

length

of

characteristic

spectrum

For transition n2 = 2 to n1 = 1 (KD line),

x

3Rc = 2.47 × 1015 Hz We have a = 4 Illustration 4 : KD wavelength emitted by an atom of atomic number Z = 11 is O. Determine the atomic number for an atom that emits KD radiation with wavelength 4O. Soln.: According to Moseley’s law, X = a2(Z – b)2, where X = frequency = ?

For KD line, b = 1 For one atom,

c

or

λ1 λ2

=

λ

c λ

= a2 (Z − 1)2

2

λ1

For other atom,

c

= a (Z1 − 1)2 c

λ2

= a2 (Z 2 − 1)2

a2 (Z 2 − 1)2 a2 (Z1 − 1)2

(Z2 – 1)2 = 25

or

λ 4λ

=

(Z 2 − 1)2 (11 − 1)2

or Z2 – 1 = 5

or

=

x

x

⎛ 1 hc 1⎞ ΔE = hυ = = Rhc(Z − b)2 ⎜ 2 − 2 ⎟ λ ⎝ n1 n2 ⎠

or

Composition and Properties of Nuclei x The nucleus of an atom contains protons and neutrons which are collectively known as nucleons. The number of protons in a nucleus is called its atomic number and is denoted by Z. The total number of protons and neutrons in a nucleus is called its mass number and is denoted by A. Number of protons in a nucleus = Z Number of nucleons in a nucleus = A Number of neutrons in a nucleus = N = A – Z. x Nuclide : It is a specific nucleus of an atom which is characterised by its atomic number Z and mass number A. It is represented by ZXA where X is the chemical symbol of the element. x Nuclear radius : Nuclear radius, R = R A1/3 where R is a constant and A is the mass number. ¾ Nuclear radius is measured in fermi. 1 fm = 10–15ȹ

x

x

>E3<59

(Z 2 − 1)2 100

Z2 = 6.

x

mass of nucleus

3m volume of nucleus 4 πR03 ¾ Nuclear density is independent of A and is order of the 1017 kg m–3. Isotopes : Isotopes of an element are the atoms of the element which have the same atomic number but different mass numbers. e.g. H1, H2, H3, are the three isotopes of hydrogen. Isobars : Isobars are the atoms of different elements which have the same mass number but different atomic numbers. e.g. Na22 and Ne22. Isotones : Isotones are the nuclides which contain the same number of neutrons e.g. Cl37 and K39. Nuclear density : ρ =

Nuclear Binding Energy x Mass Defect : The difference in mass of a nucleus and its constituent nucleons is called the mass defect of that nucleus. Mass defect, 'M = Zmp + (A – Z)mn – M, where M is the mass of a given nucleus. x Binding energy : ¾ The energy equivalent of the mass defect of a nucleus is called its binding energy. Eb = 'Mc2 = [Zmp + (A – Z)mn – M]c2 If masses are expressed in atomic mass units, 'Eb = [Zmp + (A – Z)mn – M] × 931.5 MeV Eb ¾ Binding energy per nucleon, ΔEb n = . A It is the average energy needed to separate a nucleus into its individual nucleons. PHYSICS FOR YOU | DECEMBER ‘17

53

Radioactivity x Radioactivity was discovered in 1896 by Antoine Henri Becquerel. x Radioactivity is the spontaneous disintegration of nuclei of some nuclides (called radionuclides) with the emission of alpha particles or beta particles, some accompanied by a gamma ray emission. x Law of radioactive decay : dN = −λN ( t ) dt

x

N ( t ) = N 0 e − λt

where O is the decay constant or disintegration constant, N(t) is the number of nuclei left undecayed at the time t, N0 is the number of radioactive nuclei at t = 0. Half-life of a radioactive substance is given by T1 / 2 =

x

or

ln 2 0.693 = λ λ

Mean life or average life of a radioactive substance is given by 1 τ= = = 1.44 T1/ 2 λ 0.693

x

x

Activity : The number of disintegrations occurring in a radioactive substance per second and it is given by R = –dN/dt. ¾ The SI unit of activity is becquerel. 1 becquerel = 1 Bq = 1 decay/second. Activity law : R(t) = R e–Ot where R = ON is the decay rate at t = 0 and R = NO. Fraction of nuclei left undecayed after n half live is

( ) = ( 21 )

N 1 = N0 2

x

x

54

n

t /T

x

The number of undecayed nuclei present after n n

mean life is N = (0.37)n N0 = ⎛⎜ ⎞⎟ N0. ⎝e⎠ x Probability of nucleus for survival in time t, N e− λt N = = e− λt P (survival) = N N x Probability of a nucleus to disintegrate in time t, P (disintegration) = 1 – P (survival) = 1 – e–Ot Illustration 5 : A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 year. In how many years one-fourth of the material remains? Soln.: Since

T1/ 2

x

N N N N t1/ 2 t1/ 2 t1/ 2 t1/ 2 → 0 ⎯⎯⎯ → 20 ⎯⎯⎯ → 30 .... ⎯⎯⎯ → n0 N 0 ⎯⎯⎯ 2 2 2 2

1/ 2

Number of nuclei present after n half lifes N0 . = (2)n PHYSICS FOR YOU | DECEMBER ‘17

N0

= e −λt

?

There is a simultaneous emission of two particles.

∴

N N0

= e −( λ1 + λ2 )t or

4 = e(O1 + O2)t or

N0 4 N0

= e −( λ1 + λ2 )t

ln 4 = (λ1 + λ 2 )t ln e

0.693 0.693 and λ2 = 1620 810 1 ⎤ ⎡ 1 ∴ 1.386 = 0.693 ⎢ + t ⎣1620 810 ⎥⎦ Now λ1 =

or

t =

1.386 × 1620 0.693 × 3

or t = 1080 years

WEST BENGAL at

(' t = nT 2)

If a nuclide can decay Process 1 simultaneously by O1 T1 W1 O two different process T which have decay W 2 O2 T2 W2 constant O1 and O2, Process 2 half life T1 and T2 and mean lives W1 and W2 respectively then TT T= 1 2 O O1 O2 T1 + T2

N

•

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x

Properties of D, E and J-rays Features

-particles

-particles Helium nucleus or doubly Fast moving electron (E+ or E–) ionised helium atom (2He4) + 2e – e or e 4 mp (mp = mass of proton me (me = mass of electron = 1.67 × 10–27 kg) = 9.1 × 10–31 kg) 1% to 99% of speed of light | 107 m s–1

Identity Charge Mass Speed

-rays Photons (electromagnetic waves) Zero Masselss Speed of light

Range of kinetic energy

4 MeV to 9 MeV

Penetration power (J > E > D)

1 (Stopped by a paper)

Ionisation power (D > E > J) Effect of electric or magnetic field Energy spectrum Mutual interaction with matter

10,000

All possible values between Between a minimum a minimum certain value to value to 2.23 MeV 1.2 MeV 100 10,000 (100 times of D) (100 times of E upto 30 cm of iron (or Pb) sheet) 100 1

Deflected

Deflected

Not deflected

Line and discrete Produces heat

Continuous Produces heat

Line and discrete Produces photo-electric effect, Compton effect, pair production

Equation of decay

ZX

A

D-decay

o Z–2Y A–4 + 2He4

ZX ZX

Nuclear reactions x The process by which the identity of a nucleus is changed when it is bombarded by an energetic particle is called nuclear reaction. X

+

a

⎯⎯→

(Parent nucleus) (Incident particle)

x

x

Y

+

b

+ Q

(Product nucleus) (daughter nucleus) (Energy)

Nuclear fission : A heavy nucleus breaks into two light nuclei of comparable masses. Binding energy per nucleon is greater for two lighter fragments than it is for the original nucleus. Nuclear fusion : Two light nuclei combine to form a heavier nucleus. Binding energy per nucleon is greater in the final nucleus than it is in the two original nuclei.

A E–-decay A E+-decay

Z+1Y

A

+ –1e0 + υ (ZXA)* J-decay

Z–1Y

A

+ 1e + X

ZX

A

+J

0

then in the reaction : p + 73 Li → 2 24 He , what will be the energy of proton? Soln.: Binding energy of 73 Li = 7 × 5.60 = 39.2 Me V Binding energy of 42He = 4 × 7.06 = 28.24 MeV ? Energy of proton = Energy of [2(42He) – 37Li] = 2 × 28.24 – 39.2 = 17.28 MeV.

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Illustration 6 : If the binding energy per nucleon of 37Li and 42He nuclei are 5.60 MeV and 7.06 MeV respectively, PHYSICS FOR YOU | DECEMBER ‘17

55

1. The potential energy of particle of mass m varies as ⎧ E for 0 ≤ x ≤ 1 U ( x) = ⎨ 0 ⎩0 for x > 1

The de-Broglie wavelength of the particle in the range 0 ≤ x ≤ 1 is O1 and that in the range x > 1 is O2. If the total energy of the particle is 2E0, find O1/O2. (a)

2

(b)

3

(c)

1 2

(d)

2 3

2. The binding energy per nucleon for deuteron (1H2) and helium (2He4) are 1.1 MeV and 7.0 MeV respectively. The energy released when two deuterons fuse to form a helium nucleus is (a) 36.2 MeV (b) 23.6 MeV (c) 47.2 MeV (d) 11.8 MeV 3. A particular hydrogen-like ion emits radiation of frequency 2.467 × 1015 Hz when it makes transition from n = 2 to n = 1. What will be the frequency of the radiation emitted in a transition from n = 3 to n = 1? (a) 1.42 × 1012 Hz (b) 2.92 × 1015 Hz 12 (c) 3.64 × 10 Hz (d) 3.64 × 1015 Hz 4. Light of wavelength O strikes a photo-sensitive surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wavelength must be changed to Oc where λ (b) Oc = 2O (a) Oc = 2 λ (c) < Oc < O (d) Oc > O 2 5. A nucleus at rest splits into two nuclear parts having radii in the ratio 1 : 2. Their velocities are in the ratio of (a) 8 : 1 (b) 6 : 1 (c) 4 : 1 (d) 2 : 1 6. All electrons ejected from a surface by incident light of wavelength 200 nm can be stopped before travelling 1 m in the direction of uniform electric field of 4 N C–1. The work function of the surface is (a) 4 eV (b) 6.2 eV (c) 2 eV (d) 2.2 eV 56

PHYSICS FOR YOU | DECEMBER ‘17

7. A triply ionized beryllium (Be3+) has the same orbital radius as the ground state of hydrogen. Then the quantum state n of Be3+ is (a) n = 1 (b) n = 2 (c) n = 3 (d) n = 4 8. A sample of radioactive material has mass m, decay constant O, molecular weight M. The activity of the sample after time t will be (Avogadro constant = NA) ⎛ mN A ⎞ −λt e (a) ⎜ ⎝ M ⎟⎠

⎛ mN A λ ⎞ − λt (b) ⎜ e ⎝ M ⎟⎠

⎛ mN A ⎞ − λt (c) ⎜ e ⎝ M λ ⎟⎠

(d)

m (1 − e − λt ) λ 9. An electron is in an excited state in a hydrogen like atom. It has a total energy of –3.4 eV. The kinetic energy of the electron is E and its de-Broglie wavelength is O. Then (a) E = 6.8 eV, O = 6.6 × 10–10 m (b) E = 3.4 eV, O = 6.6 × 10–10 m (c) E = 3.4 eV, O = 6.6 × 10–11 m (d) E = 6.8 eV, O = 6.6 × 10–11 m 10. The count rate from 100 cm3 of a radioactive liquid is c. Some of this liquid is now discarded. The count rate of the remaining liquid is found to be c/10 after three half-lives. The volume of the remaining liquid in cm3, is (a) 20 (b) 40 (c) 60 (d) 80 11. The energy released by the fission of one uranium atom is 200 MeV. The number of fissions per second required to produce 3.2 W of power is (Take 1 eV = 1.6 × 10–19 J) (a) 107 (b) 1010 (c) 1015 (d) 1011 12. The KD X-rays arising from a cobalt (Z = 27) target have a wavelength of 179 pm. The wavelength of KD X-rays arising from a nickel (Z = 28) target is (a) > 179 pm (b) < 179 pm (c) = 179 pm (d) None of these 13. The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T(K) and mass m, is

(a) (c)

h 3mkT 2h mkT

(b) (d)

2h 3mkT h mkT

[NEET 2017]

14. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is (a) 1 (b) 4 (c) 0.5 (d) 2 [NEET 2017] 15. An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If Omin is the smallest possible wavelength of X-ray in the spectrum, the variation of logOmin with logV is correctly represented in

(b)

(c)

(d) [JEE Main Offline 2017] 16. Some energy levels of a molecule are shown in the λ figure. The ratio of the wavelengths r = 1 is given λ2 by

4 3 3 (c) r = 4

18. The maximum velocity of the photoelectrons emitted from the surface is v when light of frequency n falls on a metal surface. If the incident frequency is increased to 3n, the maximum velocity of the ejected photoelectrons will be (a) more than 3v (b) less than 3v (c) v (d) equal to 3v [JEE Main Online 2017] 19. A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used, the approximate number of photons in the pulse are [Take Planck’s constant h = 6.62 × 10–34 J s] (a) 1019 (b) 1022 (c) 1018 (d) 1020 [JEE Main Online 2017]

(a)

(a) r =

sometime t, the ratio of the number of B to that of A is 0.3. Then t is given by log(1.3) T log 2 (b) t = T (a) t = log 2 2 log(1.3) T (c) t = T log(1.3) (d) t = log(1.3) [JEE Main Offline 2017]

2 3 1 (d) r = 3 [JEE Main Offline 2017] (b) r =

17. A radioactive nucleus A with a half-life T, decays into a nucleus B. At t = 0, there is no nucleus B. At

20. According to Bohr’s theory, the time averaged magnetic field at the centre (i.e., nucleus) of a hydrogen atom due to the motion of electrons in the nth orbit is proportional to (n = principal quantum number) (a) n–2 (b) n–3 (c) n–4 (d) n–5 [JEE Main Online 2017] SOLUTIONS 1. (a) : de-Broglie wavelength of a particle of mass m is h λ= 2mK where K is the kinetic energy of the particle. For 0 ≤ x ≤ 1, U(x) = E0 As, total energy = Kinetic energy + Potential energy ? 2E0 = K1 + E0 (∵ Total energy = 2E0 (Given)) or K1 = E0 h h = ? λ1 = …(i) 2mK1 2mE0 For x > 1, U(x) = 0 ? K2 = 2E0 h h = ? λ2 = …(ii) 2mK 2 2m(2 E0 ) Divide (i) by (ii), we get λ1 = λ2

h 2mE0

×

2m(2 E0 ) h

= 2

PHYSICS FOR YOU | DECEMBER ‘17

57

2. (b) : The given reaction is 21H2 o 2He4 + Q The energy released in the reaction is Q = BE of 2He4 – 2(BE of 1H2) = (4 × 7.0 MeV) – 2(2 × 1.1 MeV) = 23.6 MeV 3. (b) : The frequency of radiation emitted is given by c 1⎞ ⎛ 1 − X= =K ⎜ 2 2⎟ λ ⎝ n1 n2 ⎠ ⎛1 1⎞ Thus, 2.467 × 1015 Hz = K ⎜ − ⎟ ⎝ 12 22 ⎠ 4 or, K = × 2.467 × 1015 Hz. 3 The frequency of the radiation emitted in the transition n = 3 to n = 1 is 8 4 ⎡1 1⎤ 8 Xc = K ⎢ − ⎥ = K = × × 2.467 × 1015 Hz 2 2 9 9 3 ⎣1 3 ⎦

= 2.92 × 1015 Hz. 4. (c) : Kinetic energy of photoelectrons is, hc hc – I0 and 2E = – I0 E= λ λ′ ⎛ 1 + φ0 / E ⎞ λ ′ E + φ0 = Oc = O ⎜ λ 2 E + φ0 ⎝ 2 + φ0 / E ⎟⎠ ⎛ 1 + φ0 / E ⎞ 1 λ > ? Oc > Since ⎜ ⎟ 2 ⎝ 2 + φ0 / E ⎠ 2

5. (a) : Let A1 and A2 be the mass numbers of the two nuclear parts. Their radii are given by R1 = R0 (A1)1/3 and R2 = R0 (A2)1/3 Dividing, we get 1/3

3

3 A1 ⎛ R1 ⎞ 1 ⎛ 1⎞ or =⎜ ⎟ =⎜ ⎟ = ⎠ ⎝ A2 ⎝ R2 ⎠ 2 8 m1 1 Hence, the ratio of their masses is = m2 8 From the principle of conservation of momentum The magnitude of p1 = magnitude of p2 or m1v1 = m2v2, v1 m2 8 ? = = v2 m1 1

R1 ⎛ A1 ⎞ = R2 ⎜⎝ A2 ⎟⎠

6. (d) : The electron ejected with maximum speed vmax are stopped by electric field E = 4 N C–1 after travelling a distance d = 1 m 1 2 mv = eEd = 4 eV ? 2 max 1240 The energy of incident photon = = 6.2 eV 200 From equation of photoelectric effect 58

PHYSICS FOR YOU | DECEMBER ‘17

?

1 2 1 mvmax = hX – I0 or I0 = hX – mv2max 2 2 I0 = 6.2 – 4 = 2.2 eV

7. (b) : Radius of nth orbit in hydrogen like atoms is a n2 rn = 0 Z where a0 is the Bohr’s radius For hydrogen atom, Z = 1 ? r1 = a0 (∵ n = 1 for ground state) For Be3+, Z = 4 a n2 ∴ rn = 0 4 According to given problem, r1 = r n n2a0 a0 = n=2 4 8. (b) : Activity = number of disintegrations per unit time dN ? = O . N, dt where N = the total number of nuclei. Also, N = number of moles × NA = ⎛⎜ m ⎞⎟ NA. ⎝ M⎠ Activity after time t = initial activity × e–Ot. ⎛ λN A m ⎞ − λt = ON × e–Ot = ⎜ e ⎝ M ⎟⎠ 9. (b) : The potential energy = – 2 × kinetic energy = – 2E. ? Total energy = – 2E + E = – E = – 3.4 eV or E = 3.4 eV. Let p = momentum and m = mass of the electron. p2 ∴ E= or p = 2mE . 2m h h de-Broglie wavelength, λ = = p 2mE On substituting the values, we get 6.63 × 10 −34 λ= 2 × 9.1 × 10 −31 × 3.4 × 1.6 × 10 −19 = 6.6 × 10–10 m c 100 1 c 3 After 3 half-lives, CR for 1 cm of liquid = × 8 100 Let the volume of the remaining liquid = V cm3. c c c ? CR of this liquid = V × ; =V × 800 800 10 or V = 80.

10. (d) : Initial count rate (CR) for 1 cm3 of liquid =

11. (d) : Energy released per fission is E = 200 MeV = 200 × 106 × 1.6 × 10–19 J = 3.2 × 10–11 J Power, P = 3.2 W P Number of fissions per second = E 3. 2 W = = 1011 −11 3.2 × 10 J 1 12. (b) : OKD v (Z − 1)2 2 2 λ ⎛ 27 − 1⎞ ⎛ Z − 1⎞ Ni = ⎜ Co ⎟ = ⎜⎝ ⎟⎠ 28 − 1 λ Co ⎝ Z Ni − 1 ⎠ 2

17.

2

⎛ 26 ⎞ ⎛ 26 ⎞ ONi = ⎜ ⎟ × OCo = ⎜ ⎟ × 179 ⎝ 27 ⎠ ⎝ 27 ⎠ = 165.9 pm < 179 pm. 13. (a) : Kinetic energy of a neutron in thermal equilibrium with heavy water at a temperature T is given as K = 3 kT

...(i)

2

Also momentum (p) is, p = From eqn. (i)

2mK

3 2m . kT = 3mkT 2 Required de-Broglie wavelength is given as h h O= = p 3mkT 14. (b) : The wavelength of last line of Balmer series 1 ⎞ R ⎛ 1 1 = R⎜ 2 − 2 ⎟ = ⎝ ∞ ⎠ 4 2 λB The wavelength of last line of Lyman series 1 ⎞ ⎛1 1 = R⎜ 2 − 2 ⎟ = R ⎝ λL 1 ∞ ⎠ λB 4 = =4 ? 1 λL 15. (a) : Minimum possible wavelength of X-rays is hc ⎛ hc ⎞ λ min = ; log(λ min ) = log ⎜ ⎟ − log V ⎝ e⎠ eV This is the equation of a straight line with negative slope and positive intercept on the y-axis (logOmin). 16. (d) : We know, hc 1 λ= i.e. λ ∝ E energy difference hc hc Now, λ1 = ...(i) = −E − (−2E) E

hc hc ...(ii) = 4 ⎞ ⎛E⎞ ⎛ −E − ⎜ − E ⎟ ⎜ ⎟ ⎝ 3 ⎠ ⎝3⎠ Dividing eqn. (i) by eqn. (ii), we get λ1 1 = λ2 3 (b) : Let NA and NB be the number of molecules of A and B after time t. N Also, after time t, B = 0.3 NA Also, let N0 be the total number of nucleus initially. After time t, NA + NB = N0 N NA + 0.3NA = N0 ? N A = 0 1. 3 Also, rate of disintegration of A NA = N0 e–Ot 1 N0 = e − λt or ln(1.3) = λt ⇒ = N 0e − λt ; 1 . 3 1. 3 T ln(1.3) T log(1.3) ln(1.3) ∴ t= = or t = ln(2) log 2 λ 1 (a) : E1 = mv2 = hn – I 2 If incident frequency is increased to 3n 1 E2 = mvc2 = 3hn – I = 3(hn – I) + 2I 2 1 = 3 × mv2 + 2I 2 vc2 = 3v2 + 4I/m; vc > v 3 (d) : Here, O = 660 nm = 660 × 10–9 m t = 60 ms = 60 × 10–3 s; P = 0.5 kW = 500 W h = 6.62 × 10–34 J s, n = ? E nhc P λt As, P = = ; ∴ n= λt t hc 500 × 660 × 10 −9 × 60 × 10 −3 n= 6.62 × 10 −34 × 3 × 108 ? n | 1020 μ I (d) : Magnetic field at the centre, Bn = 0 2rn For a hydrogen atom, radius of nth orbit is given by λ2 =

18.

p=

19.

20.

⎛ n2 ⎞ ⎛ h ⎞ 2 4 πε0 rn = ⎜ ⎟ ⎜ ⎟ ? rn v n2 2 ⎝ ⎠ π 2 m e ⎝ ⎠ ev e e I= = = n T 2πrn / vn 2πrn Also, vn v n–1 ? I v n–3 Hence, Bn v n–5 PHYSICS FOR YOU | DECEMBER ‘17

VV 59

EXAM

ss cla i xi

PREP 2018 CHAPTERWISE MCQS FOR PRACTICE WAVE OPTICS

1. Light from sodium lamp is made to pass through two polaroids placed one after the other in the path of light. Taking the intensity of the incident light as 100%, the intensity of the outcoming light that can be varied in the range (a) 0% to 100% (b) 0% to 50% (c) 0% to 25% (d) 0% to 75% 2. In a double slit experiment, the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. The angular width of the fringe if entire experimental apparatus 4⎞ ⎛ is immersed in water is ⎜ Take μ water = ⎟ ⎝ 3⎠ (a) 0.15°

(b) 1°

(c) 2°

(d) 0.3°

3. In a single slit diffraction experiment, the slit width is 2.5 O, where O is the wavelength of light used. Then on either side of the central maximum there are (a) 4 minima, 2 secondary maxima (b) 2 minima, 2 secondary maxima (c) 2 minima, 4 secondary maxima (d) 2 minima, 1 secondary maxima. 4. A beam of natural light falls on a system of 5 polaroids, which are arranged in succession such that the pass axis of each polaroid is turned through 60° with respect to the preceding one. The fraction of the incident light intensity that passes through the system is 1 1 1 1 (a) (b) (c) (d) 64 256 512 32 5. In Young’s double-slit experiment, the distance between the centres of adjacent fringes is 0.10 mm. 60

PHYSICS FOR YOU | DECEMBER ‘17

If the distance of the screen from the slits is doubled, the distance between the slits is halved and the wavelength of light is changed from 6.4 × 10–7 m to 4.0 × 10–7 m, then the new distance between the fringes will be (a) 0.10 mm (b) 0.15 mm (c) 0.20 mm (d) 0.25 mm 6. In a Fraunhofer diffraction experiment at a single slit using a light of wavelength 400 nm, the first minimum is formed at an angle of 30°. The direction T of the first secondary maximum is given by (a) sin–1 (1/4) (b) tan–1 (2/3) –1 (c) sin (2/3) (d) sin–1 (3/4) 7. The two slits are 1 mm apart from each other and illuminated with a light of wavelength 5 × 10–7 m. If the distance of the screen is 1 m from the slits, then the distance between third dark fringe and fifth bright fringe is (a) 1.5 mm (b) 0.75 mm (c) 1.25 mm (d) 0.625 mm 8. A double-slit apparatus is immersed in liquid of refractive index Pm. It has slit separation d and distance between plane of slits and screen D as D >> d. The slits are illuminated by a parallel beam of monochromatic light of wavelength O0. The smallest thickness of a sheet of refractive index Pp to bring adjacent minima on the axis is (μ p − μm )λ 0 λ0 (b) (a) 2(μ p − μm ) 2 (c)

λ0 (μ p − μm )

(d) (Pp – Pm)O0

9. In a Young’s double slit experiment, the source is red light of wavelength 7 × 10–7 m. When a thin glass plate of refractive index 1.5 at this wavelength is put in the path of one of the interfering beams, the central bright fringe shifts to the position previously occupied by 5th bright fringe. The thickness of the plate is (a) 4 Pm (b) 5 Pm (c) 6 Pm (d) 7 Pm

DUAL NATURE OF RADIATION AND MATTER

16. A light of wavelength 600 nm is incident on a metal surface. When light of wavelength 400 nm is incident, the maximum kinetic energy of the emitted photoelectrons is doubled. The work function of the metal is (a) 1.03 eV (b) 2.11 eV (c) 4.14 eV (d) 2.43 eV

10. Two coherent light sources S1 and S2 with separation 2O are placed on the x-axis symmetrically about the origin. They emit light of wavelength O. The numbers of maxima on a circle of large radius, lying in the x-y plane with centre at origin is (a) 4 (b) 6 (c) 8 (d) 12

17. A source S1 is producing 1015 photons per second of wavelength 5000 Å. Another source S2 is producing 1.02 × 1015 photons per second of wavelength 5100 Å. Then, (power of S2)/ (power of S1) is equal to (a) 1.00 (b) 1.02 (c) 1.04 (d) 0.98

11. In Young’s double slit experiment, when sodium light of wavelength 5893 Å is used, then 62 fringes are seen in the field of view. Instead of sodium light, if violet light of wavelength 4358 Å is used, then the number of fringes than will be seen in the field of view will be (a) 54 (b) 64 (c) 74 (d) 84

18. The de Broglie wavelength of an electron moving with a velocity c/2 (c = velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is (a) 1 : 4 (b) 1 : 2 (c) 1 : 1 (d) 2 : 1

12. What is the minimum thickness of a thin film required for constructive interference in the reflected light from it, if the refractive index of the film = 1.5, wavelength of the light incident on the film = 600 nm? (a) 100 nm (b) 300 nm (c) 50 nm (d) 200 nm 13. Two beams of light of intensity I1 and I2 interfere to give an interference pattern. If the ratio of maximum intensity to that of minimum intensity is 25/9, then I1/I2 is (a) 5/3 (b) 4 (c) 81/625 (d) 16 14. Interference was observed in interference chamber when air was present, now the chamber is evacuated and if the same light is used, a careful observer will see (a) no interference (b) interference with bright bands (c) interference with dark bands (d) interference in which width of the fringe will be slightly increased 15. In Young’s double slit experiment, the central bright fringe can be identified (a) by using white light instead of monochromatic light (b) as it is narrower than other bright fringes (c) as it is wider than other bright fringes (d) as it has a greater intensity than the other bright fringes

19. The additional energy that should be given to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is (a) 2 times the initial kinetic energy (b) 3 times the initial kinetic energy (c) 0.5 times the initial kinetic energy (d) 4 times the initial kinetic energy 20. The ratio of the de Broglie wavelengths of an electron of energy 10 eV to that of person of mass 66 kg travelling at a speed of 100 km h–1 is of the order of (a) 1034 (b) 1027 17 (c) 10 (d) 10–10 21. A monochromatic light beam of wavelength 6000 Å and intensity 10 mW cm–2 fall on a surface. The number of photons striking the surface per second per cm2 are (a) 3 × 1016 (b) 2 × 1016 16 (c) 4 × 10 (d) 5 × 1016 22. When a certain metal surface is illuminated with light of frequency X, the stopping potential for photoelectric current is V0. When the same υ surface is illuminated by light of frequency , the 2 V stopping potential is 0 . The threshold frequency 4 for photoelectric emission is υ υ 2υ 4υ (b) (c) (d) (a) 6 3 3 3 PHYSICS FOR YOU | DECEMBER ‘17

61

23. Figure shows the variation Photo current of photocurrent with b c a anode potential for a photo-sensitive surface for three different radiations. Anode potential Let Ia, Ib, and Ic be the intensities and Xa, Xb and Xc be the frequencies for the curves a, b and c respectively. (a) Xa = Xb and Ia z Ib (b) Xa = Xc and Ia = Ic (c) Xa = Xb and Ia = Ib (d) Xb = Xc and Ib z Ic 24. In a photoelectric experiment, electrons are ejected from metals X and Y by light of frequency X. The potential difference V are required to stop the electrons is measured for various frequencies. If Y has a greater work function than X, which graph illustrates the expected results? V

V

X Y

(a)

V

Y

Y

(b) X

(0, 0)

X

V

X

(c)

X

(0, 0) Y

X

(d) (0, 0)

X

(0, 0)

X

25. When a metallic surface is illuminated by a monochromatic light of wavelength O, then the potential difference required to stop the ejection of electrons is 3V0. When the same surface is illuminated by the light of wavelength 2O, then the potential difference required to stop the ejection of electrons is V0. Then for photoelectric effect, the threshold wavelength for the metal surface will be (a) 6O (b) 4O/3 (c) 4O (d) 8O 26. If the maximum kinetic energy of emitted photo electrons from a metal surface of work function 2.5 eV, is 1.7 eV. If wavelength of incident radiation is halved, then stopping potential will be (a) 2.5 V (b) 6.7 V (c) 5 V (d) 1.1 V 27. The ratio of de Broglie wavelength of D-particle to that of a proton being subjected to the same magnetic field so that the radii of their paths are equal to each other assuming the field induction G vector B is perpendicular to the velocity vectors of the D-particle and the proton is (a) 1 : 1 (b) 1 : 4 (c) 1 : 2 (d) 2 : 1 62

PHYSICS FOR YOU | DECEMBER ‘17

28. A particle moves in a closed orbit around the origin, due to a force which is directed towards the origin. The de Broglie wavelength of the particle varies cyclically between two values O1 and O2 with O1 > O2. Which of the following statements is true? (a) The particle could be moving in a circular orbit with origin as centre. (b) The particle could be moving in an elliptical orbit with origin as its focus. (c) When the de Broglie wavelength is O 1, the particle is nearer the origin than when its value is O2. (d) Both (a) and (c) 29. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly (a) 3.7 eV (b) 3.2 eV (c) 2.8 eV (d) 2.5 eV 30. Red, blue, green and violet colour lights are one by one made incident on a photocathode. It is observed that only one colour light produces photoelectrons. That light is (a) red (b) blue (c) green (d) violet SOLUTIONS 1. (b) : Let I0 be the intensity of incident light. As the light coming from sodium lamp is unpolarised, so the intensity of light emerging from the first polaroid is I I1 = 0 2 If T is the angle between two polaroids, then the intensity of light emerging from the second polaroid is I I2 = I1 cos2 θ = 0 cos2 θ 2 But I0 = 100% (given) ? I2 = 50% cos2T Since T varies from 90° to 0°, so the intensity of the out coming light can be varied from 0% to 50% . 2. (a) : Since a fringe of width E is formed on the screen at distance D from the slits, so the angular fringe width β Dλ / d λ Dλ ⎤ ⎡ θ= = = ⎢⎣' β = d ⎥⎦ D D d

λ θ If the wavelength in water be Oc and the angular fringe width be Tc, then λ′ λ λ′ or = d= θ′ θ θ′ λ⎤ λ′ λ/μ ⎡ .θ ' λ′ = ⎥ or θ′ = . θ = ⎢ μ⎦ λ λ ⎣ 0. 2 ° = 0.15° ? Tc = 4/3 3. (d) : Given, width of the slit, a = 2.5O For minima, a sinT = nO nλ nλ n = sin θ = sin θ = a 2.5λ 2.5 Since maximum value of sinT is 1. ? n = 1, 2 Thus only 2 minima can be obtained on the either side of central maximum. For secondary maxima, (2n + 1) λ 2n + 1 λ = a sin θ = (2n + 1) sin θ = 2a 2 × 2.5 2 2n + 1 sin θ = 5 Since maximum value of sinT is 1. ? n = 1, thus only 1 secondary maximum can be obtained on the either side of central maximum. 4. (d) : Let I0 be the intensity of incident light. Then the intensity of light from the 1st polaroid is I I1 = 0 2 Intensity of light from the 2nd polaroid is I ⎛ 1 ⎞2 I I2 = I1cos2 60° = 0 ⎜ ⎟ = 0 2 ⎝2⎠ 8 rd Intensity of light from the 3 polaroid is I ⎛ 1 ⎞2 I I 3 = I 2 cos 2 60° = 0 ⎜ ⎟ = 0 8 ⎝2⎠ 32 th Intensity of light from the 4 polaroid is I ⎛ 1 ⎞2 I I 4 = I 3 cos 2 60° = 0 ⎜ ⎟ = 0 32 ⎝ 2 ⎠ 128 Intensity of light from 5th polaroid is ⇒ d=

I ⎛ 1 ⎞2 I I 5 = I 4 cos 2 60° = 0 ⎜ ⎟ = 0 ⎝ ⎠ 128 2 512 Therefore, the fraction of the incident light that passes through the system is I5 1 = I 0 512

5. (d) : Distance between the centres of adjacent fringes = Fringe width Fringe width, β =

λD d

...(i)

where O is the wavelength of light used, D is the distance of the screen from the slits, d is the distance between the slits. ?

β′ =

λ ′D ′ d′

...(ii)

Divide (ii) by (i), we get β′ ⎛ λ ′ ⎞ ⎛ D ′ ⎞ ⎛ d ⎞ = ⎜ ⎟⎜ ⎟⎜ ⎟ β ⎝ λ ⎠⎝ D ⎠⎝d′⎠ Substituting the given values, we get β ′ ⎛ 4.0 × 10 −7 ⎞ ⎛ 2 D ⎞ ⎛ d ⎞ = ⎜ ⎟ β ⎝⎜ 6.4 × 10 −7 ⎟⎠ ⎝ D ⎠ ⎜⎝ (d / 2) ⎠⎟ β′ = 2.5 orEc = 2.5E = 2.5 u0.10 mm = 0.25 mm β

6. (d) : sin θ =

(2n + 1) λ 2a

For the (1st) secondary minima, a sin T = 1O. i.e., O/a = sin T = sin 30° = 1/2 ? For first secondary maxima 2(1) + 1 1 3 ⋅ = sin T = 2 2 4 3 ⎛ ⎞ ∴ θ = sin −1 ⎜ ⎟ ⎝4⎠ 7. (c) : Given O= 5 × 10–7 m, D = 1 m, d = 1 mm nD λ Distance of nth bright fringe from the centre = d where n = 1, 2, 3, ..... Dλ So the distance of 5th bright fringe = d Distance of nth dark fringe from the centre 1 ⎞ Dλ ⎛ , where n = 1, 2, 3, 4 = ⎜n − ⎟ ⎝ 2⎠ d ⎛ 1 ⎞ Dλ 5 Dλ = 3rd dark fringe = ⎜ 3 − ⎟ ⎝ 2⎠ d 2 d 5 ⎞ Dλ 5 Dλ ⎛ = Distance between them = ⎜⎝ 5 − ⎟⎠ 2 d 2 d =

5 ×1 × 5 × 10 −7 = 12.5 × 10 −4 m = 1.25 mm 2 1 × 10 −3 PHYSICS FOR YOU | DECEMBER ‘17

63

8. (a) : For Young’s double slit experiment, the position of minima is 1 ⎞ Dλ ⎛ y = ⎜n + ⎟ ⎝ 2⎠ d Adjacent minima is the 1st minima or n = 0 1 ⎞ Dλ ⎛ Dλ y1 = ⎜ 0 + ⎟ = ⎝ ⎠ 2 d 2d When immersed in liquid, O =

λ0 μm

⎛ Dλ 0 ⎞ y1 = ⎜ ⎝ 2μmd ⎟⎠ Now fringe shift due to introduction of sheet on the path of one of the beams is E. D E = (μ −1)t d The requirement is, minima must come on the axis. Dλ 0 D ⎛ μp ⎞ E = y1 or ⎜ − 1⎟ t = 2μ m d d ⎝ μm ⎠ λ0 t= 2(μ p − μm ) D (μ −1)t d 5Dλ The position of 5th bright fringe = d Since central bright fringe shifts to the 5th maxima position, 5Dλ D (μ −1)t = d d ⎛ 5λ ⎞ 5 × 7 × 10 −7 m or t = ⎜ = = 7 Pm ⎟ ⎝ μ − 1⎠ (1.5 − 1)

9. (d) : Fringe shift, E =

10. (c) : S1P =

2

2

r + λ + 2r λ cos θ

S2P = r 2 + λ 2 − 2r λ cos θ S1P2 – S2P2 = 4rO cos T 4r λ cos θ S1P – S2P = (S1P + S2 P ) Since r >> O

P

r S1

2l

S2

4r λ cos θ = 2O cos T 2r For P to have a maxima, 2O cos T = nO n cos T = 2 The values of n satisfying the above equation are, n = 0, r 1, r 2. When cos T = 0 T = 90°, 270° S1P | S2P | r S1P – S2P =

64

PHYSICS FOR YOU | DECEMBER ‘17

1 T = 60°, 300° 2 1 cos T = − T = 120°, 240° 2 cos T = + 1 T = 0° cos T = – 1 T = 180°. The number of maxima on the circle is 8. (c) is correct. 11. (d) : Here, O1 = 5893 Å, n1 = 62 fringes O2 = 4358 Å Dλ1 Dλ 2 = n2 ∵ n1 or n1O1 = n2O2 d d nλ 62 × 5893 n2 = 1 1 = | 84 λ2 4358 12. (a) : The condition for constructive interference in a thin film of thickness t and refractive index P in the reflected system, for normal incidence is 1⎞ ⎛ 2 μt = ⎜ n + ⎟ λ ⎝ 2⎠ where n = 0, 1, 2, 3, …… For minimum thickness, n = 0 cos T =

∴ 2 μt =

λ ; 2

t=

−9

λ 600 × 10 = 4μ 4 × 1. 5

= 100 nm

2

⎛a +a ⎞ I 25 25 13. (d) : max = or ⎜ 1 2 ⎟ = I min 9 9 ⎝ a1 − a2 ⎠ where a1 and a2 denotes amplitudes. a1 + a2 5 = or 5a1 – 5a2 = 3a1 + 3a2 a1 − a2 3 2 ⎛ a1 ⎞ a1 2a1 = 8a2 or = 4 or ⎜ ⎟ = 16 a ⎝a ⎠ 2

2

? I1/I2 = 16 (∵ Intensity v (amplitude)2) 14. (d) : The refractive index of air is slightly more than 1. When chamber is evacuated, refractive index decreases and hence the wavelength increases and fringe width also increases. 15. (a) : When white light is used, central fringe will be white with red edges, and on either side of it, we shall get few coloured bands and then uniform illumination. 16. (a) : Here, O = 600 nm, Oc = 400 nm, Kcmax = 2 Kmax According to Einstein’s photoelectric equation hc − φ0 λ hc and 2K max = − φ0 λ′ K max =

… (i) … (ii)

Dividing (ii) by (i) we get ⎛ hc ⎞ ⎜⎝ λ′ ⎟⎠ − φ0 2hc hc or 2= − 2φ 0 = − φ 0 λ λ′ ⎛ hc ⎞ ⎜⎝ λ ⎟⎠ − φ0

19. (b) : de Broglie wavelength of an electron having kinetic energy K is h λ= where me is the mass of an electron 2me K λ2 =

or

⎛2 1 ⎞ hc ⎜ − ⎟ = φ0 ⎝ λ λ′ ⎠

?

⎛ 2 1 ⎞ φ0 = 1240 eV nm ⎜ − ⎟ = 1.03 eV ⎝ 600 nm 400 nm ⎠

(Take hc = 1240 eV nm)

2

17. (a) : For a source S1, wavelength, O1 = 5000 Å Number of photons emitted per second, N1 = 1015 Energy of each photon, E1 =

hc λ1 N1hc λ1

Power of source S1, P1 = E1N1 =

For a source S2,wavelength, O2 = 5100 Å Number of photons emitted per second, N2 = 1.02 × 1015 Energy of each photon, E2 =

hc λ2

Power of source S2, P2 = N2E2 =

?

N 2 hc

λ2 N 2hc Power of S2 P2 λ N λ = = 2 = 2 1 Power of S1 P1 N1hc N1 λ 2 λ1

=

(1.02 × 1015 ) × (5000 × 10−10 ) 15

(10 ) × (5100 × 10

1 h2 , h2 K∝ or K = 2me K λ2 2me λ 2

−10

)

=

...(i)

2

1 ⎛ c⎞ 1 K e = me ⎜ ⎟ = me c 2 ⎝ ⎠ 2 2 8 Kinetic energy of a photon, K ph = m c2 hc hc ∴ K ph = = = e 2h λe 2 me c Ke 1 2 1 ∴ = m c2 × = 2 4 K ph 8 e me c

hc λ ph

(' λ e = λ ph )

2 ⎛ 1 nm ⎞ K′ ⎛ λ ⎞ =⎜ ⎟ =⎜ ⎟ =4 K ⎝ λ′ ⎠ ⎝ 0.5 nm ⎠

Kc = 4K Additional energy given = Kc – K = 4K – K = 3K 20. (b) : For an electron, mass me = 9.11 × 10–31 kg, Kinetic energy, K = 10 eV = 10 × 1.6 × 10–19 J h de Broglie wavelength, λ e = ...(i) 2me K For the person, mass m = 66 kg, 5 Speed, v = 100 km h–1 = 100 × m s −1 18 h de Broglie wavelength, λ = ...(ii) mv Divide (i) by (ii), we get λe mv h mv = × = λ h 2me K 2me K =

51 =1 51

18. (a) : de Broglie wavelength of an electron, h h 2h λe = = = pe ⎛ c ⎞ me c me ⎜ ⎟ ⎝ 2⎠ Kinetic energy of an electron,

∴

66 × 100 ×

5 18

2 × 9.11 × 10−31 × 10 × 1.6 × 10−19

= 1.07 × 1027

21. (a) : Here, intensity I = 10 mW cm–2 Wavelength O = 6000 Å Energy of photon, hc (6.63 × 10−34 J s) × (3 × 108 m s −1 ) = E= λ (6000 × 10−10 m) Number of photons striking the surface per second per m2 is I (10 × 10) × (6000 × 10−10 ) N= = = 3 × 1020 E (6.63 × 10−34 ) × (3 × 108 ) ? Number of protons striking the surface per second per cm2 is 3 × 1016. 22. (b) : According to Einstein’s photoelectric equation Kmax = hX – I0 where X is the frequency of incident light and I0 is the work function of the metal. As Kmax = eV0 where V0 is the stopping potential Therefore, eV0 = hX – I0 ...(i) V0 υ = h − φ0 and e ...(ii) 4 2 PHYSICS FOR YOU | DECEMBER ‘17

65

From (i) and (ii), we get φ hυ hυ hυ φ 0 hυ − − = − φ 0 or φ 0 − 0 = 4 2 4 4 4 2 3 hυ h υ φ0 = or φ 0 = 4 4 3 φ hυ 1 υ ? Threshold frequency, υ0 = 0 = × = h 3 h 3 23. (a) : Intensity effects the saturation current. Frequency effects the stopping potential. ⎛ φ ⎞ ⎛ eV ⎞ X = ⎜ 0⎟ +⎜ 0⎟ ⎝ h⎠ ⎝ h ⎠ As stopping potential of a and b is same, Xa = Xb. As their saturation currents are different. Ia z Ib. 24. (a) : eV0 = hX – I0 ⎛φ ⎞ ⎛ h⎞ V0 = ⎜ ⎟ υ − ⎜ 0 ⎟ ⎝ e⎠ ⎝ e⎠ V0 vs X graph is a straight line with constant h slope = e X and Y will be parallel. ⎛φ ⎞ When V0 = 0, X0 = ⎜ 0 ⎟ . ⎝ h⎠ As I0Y > I0X X0Y > X0x. It means the X intercept of Y graph will be larger than the X-intercept of X graph. 25. (c) : Using Einstein photoelectric equation hX = I0 + Kmax hc = I0 + e(3V0) ...(i) λ hc Also, = I0 + eV0 ...(ii) 2λ On solving eqn. (i) and (ii), we get, hc ⎛ 3 ⎞ hc ⎜⎝ 2 − 1⎟⎠ λ = 3I0 – I0 or I0 = 4λ hc But I0 = , where O0 is the threshold wavelength, λ0 hence O0 = 4O. 26. (b) : Energy of incident photons, hX = I0 + Kmax = 2.5 + 1.7 = 4.2 eV λ ∵ XO = c and υ′ = c ⇒ υ′ = 2 υ 2 According to Einstein’s photoelectric equation, eV0 = hXc – I0 = h(2X) – I0 = 2 × 4.2 – 1.7 V0 = 6.7 V 66

PHYSICS FOR YOU | DECEMBER ‘17

27. (c) : ∵ de-Broglie wavelength, λ = ∴

h p

λα pp q p 1 λ 1 = = = ∴ α = [∵ p = mv = qBr] λ p pα qα 2 λp 2

28. (b) : The de Broglie wavelength of the particle v2 can be varying cyclically between two values O1 and O2, if particle is moving A B O in an elliptical orbit with origin as its one focus. v1 Refer figure. Let v1, v2 be the speed of particle at A and B respectively and origin is at focus O. If O1, O2 are the de Broglie wavelengths associated with particle while moving at A and B respectively. h h λ1 v2 and λ2 = ∴ = mv1 mv2 λ2 v1 Since O1 > O2 ? v2 > v1 From figure we note that origin O is close to B than A. 29. (a) : For O = 248 nm, v = u1 1 hc ? K1 = mu12 = −φ 2 248 For O = 310 nm, v = u2 Then, λ1 =

hc 1 −φ K 2 = mu22 = 2 310 hc 2 5−φ u12 248 − φ ⎛ 2⎞ ⇒ ⎜ ⎟ = ⇒ = ⎝ 1⎠ hc 4−φ u22 −φ 310

?

or I = 3.67 eV | 3.7 eV 30. (d) : The energy of incident light is E = hX As Xviolet > Xblue > Xgreen > Xred ? Eviolet > Eblue > Egreen > Ered Since the incident energy is maximum for violet colour, therefore violet light produces photoelectrons. 477*3(::?0 ANSWER KEY 1. (b) 6. (a) 11. (c) 16. (d) 21. (b,d) 26. (1)

2. 7. 12. 17. 22. 27.

(c) (c) (a) (c) (a,d) (b)

3. 8. 13. 18. 23. 28.

(d) (b) (c) (b) (b,c) (c)

4. 9. 14. 19. 24. 29.

(c) (d) (c) (b) (2) (c)

5. 10. 15. 20. 25. 30.

(a) (c) (a) (a,b,d) (4) (a)

Series 7 CHAPTERWISE PRACTICE PAPER SEMICONDUCTOR ELECTRONICS : MATERIALS, DEVICES AND SIMPLE CIRCUITS | COMMUNICATION SYSTEMS

Time Allowed : 3 hours

Maximum Marks : 70 GENERAL INSTRUCTIONS

(i) (ii) (iii) (iv) (v) (vi) (vii)

All questions are compulsory. Q. no. 1 to 5 are very short answer questions and carry 1 mark each. Q. no. 6 to 10 are short answer questions and carry 2 marks each. Q. no. 11 to 22 are also short answer questions and carry 3 marks each. Q. no. 23 is a value based question and carries 4 marks. Q. no. 24 to 26 are long answer questions and carry 5 marks each. Use log tables if necessary, use of calculators is not allowed.

SECTION - A

1. Why are elemental dopants for Silicon or Germanium usually chosen from group XIII or group XV? 2. How does the effective power radiated by an antenna vary with wavelength? 3. Can the potential barrier across a p-n junction be measured by simply connecting a voltmeter across the junction? 4. What is the importance of modulation index in an AM wave? 5. Under what condition a transistor works as an open switch? SECTION - B

6. What is difference between electron current and hole current? 7. Calculate the emitter current and collector current for a CE transistor circuit, if base current IB = 20 PA and dc current gain of transistor Edc = 100.

OR In a transistor, emitter-base junction is forward biased while the base-collector junction is reverse biased. Why? 8. Identify the gates P and Q and write the truth table for the logic gate formed due to combination of P and Q, as shown in the following figure.

9. What do the acronyms LASER and LED stand for? Name the factor which determines (i) frequency, and (ii) intensity of light emitted by LED. 10. An AC input of 50 Hz frequency is fed to a (i) half-wave rectifier, and (ii) full-wave rectifier. What is the frequency of rectified dc output, and why? SECTION - C

11. What is space wave propagation? If the sum of the heights of transmitting and receiving antenna in line of sight of communication is fixed at h, show PHYSICS FOR YOU | DECEMBER â€˜17

67

12.

13.

14.

15.

16.

17.

that the range is maximum when the two antenna h have a height each. 2 The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of arsenic and 5 × 1020 per m3 atoms of indium. Calculate the number of electrons and holes. Given that ni = 1.5 × 1016 m–3. Is the material n-type or p-type? A change of 8.0 mA in the emitter current brings a change of 7.9 mA in the collector current. How much change in the base current is required to have the same change 7.9 mA in the collector current ? Find the value of D and E. In a Zener regulated power supply, a Zener diode with VZ = 6.0 V is used for regulation. The load current is to be 4.0 mA and the unregulated input is 10.0 V. What should be the value of series resistor RS? OR With the help of a suitable diagram, explain the formation of depletion region in a p-n junction. How does its width change when the junction is (i) forward biased, and (ii) reverse biased? Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams. Draw a schematic diagram showing the (i) ground wave (ii)sky wave and (iii) space wave propagation modes for electromagnetic waves. Write the frequency range for each of the following: (i) Standard AM broadcast (ii) Television (iii) Satellite communication In the given figures, which of the diodes are forward biased, and which are reverse biased and why?

(i)

18. The input resistance of a silicon transistor is 665 :. Its base current is changed by 15 PA which results in change in collector current by 2 mA. This transistor is used as common emitter amplifier with a load resistance of 5 k: Calculate (i) current gain (Eac) (ii) change in emitter current ('IE) and (iii) voltage gain (AV) of the amplifier. 19. Write the truth table for the circuit shown. Show that it represents a XOR gate.

20. A 2 V battery may be connected D1 10 Ω across the points A and B as 20 Ω D2 shown in figure. Assume that the resistance of each diode is zero A B in forward bias and infinity in reverse bias. Find the current supplied by the battery if the positive terminal of the battery is connected to (i) the point A (ii) the point B. 21. Explain briefly the following terms used in communication system: (i) Transducer (ii) Repeater (iii) Amplification 22. Two signals A and B shown in the given figure are used as two inputs of (i) AND gate, (ii) NOR gate and (iii) NAND gate. Obtain the output in each of the three cases.

(ii) SECTION - D

(iii)

(iv)

(v)

68

PHYSICS FOR YOU | DECEMBER ‘17

23. Teena went out for shopping with her mother. During purchase of vegetables, she noticed that the vendor used a digital weighing machine. On another shop, she noticed that the vendor was using an ordinary weighing machine. She remembered having studied about logic gates where, digital codes are used.

(i) What do you mean by logic gate? Mention the basic universal gates. (ii) Draw symbols for OR, AND and NOT gates. (iii) What is the value, in your opinion, that Teena created by the given incident? SECTION - E

24. Draw a block diagram of a simple amplitude modulation. Explain briefly how amplitude modulation is achieved. OR Mention three different modes of propagation used in communication system. Explain with the help of a diagram how long distance communication can be achieved by ionospheric reflection of radio waves. 25. Draw a circuit diagram to study the characteristics of an n-p-n transistor in common emitter configuration. Sketch typical (i) input characteristics, (ii) output characteristics for such a configuration. Explain how the current gain of the transistor can be calculated from output characteristics. OR Give a circuit diagram of a common base amplifier using an n-p-n transistor. Define common base current amplification (D) and common emitter current amplification (E). Deduce a relation between D and E. 26. (i) With the help of circuit diagrams, distinguish between forward biasing and reverse biasing of a p-n junction diode. (ii) Draw V – I characteristics of a p-n junction diode in (a) forward bias, (b) reverse bias. OR Explain, with the help of a circuit diagram the working of a p-n junction diode as a rectifier. SOLUTIONS

1. Size of the dopant atom should be compatible in the pure semiconductor and contribute a charge carrier by forming covalent bond with Si or Ge atoms. Elemental dopants from group XIII and group XV fulfil this condition. 2. The effective power radiation by an antenna of the given length is inversely proportional to the square of the wavelength of the carrier wave. 1 Mathematically, P ∝ . λ2

3. No, the voltmeter should have a very high resistance as compared to the resistance of p-n junction, which is not possible in this case. 4. The modulation index in an AM wave determines the strength and quality of the transmitted signal. In fact, in an AM modulated wave, the information signal is carried by the side bands only and their amplitude depends on modulation index. 5. A transistor works as an open switch in the cut-off state, i.e., when both the emitter and collector are reverse biased. 6. When an electric field is applied across a semiconductor, its free electrons drift in the conduction band in the opposite direction of the field. The motion of these electrons constitutes an electron current. At the same time, the bound electrons in the valence band drift from one vacancy to the next in the opposite direction of the field. This motion of the bound electrons is equivalent to the motion of the holes in the direction of the applied field. The motion of the holes constitutes a hole current. 7. As per question, IB = 20 PA and Edc = 100 ? Collector current, IC = Edc . IB = 100 × 20 PA = 2000 PA = 2.0 mA and emitter current, IE = IC + IB = 2000 PA + 20 PA = 2020 PA = 2.02 mA OR Emitter-base junction is forward biased so that majority charge carriers (say, electrons, for a n-p-n transistor) from heavily doped emitter may easily cross over to the junction to the base side. These electrons are minority charge carriers in base region. The base-collector junction is reverse-biased so that these charge carriers may further cross over to collector region. Because for drift of minority carriers, a reverse bias is needed. Solution Senders of Physics Musing SET-51

1. Ankit Pandey, Delhi 2. Shurabhit Gupta, Bilaspur (Chhattisgarh) 3. Meenakshi Iyer, Chennai (T.N.)

PHYSICS FOR YOU | DECEMBER ‘17

69

8. The gate P is an OR gate and gate Q is an AND gate. The truth table for the logic gate is given in the following table : A

B

C=A+B

Y = A.C

0

0

0

0

0

1

1

0

1

0

1

1

1

1

1

1

10. (i) The frequency of rectified output of a half-wave rectifier is exactly the same (i.e., 50 Hz) as that of ac input. It is because in rectified output, we get output only once (for half period) in 1 cycle of ac input. (ii) The frequency of rectified output of a full-wave rectifier is doubled (i.e., 100 Hz) as that of ac input. It is because during one cycle of ac, the rectified output rises from zero to its maximum level and falls back to zero two times (once in each half cycle). 11. In space wave propagation, the radio wave travels in straight line from transmitting antenna to the receiving antenna. Television broadcast, microwave links and satellite communication are some examples of communication systems that use space wave method of propagation. Distance between the transmitting and receiving antenna in line of sight communication is given by

Differentiating it w.r.t. hT, we get d(dm ) 1 1 = 2R × − 2R × dh T 2 hT 2 h−hT 70

PHYSICS FOR YOU | DECEMBER ‘17

2R ⎛ ⎜ 2 ⎜⎝ 1 hT

−

h h h hR = h – hT = h − = 2 2 2 12. We know that for each atom doped of arsenic one free electron is received. Similarly for each atom doped of indium a vacancy is created. So, the number of free electrons introduced by pentavalent impurity added, ne = NAs = 5 × 1022 m–3 The number of holes introduced by trivalent impurity added, nh = NIn = 5 × 1020 m–3 We know the relation ne nh = ni2 ...(i) Now the number of net electrons available nec = ne – nh = 5 × 1022 – 5 × 1020 = 4.95 × 1022 ...(ii) Now using equation (i), net holes, 16 2 n 2 (1.5 × 10 ) n′h = i = = 4.5 × 109 m −3 n′e 4.95 × 1022 So, nec >> nch, the material is of n-type. or h – hT = hT or h T =

(ii) The intensity of light depends on the doping level of the semiconductor used.

Given that hT + hR = h or hR = h – hT Using it in equation (i), we get dm = 2RhT + 2R(h − hT )

or or

9. LASER stands for light amplification by stimulated emission of radiation; LED stands for light emitting diode. (i) The frequency of light emitted by an LED depends on the band gap of the semiconductor used in LED.

dm = 2RhT + 2RhR

d(dm ) =0 dh T ⎞ 1 1 − ⎟=0 hT h − h T ⎟⎠ 1 1 1 = 0 or = h − hT hT h−hT

For maximum range,

…(i)

13. We have, IE = IB + IC or 'IE = 'IB + 'IC From the question, when 'IE = 8.0 mA, 'IC = 7.9 mA. Thus, 'IB = 8.0 mA – 7.9 mA = 0.1 mA. So a change of 0.1 mA in the base current is required to have a change of 7.9 mA in the collector current. 7.9 mA ΔI α= C = ≈ 0.99 8.0 mA ΔI E 7.9 mA ΔIC = = 79 ΔI B 0.1 mA 14. The value of RS should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e., IZ = 20 mA. The total current through RS is, therefore, 24 mA. The voltage drop across RS is 10.0 – 6.0 = 4.0 V 4. 0 V Hence, RS = = 167 : (24 × 10−3 )A β=

The nearest value of carbon resistor is 150 :. So, a series resistor of 150 : is appropriate. Note that slight variation in the value of the resistance does not matter, what is important is that the current IZ should be sufficiently larger than IL.

OR

When p-n junction is formed, then at the junction, free electrons from n-type diffuse over to p-type, and hole from p-type over to n-type. Due to this a layer of positive charge is built on n-side and a layer of negative charge is built on p-side of the p-n junction. This layer sufficiently grows up within a very short time of the junction being formed, preventing any further movement of charge carriers (electrons and holes) across the p-n junction. This space - charge region, developed on either side of the junction is known as depletion region as the electrons and holes taking part in the initial movement across the junction deplete this region of its free charges. Width of depletion region layer (i) decreases when the junction is forward biased and (ii) increases when it is reverse biased.

Two distinguishing features : (i) In conductors, the valence band and conduction band tend to overlap (or nearly overlap) while in insulators they are separated by a large energy gap and in semiconductors they are separated by a small energy gap. (ii) The conduction band of a conductor has a large number of electrons available for electrical conduction. However, the conduction band of insulators is almost empty while that of the semiconductor has only a (very) small number of such electrons available for electrical conduction.

15.

PHYSICS FOR YOU | DECEMBER â€˜17

71

16. The diagram given is showing various propagation modes for electromagnetic waves.

Frequency ranges : (i) Standard AM broadcast : 540-1600 kHz (ii) Television : Very high frequencies 54-72 MHz TV 76-88 MHz Ultra high frequencies 174-216 MHz TV 420-890 MHz (iii) Satellite communication : 5.925-6.425 GHz (Uplink) 3.7-4.2 GHz (Downlink) 17. (i) Reverse biased as p is at a lower potential (zero) than n (+5 V). (ii) Forward biased as p is at a higher potential (+10 V) than n (+5 V) (iii) Reverse biased as p is at a lower potential (–10 V) than n (zero). (iv) Forward biased as p is at a higher potential (–5 V) than n (–12 V). (v) Forward biased as p is at a higher potential (zero) than n (–10 V). 18. Rin = 665 :, 'IB = 15 × 10–6 A, 'IC = 2 × 10–3 A RL = 5 × 103 : 2 × 10−3 ΔI = 133.33 (i) Current gain, βac = C = ΔI B 15 × 10−6 (ii) Change in emitter current, 'IE = 'IB + 'IC = (15 × 10–6) + (2 × 10–3) = 2.015 × 10–3 A = 2.015 mA R 5 × 103 (iii) Voltage gain, AV = βac ⋅ L = 133. 33 × Rin 665 = 1002.5 19. The given circuit contains an OR, a NAND and an AND gate, as shown.

72

PHYSICS FOR YOU | DECEMBER ‘17

The inputs A and B are fed to OR gate so that its output is Yc = A + B Also, the inputs A and B are fed to NAND gate so that its output is Ys = AB Then Yc and Ys are fed to AND gate so that the output is Y = Yc.Ys = (A + B).(AB) The logic table for the given circuit is OR gate Inputs

NAND gate

Output

Inputs

AND gate

Output

Inputs

Output

A

B Yc = A + B

A

B

Ys = AB

Yc

0

0

0

0

0

1

0

1

0

0

1

1

0

1

1

1

1

1

1

0

1

1

0

1

1

1

1

1

1

1

1

1

0

1

0

0

Ys Y = Yc.Ys

Hence the truth table for the given circuit is Inputs

Output

A

B

Y

0

0

0

0

1

1

1

0

1

1

1

0

which is the truth table of XOR gate. 20. (i) When the positive terminal of the battery is connected to the point A, the diode D1 is forward biased and D2 is reverse-biased. The resistance of the diode D1 is zero, and it can be replaced by a resistanceless wire. Similarly, the resistance of the diode D2 is infinity, and it can be replaced by a broken wire. The equivalent circuit is shown in figure. The current supplied by the battery is 2 V/10 : = 0.2 A. (ii) When the positive terminal of the battery is connected to the point B, the diode D2 is forward biased and D1 is reverse biased. The equivalent circuit is shown in figure. The current through the battery is 2 V/20 : = 0.1 A.

PHYSICS FOR YOU | DECEMBER â€˜17

73

21. (i) Transducer : Any device that convert one form of energy into another can be termed as a transducer. In electronic communication systems, we usually come across devices that have either their inputs or outputs in the electrical form. An electrical transducer is a device that converts some physical variable (pressure, displacement, force, temperature, etc.) into corresponding variations in the electrical signal at its output. (ii) Repeater: A repeater is a combination of a receiver and a transmitter. A repeater picks up the signal from the transmitter, amplifies and retransmits it to the receiver, sometimes with a change in carrier frequency. Repeaters are used to extend the range of communication system. A communication satellite is essentially a repeater station in space.

23. (i) A gate is a digital circuit that follows certain logical relationship between the input and output voltages. Therefore, they are generally known as logic gate. NAND and NOR gates are universal gates. (ii) OR gate AND gate NOT gate (iii) Concentration and observation in the class room, retaining capacity, co-relating of what was taught with the real life incident. 24. Refer to point 10.2 (6), page no. 638 (MTG Excel in Physics). OR Refer to point 10.3 (1, 2, 3), page no. 641 (MTG Excel in Physics). 25. Refer to point 9.4 (7), page no. 594 (MTG Excel in Physics). OR Refer to point 9.4 (4, 6), page no. 593 (MTG Excel in Physics).

(iii) Amplification: It is the process of increasing the amplitude of a signal using an electronic circuit called the amplifier. Amplification is necessary to compensate for the attenuation of the signal in communication systems. The energy needed for additional signal strength is obtained from a dc power source. Amplification is done at a place between the source and destination wherever signal strength becomes weaker than the required strength. 22.

26. Refer to point 9.3 (4, 5), page no. 588 (MTG Excel in Physics). OR Refer to point 9.3 (6), page no. 588 (MTG Excel in Physics).

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74

PHYSICS FOR YOU | DECEMBER ‘17

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MPP-8

Class XII

T

his specially designed column enables students to self analyse their extent of understanding of speciﬁed chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.

Dual Nature of Matter and Radiation Time Taken : 60 min

Total Marks : 120 NEET / AIIMS

3.

If the de-Broglie wavelength of an electron is 1 Å, what is its kinetic energy? (a) 151 eV (b) 241 eV (c) 96 eV (d) 184 eV

4.

The work functions of four materials M1, M2, M3 and M4 are 1.9 eV, 2.5 eV, 3.6 eV and 4.2 eV, respectively. Which of these material is (are) useful in a photocell to detect visible light? (a) M1 only (b) M1 and M2 (c) M3 only (d) All

5.

Maximum velocity of the photoelectron emitted by a metal is 1.8 u106 m s–1. Take the value of specific charge of the electron is 1.8 u1011 C kg–1. Then the stopping potential in volt is (a) 1 (b) 3 (c) 9 (d) 6

6.

When monochromatic radiation of intensity I falls on a metal surface, the number of photoelectrons and their maximum kinetic energy are N and T respectively. If the intensity of radiation is 2I, the number of emitted electrons and their maximum kinetic energy are respectively (a) N and 2T (b) 2N and T (c) 2N and 2T (d) N and T

7.

For photoelectric emission from certain metal the cutoff frequency is X. If radiation of frequency 2X impinges on the metal plate, the maximum possible velocity of the emitted electron will be (m is the electron mass)

Only One Option Correct Type 1.

2.

The stopping potential for photoelectrons from a metal surface is V1 when monochromatic light of frequency X1 is incident on it. The stopping potential becomes V2 when monochromatic light of another frequency is incident on the same metal surface. If h be the Planck’s constant and e be the charge of an electron, then the frequency of light in the second case is e e (a) υ1 − (V2 + V1) (b) υ1 + (V2 + V1) h h e e (c) υ1 − (V2 − V1) (d) υ1 + (V2 − V1) h h Which of the following facts about the photoelectric effect can be understood without invoking the quantum concept of light propagation? (a) The rate of photoelectrons emission, when they are emitted, increases with the intensity of light used. (b) There is a threshold frequency, below which no photoelectrons are emitted, no matter how long the light is thrown on the metallic surface. (c) Once the frequency of light is more than the threshold frequency, photoelectrons are emitted almost instantaneously, no matter how weak the light intensity is. (d) For each frequency of light, exceeding the threshold frequency, there is a maximum kinetic energy of the emitted electrons.

(a)

hυ 2hυ (b) 2 (c) m m

hυ (2m)

(d)

PHYSICS FOR YOU | DECEMBER ‘17

hυ m 75

8.

An electron of mass me and a proton of mass mp are moving with the same speed. The ratio of their de-Broglie’s wavelengths

1 (d) 918 1836 The de-Broglie wavelength of a neutron at 27 °C is O. What will be its wavelength at 927 °C? (a) O/2 (b) O/3 (c) O/4 (d) O/9

(a) 1 9.

λe is λp

(b) 1836

(c)

(c)

13. Assertion : Stopping potential depends upon the frequency of incident light but is independent of the intensity of the light. Reason : The maximum kinetic energy of the photoelectrons is directly proportional to the stopping potential.

(b)

14. Assertion : Though light of a single frequency (monochromatic light) is incident on a metal, the energies of emitted photoelectrons are different. Reason : The energy of electrons just after they absorb photons incident on the metal surface may be lost in collision with other atoms in the metal before the electron is ejected out of the metal.

(d)

15. Assertion : The de-Broglie wavelength of a molecule (in a sample of ideal gas) varies inversely as the square root of absolute temperature. Reason : The de-Broglie wavelength depends on the kinetic energy of a body only.

10. The log-log graph between the energy E of an electron and its de-Broglie wavelength O will be

(a)

(c) If assertion is true but reason is false. (d) If both assertion and reason are false.

JEE MAIN / JEE ADVANCED Only One Option Correct Type

11. An AIR station is broadcasting the waves of wavelength 300 m. If the radiating power of the transmitter is 10 kW, then the number of photons radiated per second is (a) 1.5 × 1029 (b) 1.5 × 1031 33 (c) 1.5 × 10 (d) 1.5 × 1035 12. A point source of light is used in an experiment on photoelectric effect. Which I I of the following curves best II III represents the variation of photo current (I) IV with distance (s) of the s source from the emitter? (a) I (b) II (c) III (d) IV Assertion & Reason Type Directions : In the following questions, a statement of assertion is followed by a statement of reason. Mark the correct choice as : (a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. 76

PHYSICS FOR YOU | DECEMBER ‘17

16. When radiation of wavelength O is incident on a metallic surface, the stopping potential is 4.8 V. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 V. Then the threshold wavelength for the surface is (a) 2O (b) 4O (c) 6O (d) 8O 17. Light of wavelength 2475 Å o B is incident on Barium. e Photoelectrons emitted describe a circle of radius 100 cm by a Barium magnetic field of flux density 1 × 10–5 T. Work function of the Barium is 17 e = 1.7 × 1011 C kg–1) (Given m (a) 1.8 eV (b) 2.1 eV (c) 4.5 eV (d) 3.3 eV 18. In a photoemissive cell with executing wavelength O, the fastest electron has speed v. If the exciting wavelength is changed to 3O/4, the speed of the fastest emitted electron will be

(a) v(3/4)1/2 (b) v(4/3)1/2 1/2 (c) less than v(4/3) (d) greater than v(4/3)1/2 19. In the following arrangement y = 1.0 mm, d = 0.24 mm and D = 1.2 m. The work function of the material of the emitter is 2.2 eV. The stopping potential V needed to stop the photo current will be

(a) 0.9 V (b) 0.5 V (c) 0.4 V (d) 0.1 V More than One Options Correct Type 20. Which of the following are correct according to wave theory of light regarding photoelectric effect? (a) Amplitude of vibration of an electron is directly proportional to the maximum amplitude of incident wave. (b) Average kinetic energy of photoelectrons is directly proportional to the square of their vibrational amplitude. (c) Average kinetic energy of photoelectrons is proportional to the intensity of incident light. (d) Stopping potential for the most energetic photoelectrons is proportional to the intensity of incident light. 21. When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have a maximum kinetic energy EA eV and de-Broglie wavelength OA. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.70 eV is EB = (EA – 1.50) eV. If the de-Broglie wavelength of the photoelectrons is OB = 2OA, then (a) the work function of A is 2.25 eV (b) the work function of B is 4.20 eV (c) EA = 2.0 eV (d) EB = 2.75 eV 22. When a point light source of power W emitting monochromatic light of wavelength O is kept at a distance a from a photo-sensitive surface of work function I and area S, we will have

(a) number of photons striking the surface per unit time as W OS/4 Shca2 (b) the maximum energy of the emitted photoelectrons as (1/O) (hc – OI) (c) the stopping potential needed to stop the most energetic emitted photoelectrons as (e/O)(hc – OI) (d) photo-emission only if O lies in the range 0 ≤ O ≤ (hc/I) 23. When a metal is irradiated by a light of O = 4000 Å, all the photoelectrons emitted are bent in a circle of radius 50 cm by a magnetic field of flux density 5.26 × 10–6 T acting perpendicular to plane of emission of photoelectrons. Then, (a) the kinetic energy of fastest photoelectron is 0.6 eV (b) work function of the metal is 2.5 eV (c) the maximum velocity of photoelectric effect is 0.46 × 106 m s–1 (d) the stopping potential for photoelectric effect is 0.8 V Integer Answer Type 24. A 100 W light bulb is placed at the centre of a spherical chamber of radius 20 cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. The pressure exerted by the light on the surface on the surface of the chamber is 4.0 × 10–x Pa. Find x. 25. A light beam of wavelength 400 nm is incident on a metal plate of work function 2.2 eV. Assuming that 10% of the extra energy is lost to the metal in each collision, find the maximum number of collisions the electron can suffer before it becomes unable to come out of the metal. 26. In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter so that the current reaches its saturation value. Assuming that on the average one out of every 106 photons is able to eject a photoelectron, the photocurrent in the circuit is 1.6 × 10–x A. Find x. Comprehension Type A bound electron can absorb a photon (as it does in case of photoelectric effect) but a free electron cannot absorb a photon. According to quantum theory of light, photons behave like particles except for their lack of rest mass. PHYSICS FOR YOU | DECEMBER ‘17

77

We treat electrons and photons both as particles (if both are billiard balls). Let an X-ray photon strikes an electron.

The calculations of momentum and energy show that in such collisions, a photon loses a ceratin amount of energy and an electron gains that part of energy. This is called Compton effect. If O is wavelength of incident photon and Ocis wavelength of scattered photon, then Compton shows that, h λ′ − λ = (1 − cos φ) mc where h = Planck’s constant, m = mass of electron, c = speed of light in vacuum, I = angle between directions of incident and scattered photon. The Compton effect is the chief means by which the X-rays lose energy, when pass through matter. 27. X-rays of wavelength 10 pm are scattered after colliding with an electron. Maximum wavelength present in scattered X-rays spectrum will be (b) 10.7 pm (a) 10 pm (c) 14.9 pm (d) 5.1 pm

Column-I

Column-II

(A) If X is increased keeping I and I constant

(P) Stopping potential increases

(B) If I is increased keeping X and I constant

(Q) Saturation photocurrent increases

(C) If the distance between anode and cathode is increased

(R) Maximum KE of the photoelectrons increases

(D) If Iis decreased keeping X and I constant

(S) Stopping potential remains the same D P, R P, R P, R Q

(a) (b) (c) (d)

A P, Q P, R Q, R R

B P, S Q, S Q, S S

C S S P, R P, R

30. If OD, OP, OD, OT are the de-Broglie wavelengths of D-particle, proton, deutron and tritium respectively when they are accelerated by 100 V, 400 V, 200 V and 600 V, respectively. Match the entries of Column I with the entries in Column II and choose the correct option.

28. Maximum kinetic energy of recoiling electron is (a) 40.8 keV (b) 40.8 MeV (c) 40.8 eV (d) 40.8 J Matrix Match Type 29. With respect to photoelectric experiment, match the entries of Column I with the entries of Column II and choose the correct option. Here, X = frequency of incident light, I = intensity of light, I = work function of a metal

Column-I

Column-II

(A) OD/OP

(P) Less than

(B) OD/OD (C) OD/OT (D) OP/OT

(Q) Less than 3 (R) Greater than 2 (S) Less than 3

(a) (b) (c) (d)

A B P, Q P, S Q, S P, Q, S P, Q, S P, Q, S P, Q, S P, Q

C Q, S Q, R Q, S Q, R

2

D R S R, S R, S

Keys are published in this issue. Search now! -

Check your score! If your score is > 90%

EXCELLENT WORK !

You are well prepared to take the challenge of ﬁnal exam.

No. of questions attempted

……

90-75%

GOOD WORK !

You can score good in the ﬁnal exam.

No. of questions correct

……

74-60%

SATISFACTORY !

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……

< 60%

78

PHYSICS FOR YOU | DECEMBER ‘17

NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

OLYMPIAD PROBLEMS 1. Two fixed equal positive charges, each of magnitude 5 × 10–5 C are located at points A and B separated by a distance of 6 m. An equal and opposite charge (– q) moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge, when it reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 J. Calculate the distance of the farthest point D from point O at which the negative charge will reach before returning towards C. (a) 68 m (b) 72 m (c) 9 m (d) 5 m

2. In a Searle’s experiment, the diameter of the wire as measured by a screw gauge of least count 0.001 cm is 0.050 cm. The length, measured by a scale of least count 0.1 cm, is 110.0 cm. When a weight of 50 N is suspended from the wire, the extension is measured to be 0.125 cm by a micrometer of least count 0.001 cm. Find the maximum error in the measurement of Young’s modulus of the material of the wire from these data. (a) 1.09 × 1010 N m–2 (b) 2.24 × 1011 N m–2 (c) 4.89 × 109 N m–2 (d) 3.24 × 1010 N m–2 3. Two parallel vertical metallic rails AB and CD are separated by 1 m. They are connected at the two-ends by resistances R1 and R2 as shown in figure. A horizontal metallic bar L of mass 0.2 kg slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of 0.6 T perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in R1 and R2 are 0.76 W and 1.2 W respectively. Find the approximate value of terminal velocity of the bar L. (a) 10 m s–1(b) 3 m s–1 (c) 8 m s–1 (d) 1 m s–1

4. A body is projected vertically upwards from the R where R bottom of a crater of moon of depth 100 is the radius of moon with a velocity equal to the escape velocity on the surface of moon. Calculate maximum height attained by the body from the surface of the moon. R R (a) 99 R (b) 50 R (c) (d) 2 36 5. 12 cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ammeter and two cells identical with the others. The current is 3 A when the cells and battery aid each other and is 2 A when the cells and battery oppose each other. How many cells in the battery are wrongly connected? (a) 1 (b) 3 (c) 11 (d) 4 6. Two bodies A and B are connected by a light rigid bar 10 m long move in two frictionless guides as shown in figure. If B starts from rest when it is vertically below A. The velocity (in m s–1) of B when x = 6 m. As s u m e mA = mB = 200 kg and mC = 100 kg is (a) 4.8 m s–1 (b) 1.6 m s–1 –1 (c) 6.9 m s (d) 3.2 m s–1 7. The arrangement shown in figure consists of two identical uniform solid cylinders, each of mass m, on which two light threads are wound symmetrically. Find the tension in each thread during the motion. The friction in the axle of the upper cylinder is assumed to be absent. mg mg (a) mg (b) (c) (d) 2mg 2 10 PHYSICS FOR YOU | DECEMBER ‘17

79

SOLUTIONS 1. (b) : Two fixed equal positive charges each of magnitude = 5 × 10–5 C Separation between charges = 6 m An equal and opposite charge moves towards them. When moving charge is at 4 m from O then kinetic energy = 4 J From the diagram, AO = OB = 3 m, OC = 4 m, AC = BC = 5 m. Let AD = BD = r. Using law of conservation of energy, Total energy at C = Total energy at D Since at D, charged particle changes its direction. ? vD = 0 or kinetic energy at D = 0 ⎛ 9 × 109 × 5 × 10−5 × 5 × 10−5 ⎞ ∴ ⎜ −2 × ⎟+4 5 ⎝ ⎠ ⎛ 9 × 109 × 5 × 10−5 × 5 × 10−5 ⎞ = ⎜ −2 × ⎟ r ⎝ ⎠

or r = 9 m OD = (AD)2 − (OA)2 = 92 − 32 = 72 m.

2. (a) : Since, Young’s modulus Y=

FL Al

∴Y =

=

FL

=

4FL

....(i)

⎛ πd ⎞ πd 2l ⎜ 4 ⎟l ⎝ ⎠ 2

4 × 50 × 1.1 3.14 × (5.0 × 10−4 )2 × (1.25 × 10−3) 11

–2

= 2.24 × 10 N m Maximum possible fractional error, 2 Δd ΔY ΔL Δl = + + Y L l d 0.1 0.001 2 × 0.001 =

110

+

0.125

+

0.05

= 0.0009 + 0.008 + 0.04 = 0.0489 or 'Y = (0.0489) × Y = (0.0489) × (2.24 × 1011) N m–2. = 1.09 × 1010 N m–2 3. (d) : If v be the instantaneous velocity of the bar L, then, magnitude of the emf induced in dx dφ d = (BLx) = BL = BLv the bar, ε = dt dt dt where x is the distance through which the bar slides. The equivalent resistance of the resistors is given by RR R= 1 2 R1 + R2 BLv ? current through the bar, I = R

80

PHYSICS FOR YOU | DECEMBER ‘17

Since there are two forces acting on the bar, i.e., force of gravity and magnetic force, in the opposite directions. The bar will attain terminal velocity when mg 0.2 × 9.8 mg = BIL or, I = = = 3.267 A BL 0. 6 × 1 If terminal velocity attained be vt, then at steady state the emf induced in the bar = BLvt and (BLvt ) 2 = 0.76 W and Power dissipated in R1 is = R1 power dissipated in R2 is =

(BLvt ) 2 = 1.2 W R2

⎛1 1 ⎞ Adding (BLvt)2 ⎜ + ⎟ = 0.76 + 1.2 = 1.96 ⎝ R1 R2 ⎠ 1.96 1.96 1 = 0.184 : or R2I2 × = 1.96 or R = 2 = R (3.267)2 I Again (BLvt)2 = 1.96 R = 1.96 × 0.184 ƿ 0.36 0. 6 0. 6 or, BLvt = 0.6 ? vt = = 1 m s–1 = BL 0.6 × 1 4. (a) : BC is the crater of moon. B denotes bottom of crater. R Depth of crater, d = 100 Radius of moon = R Speed of particle at B = vB Escape velocity on surface of moon, ∴

ve =

2GM R

where M is mass of the moon. At highest point vA = 0 GM VA = Potential at A = − (R + h) −GM VB = Potential at B = (R − d) −GM −GM = = (1.01) R ⎞ ⎛ R ⎟ ⎜⎝ R − 100 ⎠ Using conservation of mechanical energy GMm GMm × 1.01 1 2 mv B = − + R 2 (R + h) GM R or

1 1.01⎤ ⎡ = GM ⎢− + R ⎥⎦ ⎣ R+h 1 R

= −

1 R+h

+

1.01 R

[∵ vB = ve]

or

1 R+h

=

0.01 R

=

1 100R

or 100R = R + h or h = 99R 5. (a) : Total number of cells connected in series = 12 Current in circuit when cell and battery aid each other = 3 A Current in circuit when cell and battery oppose each other = 2 A. To find the number of cells wrongly connected in the circuit. Let x cells are connected correctly and y cells be connected wrongly. According to problem (x + y) = 12 ...(i) The net emf of the battery = (x – y) H, where His the emf of one cell. Let R be the total resistance of the circuit, which remains constant. When the battery and the cells aid one another, the net emf = (x – y)H + 2H When the two oppose each other, the net emf = (x – y)H – 2H net emf Current I = total resistance (x − y)ε + 2ε ? In the first case, 3 = and in the R (x − y)ε − 2ε second case, 2 = R 3 x − y +2 Dividing, we get = 2 x − y −2 ? 3x – 3y – 6 = 2x – 2y + 4 ? x – y = 10 ...(ii) Solving (i) and (ii), we get x = 11 and y = 1 Hence only one cell is wrongly connected. 6. (c) : At the instant, when the bar is shown in the figure, x2 + y2 = l2 dx dy ∴ 2x + 2y = 0 ...(i) dt dt dx dy dy x dx ∴ x +y =0 ⇒ =− ...(ii) dt dt dt y dt dx dy where = velocity of A and = velocity of B dt dt y = l 2 − x 2 = (10)2 − (6)2 = 8 m Applying the law of conservation of energy, loss of potential energy of A, if it is going down when the rod is vertical to the position as shown in the figure = mAg [10 – 8] = 2 × 200 × 9.8, C moves down 6 m since B moves 6 m along x-axis

Loss of potential energy of C (= mgh)= 100 × 9.8 × 6 Total loss of potential energy = (200 × 9.8 × 2) + (100 × 9.8 × 6) = 9800 J ∵ Kinetic energy gained = Loss of potential energy ⇒

2 1 ⎛ dx ⎞ ⎛ dy ⎞ 1 × 200 ⎜ ⎟ + × 200 ⎜ ⎝ dt ⎠ 2 ⎝ dt ⎟⎠ 2

2

2

1 ⎛ dx ⎞ = 9800 + × 100 ⎜ ⎝ dt ⎟⎠ 2 2

⎛ dx ⎞ ⎛ dy ⎞ or 100 ⎜ ⎟ + 150 ⎜ ⎝ dt ⎠ ⎝ dt ⎠⎟ 2 ⎛ x dx ⎞ ⎛ dx + 150 ⎜ 100 ⎜ − ⎟ ⎝ dt ⎝ y dt ⎠

2

= 9800

2 ⎞ ⎟⎠ = 9800 [Using eqn. (ii)] On substituting values, we get 98 × 16 3300 2 ∴ vB = 9800 ∴ vB = = 6.9 m s −1 33 16 ? velocity of B at the required moment = 6.9 m s–1 7. (c) : Mass of each cylinder = m Let r be the radius of each solid cylinder. For the upper cylinder, the torque, W = ID ...(i) where I is moment of inertia and D is angular acceleration of upper cylinder about its stationary axis of rotation. If T be the tension in the string, then from equation (i), 4T mr 2 ...(ii) 2Tr = α or α = mr 2 For the rotation of the lower cylinder from the equation, W = I Dc 4T mr 2 =α 2Tr = .α′ or α′ = mr 2 Now for the translational motion of lower cylinder from the equation, F = ma mg – 2T = ma ...(iii) As there is no slipping of threads on the cylinders ? a = Dcr + Dr = 2Dr Now, from equations (ii) and (iii), 4T m( g − 2 .r) m( g − a) m( g − 2αr) mr T= = or T = 2 2 2 mg 2T = mg – 8T T = 10 PHYSICS FOR YOU | DECEMBER ‘17

81

P

MUSING

PHYSICS

hysics Musing was started in August 2013 issue of Physics For You. The aim of Physics Musing is to augment the chances of bright students preparing for JEE (Main and Advanced) / NEET / AIIMS / JIPMER with additional study material. In every issue of Physics For You, 10 challenging problems are proposed in various topics of JEE (Main and Advanced) / NEET. The detailed solutions of these problems will be published in next issue of Physics For You. 6JGTGCFGTUYJQJCXGUQNXGFſXGQTOQTGRTQDNGOUOC[UGPFVJGKTFGVCKNGFUQNWVKQPUYKVJVJGKTPCOGUCPFEQORNGVGCFFTGUU6JGPCOGUQHVJQUG YJQUGPFCVNGCUVſXGEQTTGEVUQNWVKQPUYKNNDGRWDNKUJGFKPVJGPGZVKUUWG We hope that our readers will enrich their problem solving skills through “Physics Musing” and stand in better stead while facing the competitive exams.

53 SINGLE OPTION CORRECT TYPE

1. Two long straight cylindrical conductors with resistivities U1 and U2 respectively are joined together as shown in figure. If current I flows through the conductors, the magnitude of the total free charge at the interface of the two conductors is I

r1

I

r2

(ρ1 − ρ2 )I ε0 2 (d) ε0 I ρ1 + ρ2

(a) zero

(b)

(c) ε0 I ρ1 − ρ2

2. In the given black box unknown emf sources and R I Black box unknown resistances are connected by an unknown method such that (i) when terminals of 10 : resistance are connected to box then 1 A current flows through it and (ii) when terminals of 18 : resistance are connected then 0.6 A current flows through it then for what value of resistance does 0.1 A current flow? (a) 118 : (b) 98 : (c) 18 : (d) 58 : 3. In the given circuit, AB is a potentiometer wire of length 40 cm and resistance per unit length 50 :m–1. As shown in the figure, the free end of an ideal voltmeter is touching the potentiometer wire. What should be the velocity of the jockey as a function of time so that reading in the voltmeter varies with time as (2sinSt)? 10 W

10 W V

A

J 4V

B

(a) (10S sinSt) cm s–1 (c) (20S sinSt) cm s–1

(b) (10S cosSt) cm s–1 (d) (20S cosSt) cm s–1

MULTIPLE OPTIONS CORRECT TYPE

4. A metal sphere of radius a is surrounded by a concentric metal sphere of inner radius b, where b > a. The space between the spheres is filled with a material whose electrical conductivity V varies with the electric field strength E as V = kE where k is a constant. A potential difference V is maintained between spheres. Choose the correct statements. (a) Current is 4Sr2kE2, where (a < r < b). (b) Current is 2Sr2kE2, where (a < r < b). (c) Potential difference between spheres is I ⎛b⎞ V= ln ⎜ ⎟ where I is total current. 4πk ⎝ a ⎠ (d) Potential difference between spheres is I ⎛b⎞ V= ln ⎜ ⎟ where I is total current. 2πk ⎝ a ⎠ 5. Consider a simple circuit shown in figure.

stands for a variable A B R resistance Rc. Rc can vary from R0 to infinity. r is I V internal resistance of the r battery (r << R << R0). Choose the correct statements. (a) Potential drop across AB is nearly constant as Rc is varied. (b) Current through Rc is nearly a constant as Rc is varied. (c) Current I depends upon Rc V (d) I > , always. r+R

By Akhil Tewari, Author Foundation of Physics for JEE Main & Advanced, Professor, IITians PACE, Mumbai.

82

PHYSICS FOR YOU | DECEMBER ‘17

R¢

INTEGER ANSWER TYPE

6. The electric potential varies in space according to the relation V = 3x + 4y. A particle of mass 10 kg and charge +1 PC starts from rest from point (2, 3.2) under the influence of this field. Find the velocity (in mm s–1) of the particle when it crosses the x-axis. Assume V and (x, y) are in S.I. units. 7. In the circuit shown, find the current through 5 : resistance.

8. A free water molecule of dipole moment p = S2 × 10–30 C m lies in xy-plane in gravity free space. The moment of inertia of the molecule about its centroidal axis perpendicular to the plane containing the molecules, that is, z-axis is 2 × 10– 48 kg m2. If the molecule is kept in a uniform electric field parallel to x-y plane having

field intensity E = 2 × 106 N C–1, the dipole oscillates about z-axis. The frequency of small oscillations is n × 1011 Hz. Write the value of n. 9. The coil of a calorimeter C has a resistance R1 = 60 :. The coil R1 is connected to the circuit as shown in figure. If rise in temperature (°C) of 240 g of water poured into the calorimeter is 5x when it is heated for 7 min during which a current flows through the coil and the ammeter shows 3A? The resistance R2 = 30 :. Then find the value of x. [Disregard the resistances of the battery and the ammeter, and the heat losses and heat capacity of the calorimeter and the resistor and take specific heat of water = 4200 J kg–1°C–1] 10. In shown circuit, there are two identical ideal cells of emf H and r = 9 :, find the value of R (in :) so that the power dissipated in resistance R is maximum. r

e r

e r

R

PHYSICS FOR YOU | DECEMBER ‘17

83

1. A toy yo-yo of total mass M = 0.24 kg consists of two disks of radius R = 2.8 cm connected by a thin shaft of radius R0 = 0.25 cm. A string of length L = 1.2 m is wrapped around the shaft. If the yo-yo is thrown downward with an initial velocity of v0 = 1.4 m s–1, what is its rotational velocity when it reaches the end of the string?

5. Twelve capacitors, each having a capacitance C, are connected to form a cube as shown. Find the equivalent capacitance between the diagonally opposite corners such as A and B.

A yo-yo falls as the string unwinds from the axle.

2. A gun of mass M1 fires a bullet of mass M2 with a horizontal speed v0. The gun is fitted with a concave mirror of focal length f facing towards the receding bullet. Find the speed of separations of the bullet and the image just after the gun was fired. 3. A moving hydrogen atom makes a head on collision with a stationary hydrogen atom. Before collision both atoms are in ground state and after collision they move together. What is the minimum value of the kinetic energy of the moving hydrogen atom, such that one of the atoms reaches one of the excitation state. 4. A rod has a uniform positive charge density O on the top half of the rod and a uniform charge density –Oon the bottom half of the rod. Find the net force on the point charge q0 placed at a distance y on axis perpendicular to the rod and passing through the centre. 84

PHYSICS FOR YOU | DECEMBER ‘17

6. A particle A having a charge of 5.0 × 10–7 C is fixed in a vertical wall. A second particle B of mass 100 g and having equal charge is suspended by a silk thread of length 30 cm from the wall. The point of suspension is 30 cm above the particle A. Find the angle of the thread with the vertical when it stays in equilibrium. 7. In an Atwood’s machine one block has a mass of 512 g and the other a mass of 463 g. The pulley, which is mounted in horizontal frictionless bearings, has a radius of 4.90 cm. When released from rest, the heavier block is observed to fall 76.5 cm in 5.11 s. Calculate the rotational inertia of the pulley. 8. An electrical circuit is shown in the figure. Calculate the potential difference across the resistor of 400 :, as will be measured by the voltmeter V of resistance 400 :.

9. A non-conducting ring of mass m and radius R has charge Q uniformly distributed over its

circumference. The ring is placed on a rough horizontal surface such that the plane of the ring is parallel to the surface. A vertical magnetic field B = B0t2 is switched on. After 2 s from switching on the magnetic field, the ring is just about to rotate about vertical axis through its center. (a) Find friction coefficient P between the ring and the surface. (b) If magnetic field is switched off after 4 s, then find the angular velocity of the ring just after switching off the magnetic field. 10. A glass vessel of volume 100 cm3 is filled with mercury and is heated from 25 °C to 75 °C. What volume of mercury will overflow? Coefficient of linear expansion of glass = 1.8 × 10–6 °C–1 and coefficient of volume expansion of mercury is 1.8 × 10–4 °C–1. SOLUTIONS 1. The free-body diagram for the yo-yo is shown. The net force acting is ∑ Fy = Mg – T Mg – T = May ... (i) (taking the downward direction to the positive), The net torque acting about the center of mass is ∑Wz = TR0 The force diagram TR0 = IDz ... (ii) (taking counterclockwise torque to be positive). We consider the string to be of negligible thickness and assume that it does not slip as it is unwinding. ? The point where the string contacts the shaft is instantaneously at rest. Hence, ay = DzR0 and Solving eqns. (i) and (ii) g ⎛ 1 ⎞ αz = R0 ⎜⎝ 1 + I / MR02 ⎟⎠ Let us assume that the thin shaft makes a negligible contribution to I as the mass and radius of the shaft are both small compared to the disks 1 2 Then the rotational inertia is I = MR 2 g ? αz = R0 + R2 / 2R0 980 = = 61.5 rad s −2. 2 0.25 + (2.8) /(2 × 0.25)

1 D t2 ...(iii) 2 z The angle through which the yo-yo rotates as the string unwinds is I – I0 = L/R0 = 480 rad, and the initial angular velocity is Z0z = v0/R0 = (1.4 m s–1)/(0.0025 m) = 560 rad s–1. Substituting these in eqn. (iii) we get, (30.75)t2 + (560)t – 480 = 0. Solving this quadratic equation, we find t | 0.82 s or –19 s. Neglecting negative value of time, we have Zz = Z0z + Dzt = 560 rad s–1 + (61.5 rad s–2)(0.82 s) | 610 rad s–1.

I = I0 + Z0zt +

2. Let v1 be the speed of gun (or mirror) just after the firing of bullet. From conservation of linear momentum, Mv M2v0 = M1v1 v1 = 2 0 ...(i) M1 Assuming y as image distance and x as object distance from pole of mirror. dy Now, = rate at which distance between mirror dt ...(ii) and bullet is increasing = v1 + v0 2⎞ ⎛ dy y dx Since = −⎜ 2 ⎟ . dt ⎝ x ⎠ dt 2 y Here, 2 = m2 = 1 x (as at the time of firing bullet is at pole). dy dx = = v1 + v0 ? ...(iii) dt dt Hence speed of separation between bullet and image will be ⎛ dx ⎞ ⎛ dy ⎞ ⎛ dx ⎞ vr = ⎜ ⎟ − ⎜ ⎟ = 2 ⎜ ⎟ = 2(v1 + v0 ) [Using (iii)] ⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠ ⎛ M ⎞ vr = 2 ⎜1 + 2 ⎟ v0 ⎝ M1 ⎠ 3. Let K be the kinetic energy of the moving hydrogen atom and Kc the kinetic energy of combined mass after collision.

Using Zz = Z0z + Dzt, if we know the time t for the yo-yo to unwind. This time can be found from equation PHYSICS FOR YOU | DECEMBER ‘17

85

From conservation of linear momentum, 2Km = 2K ′(2m)

or

K = 2Kc

...(i)

From conservation of energy, K = Kc + 'E ...(ii) K Solving eqns. (i) and (ii), we get ΔE = 2 Now, minimum value of 'E for hydrogen atom is 10.2 eV. K ≥ 10.2 eV or K ≥ 20.4 eV 2 Therefore, the minimum kinetic energy of moving hydrogen is 20.4 eV.

or 'E ≥ 10.2 eV ?

4. The force on the charge particle due to element dz marked in the figure, along the length of rod is z

r T y

x –O

dFx = dF sin θ =

°–

T

dq

90

dz +O z L

q0 L dFx

1 q0 λdz 4πε0 ( y 2 + z 2 )

dFy dF

z y2 + z2

.

We will need to take into consideration that O changes sign for the two halves of the rod. Then Fx =

=

=

L /2 q0 λ ⎛ 0 + zdz ⎞ − zdz +∫ , ⎜ ∫− L / 2 2 2 3 2 2 / 0 ( y + z 2 )3/2 ⎟ 4 πε0 ⎝ (y + z ) ⎠

q0 λ 2πε0

L /2

∫0

qλ = 0 , 2πε0 ( y 2 + z 2 )3/2

q0 λ ⎛ 1 ⎜ − 2πε0 ⎜⎝ y

zdz

−1 y2 + z2

L/2

0

⎞ ⎟ y 2 + (L / 2)2 ⎟⎠ 1

The component of force perpendicular to the rod dFy cancels due to bottom half. Hence the net force on q0 is Fq = 0

q0 λ ⎛ 1 − 2πε0 ⎜ y ⎜ ⎜⎝

⎞ ⎟ ⎛ L⎞ y 2 + ⎜ ⎟ ⎟⎟ ⎝ 2⎠ ⎠ 1

2

5. By symmetry, the capacitors a, g and d have charges Q/3, Q/6 and Q/3 respectively as shown. 86

We have, VA – VB = (VA – VE) + (VE – VF) + (VF – VB) Q 6 Q/3 Q/6 Q/3 5 Q or Ceq = = C. = + + = VA − VB 5 6C C C C 6. The situation is shown O in figure. Suppose the T3 0 X cm T point of suspension is 90° T/2 F O and let T be angle 30 cm 90° – T/2 B between the thread A and the vertical. Forces Xc mg on the particle B are (i) weight mg downward (ii) tension T along the thread and (iii) electric force of repulsion F along AB. For equilibrium, these forces should add to zero. Let XcBX be the line perpendicular to OB. Take the components of the forces along BX. θ θ θ F cos = mg cos(90° − θ) = 2mg sin cos 2 2 2 F θ ...(i) or, sin = 2 2mg 1 Now, F = (9 × 109 N m2 C–2) × (5.0 × 10–7 C)2 × AB2 θ and AB = 2(OA) sin . 2 6.25 × 10−3 9 × 109 × 25 × 10−14 Thus, F = = N ...(ii) θ θ 4 × (30 × 10−2)2 × sin2 sin2 2 2 From (i) and (ii),

PHYSICS FOR YOU | DECEMBER ‘17

6.25 × 10−3 θ sin3 = = 0.0032 2 2 × 100 × 10−3 kg × 9 ⋅ 8 θ or sin = 0 ⋅15 ⇒ θ = 2 sin −1(0.15) 2 7. Given, the heavier block (512 g) is observed to fall from rest 76.5 cm in 5.11 s; Using kinematic 1 equation, y = voyt + a yt 2 , we get 2 2 0 765 ( . m) 2y ay = 2 = = 0.059 m s −2 2 t (5.11 s)

For the heavier block, m1 = 0.512 kg, m1g – T1 = m1ay ...(i) R ay For the lighter block, m2 = 0.463 kg, ay T2 T1 T2 – m2g = m2ay ...(ii) m2 Since T1 > T2; the net force on the m2 pulley creates a torque which results m 2g in the pulley rotating towards the m 1g heavier mass. ? (T1 – T2) R = IDz = Iay /R ...(iii) where R = 0.049 m is the radius of the pulley Adding eqn. (i) and (ii) m1g – T1 + T2 – m2g = m1ay + m2ay

⎛ 1⎞ = (20/3) V ⎝ 30 ⎟⎠ = 6.67 V

∴ Voltmeter reading = I 2Q = (200) ⎜

9. (a) From Faraday’s law, the magnitude of electric R dB ⇒ E = B0Rt field is given by E = 2 dt (' B = B0 t2) This field is tangential to the ring. Torque on the ring, WF = QER = B0QR2t The ring starts rotating when torque due to electric force is greater than the torque due to maximum friction (fmax = Pmg). WF ≥ Wfric (max)

T1 – T2 = (g – ay)m1 – (g + ay)m2 Substitute (T1 – T2) in eqn. (iii) ((g – ay)m1 – (g + ay)m2)R = Iay/R ⎡⎛ g ⎞ ⎤ ⎞ ⎛ g ⎢ ⎜ − 1⎟ m1 − ⎜ + 1⎟ m2 ⎥ R2 = I ⎢⎣ ⎝ a y ⎠ ⎝ a y ⎠ ⎥⎦ Putting the values of all the quantities, ⎡ ⎛ 9.81 ⎞ ⎤ ⎛ 9.81 ⎞ + 1⎟ (0.463)⎥ (00.049)2 I = ⎢⎜ − 1⎟ (0.512) − ⎜ ⎠ ⎠ ⎝ ⎝ . . 0 059 0 059 ⎣ ⎦ I = 7.17 × (0.049)2 = 0.017 kg m2. 8. The given circuit is a balanced Wheatstone bridge. A voltmeter of 400 : is placed across a resistance of 400 :. The two are in parallel. ? combined resistance =

400 × 400 400 + 400

= 200 :

The simplified circuits is shown in the figure. Wheatstone bridge is balanced as P/Q = R/S. ? No current flows through BD. Hence, resistance of BD becomes ineffective in the balanced bridge. The voltmeter reads the potential difference across resistance Q. As P and R are equal, I1 = I2 ∴

I1 = I 2 =

10 100 + 200

=

1 30

A

? Potential difference across Q is= I2Q

For just starting the rotation, B0QRt = Pmg B QRt Hence, μ = 0 mg 2B QR Given, t = 2 s μ = 0

mg

(b) Net torque W = WF – Wf max = B0QR2t – PmgR 2 B QR mgR =W = B0QR2(t – 2) τ = B0QR2t − 0 mg ⎛ dω ⎞ = B0QR2(t − 2) or ID = B0QR2(t – 2) or mR2 ⎜ ⎝ dt ⎟⎠ ω 2B Q BQ 4 ? ∫ dω = 0 ∫ (t − 2)dt ⇒ ω = 0 0 2 m m 10. The volume of mercury at 25°C is V0 = 100 cm3. The coefficient of volume expansion of mercury JL = 1.8 × 10–4 °C–1. The coefficient of volume expansion of glass JS = 3 × 1.8 × 10–6 °C–1 = 5.4 × 10 –6 °C–1. Thus, the volume of mercury at 75°C, VLT = V0 (1 + JL 'T) and the volume of the vessel at 75°C, VST = V0 (1 + JS 'T). The volume of mercury overflown, V = VLT – VST= V0 (JL – JS)'T = (100 cm3) (1.8 × 10 –4 – 5.4 × 10–6) °C–1 × (50 °C) = 0.87 cm3 PHYSICS FOR YOU | DECEMBER ‘17

87

SOLUTION SET-52

1. (a,c) : The current velocity, v v y or v = ky, y where k is a constant. d (0, d) At y = , v = v0 ⎛ d⎞ v 2 ⎜⎝ 0, ⎟⎠ 2 P⋅(x, y) 2v0 k= x (0, 0) d 2v Current velocity at a point P, v = 0 y d dv 2v0 u ? Acceleration of river flow, a = = dt d till middle of river Time taken by boat to reach the middle of river, d T= 2u Horizontal drift at the middle of the river aT 2 v0d = s= 2 4u v d By symmetry, total drift = 2s = 0 2u v u Also x = 0 t 2 , y = ut d ud ? y2 = x v0 The path of the boat will be parabolic as seen by observer on river bank. 2. (a,b,c) : Acceleration will be zero when kxc kxc = mg 1 ? 300 × xc = 1 × 10 x ′ = m m 30 For maximum compression x, mg 1 mg (0.4 + x ) − kx 2 = 0 2 (By using conservation of energy) ? 10(0.4 + x) – 150x2 = 0 or x = 0.2 m For maximum acceleration, kx – mg = ma 300(0.2) − 10 × 1 a= = 50 m s −2 1 3. (a,d) : Suppose blocks A and B move together. Friction between blocks and surface, f c = (9 × 10 × 0.2) N = 18 N Applying Newton's second law on C, A + B and D respectively 60 – T = 6a, T – 18 – Tc = 9a, Tc – 10 = 1a 88

PHYSICS FOR YOU | DECEMBER ‘17

a On solving these question, T A f T′ –2 a = 2 m s , T = 48 N and B f′ Tc = 12 N T′ T a D To check slipping between a C A and B, we have to find 1 kg 6 kg friction force (f) in this case. If it is less than limiting static friction, then there will be no slipping between A and B. For block A, T – f = 6 × 2 As T = 48 N, f = 36 N and fsmax = 42 N. Hence blocks A and B will move together with acceleration 2 m s–2. 4. (a,b,c,d) : Taking origin at O coordinates of point P ⎛l ⎞ x = ⎜ − r ⎟ cos θ P(x, y) r ⎝2 ⎠ θ A y = r sin T O l cos θ 2 y2 x2 1 + = 2 r2 ⎛l ⎞ r − ⎜⎝ 2 ⎟⎠ l When r = equation reduces to 4 l2 x 2 + y 2 = , which is an equation of circle. 16 6Fx = 0 ? COM moves vertically down

a=

d2 x 2

+

d2 y 2

at θ = 0; a =

3g 4

dt dt 3g dθ ω = ; ω max = dt l 5. (a,c) : Let us consider that bead reaches point A with speed v. According to question, bead have enough speed to just reach B, hence the range of trajectory from A to B is 2R sin D. 2v 2 sin α cos α Rg ⇒ v2 = ? 2R sin D = cos α g Using conservation of energy at point O and A, 1 2 1 2 mv = mv + mg (R + R cos α) 2 0 2 Rg 1 2 1 mv0 = m + mgR(1 + cos α) 2 2 cos α ?

1 ⎞ ⎛ v0 = gR ⎜ 2 + 2 cos α + ⎟ ⎝ cos α ⎠

For the motion along the wire, the normal reaction on the bead is radially inward i.e., towards the centre. But for the motion along A to B, there is a possibility of change in direction of normal reaction at A and B.

6. (a,b,d) : Initial momentum of the ball, pi = 10 kg m s −1 i −1 Final momentum of the ball, p f = −10 kg m s i Impulse imparted to the wall |'p| = 20 kg m s–1 Work done is zero as point of contact does not move due to impulse of the force. As the ball rebounds will same speed hence the kinetic energy of the ball remains unchanged. 7. (4) : Let us consider that the A μS force of gravity of whole body is acting at G, such that C S Lθ O θ G θ R OG = 4a/3S, where a is the B radius and O is the centre of E θ D the semicircle. Resolving the μR W forces horizontally and vertically we get S = PR ...(i) and R + PS = W ..(ii) Solving eqns. (i) and (ii) we get S = PW/(1 + P2) ...(iii) Also taking moments about O, we get μS.OC + μR.OD = W .OG sin θ 4a 4a ⎤ ⎡ or μS.a + μR.a = W sin θ ⎢' OG = ⎥ 3π 3π ⎦ ⎣ μ 2W μW 4W sin θ [using (i) and (ii)] + = or 2 2 3π (1 + μ ) (1 + μ ) or sin θ =

2 ⎡ 3π ⎛ μ + μ 2 ⎞ ⎤ 3π ⎛ μ + μ ⎞ or θ = sin −1 ⎢ ⎜ ⎟⎥ ⎜ ⎟ 4 ⎝ 1 + μ2 ⎠ ⎢⎣ 4 ⎝ 1 + μ 2 ⎠ ⎥⎦

? n=4 ...(i) 8. (7) : When C is fixed, F − 2T = 2ma1 3T = 4ma2 ...(ii) 2Ta1 = 3Ta2 ...(iii) Solving eqns. (i), (ii) and (iii), gives a2 = 3F / 17m When all the blocks are free to move, assume ac1, ac2 and ac3 are the accelerations of blocks A, B and C respectively. F − 2T = 2ma1′ ...(iv) 3T = 4 ma2′ ...(v) T = ma3′ ...(vi) ...(vii) −2Ta1′ + 3Ta2′ + a3′T = 0 Solving eqns. (iv), (v), (vi) and (vii) a2 3F / 17m 21 3 × 7 3F = = = gives, a2′ = . ∴ a2′ 3F / 21m 17 17 21m 9. (4) : Let cross-sectional area of container be A. ? Al = V0

When the spring is stretched by x, the volumes of the left and right compartment are (l/2 – x) (l/2 + x)

⎛l ⎞ V1 = A ⎜ − x ⎟ and ⎝2 ⎠ a P1 P2 ⎛l ⎞. V2 = A ⎜ + x ⎟ ⎝2 ⎠ Volume of each compartment at equilibrium, V = A(l/2) Since process is adiabatic, P0V γ = P1V1γ = P2V2γ γ

−γ γ ⎛V⎞ ⎛ l ⎞ ⎛ 2x ⎞ 1 ⇒ P1 = P0 ⎜ ⎟ = P0 ⎜ P = − 0⎜ ⎝ l − 2 x ⎟⎠ ⎝ l ⎟⎠ ⎝ V1 ⎠

⎛ 2 γx ⎞ P1 = P0 ⎜ 1 + ⎝ l ⎟⎠

(' x << 1)

⎛ 2 γx ⎞ Similarly P2 = P0 ⎜ 1 − ⎝ l ⎟⎠ Force on piston, (P1 − P2 )A + kx = ma 4 P0 γxA 4 ? n=4 + kx = 4 ; x = 4 P γA l k+ 0 l 10. (4) : Let the angle of y 30 m s–1 projection of the particle be T. The path of the particle has P(40, 10) through the point where 10 m θ x x = 40 m, y = 10 m. Initial O 40 m – 1 v e l o c i t y, v 0 = 3 0 m s Using the equation of the path of the projectile, x2 g y = x tan θ − 2 (1 + tan2 θ) 2v0 For point P : 10 = 40 tan θ − (80 / 9)(1 + tan2 θ) ...(i) ⇒ 8 tan2 θ − 36 tan θ + 17 = 0 This is a quadratic equation in tanT with two positive roots. Therefore there are two values of T less than 90°, giving two possible angles of projection. Now we can calculate tan(D + E) which is equal tan α + tan β to , 1 − tan α tan β where tanD and tanE are the roots of eqn. (i). 36 9 Hence tan α + tan β = = (sum of roots) 8 2 17 and tan α tan β = (product of roots) 8 9/2 Therefore tan (α + β ) = = −4 1 − (17 / 8) ⇒ tan[−(α + β)] = − tan(α + β) = −(−4) = 4 PHYSICS FOR YOU | DECEMBER ‘17

VV 89

90

PHYSICS FOR YOU | DECEMBER â€˜17

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