H E A T N T H E R M O

Q.1 A steel rod is 4.000 cm in diameter at 30 ºC. A brass ring has an interior diameter of 3.992

cm at 30 ºC. In order that the ring just slides onto the steel rod, the common temperature

of the two should be nearly (steel

= 11 × 10–6/ºC and brass

= 19 × 10–6/ºC)

(A) 200 ºC (B) 250 ºC (C*) 280 ºC (D) 400 ºC

[Sol: Final diameters should be same

4(1+11×10–6 T) = 3.992 (1+(19×10–6)T)

0.008 = (75.848 – 44) × 10–6T

T = 251

final temperature should be 2800C (C) ]

Q.2 A body of mass 25 kg is dragged on a rough horizontal floor for one hour with a speed

of 2km/h. The coefficient of friction between the body and the surface in contact is 0.5

and half the heat produced is absorbed by the body. If specific heat of body is 0.1 cal g–

1 (0C)–1, then the rise in temperature of body is

(A) 50 K (B) 23.8 K (C) 100 K (D*) 11.9 K

[Sol: |friction

| = Mg×

= (0.5) (25) (10) (2000)

= 2.5 × 105 J

MST = 2

105.2 5

(25000) × (0.1 × 4.2) T = 2

105.2 5

Q.3 A composite rod made up of two rods AB and BC are joined at B. The rods are of equal length

at room temperature and have equal masses. The coefficient of linear expansion of AB is

more than that of BC. The composite rod is suspended horizontal by means of a thread at B.

When the rod is heated:

(A) It remains horizontal (B*) It tilts down on the side of AB

(C) It tilts down on the side of BC (D) Its centre of mass does not move

[Sol:

AB 2d2 = 2d (1 +

2T) d

2 = d(1+

2T)

BC 2d1 = 2d (1 +

1T) d

1 = d(1+

1T)

As 2 >

1 d

2 > d

1

The system tilts down onside AB

Ans: (B)

H E A T N T H E R M O

Q.4 At 0°C a body emits:

(A) no radiation

(B) electromagnetic radiation of single wavelength

(C*) electromagnetic radiation of all wave lengths that are emitted by it at room temperature.

(D) electromagnetic radiation of fewer wave lengths than are emitted by it at room temperature.

[Sol: The body emits radiation of all wavelengths at all

temperature only difference being Xm increases

as T decreases. Ans: (C) ]

Q.5 Figure shows V-T graph of a cyclic process. Which of the following PV graphrepresents the same process?

(A) (B) (C*) (D)

[Sol: CA V = CT Isobaric process P1 = P0AB V = V0 Isochoric process

T

P = constt.

P1 = aP0BC T = aT0 Isothermal process PV = constt.

(C)

Q.6 A solid cylindrical copper rod 0.200 m long has one end maintained at a temperature of T

Kelvin. The other end is blackened and exposed to thermal radiation from surrounding walls

maintained at 400 K. The sides of the rod are insulated, so no energy is lost or gained

except at the ends of the rod. When steady state is reached, the temperature of the blackened

end is 200 K. What is the value of T ? Take k = 170 W/mK, Stefan's constant = 3

17×10–

8 Wm–2K–4.

[Sol:)gminco(

radiationQ = )leaving(

conductionQ (at steady state at the other end at temperature T1 = 200K)

Room temperature, T0 = 400 K

A (T04 – T1

4) = KA l

TT1

H E A T N T H E R M O

T = T1 – K

l (T0

4 – TT14)

T = 200 – 1703

10172.0 8

= (4004 – 2004)

T = 200 – 5

8 = 198.4 K ]

Q.7 Two blocks A and B of the same mass are connected to a light spring and placed on a smoothhorizontal surface. B is given velocity v0 (as shown in the figure) when the spring is in naturallength. In the subsequent motion.

(A*) the maximum velocity of B will be v0

(B*) as seen from ground, A can move towards right only(C) the spring will have maximum extension when A and B both stop(D *) the spring w illbe at natural length again when B is at rest

[Sol. In CM frame the blocks will perform SHM as

VCM is constant hence both will can have maximum velocity v0/2 towards right in CMframe.

(vB)max = v0 in ground frameIn CM frame (vA)min = –v0/2 (vA)min = 0 in ground framei.e. A move only in right directionIn CM frame vB = v0/2 left in the case when spring is at natural length vB = 0 in ground frame at that instant. ]

Q.8 A particle executes SHM along a straight line with mean position at x = 0, period 20 sec andamplitude 5 cm. The time taken by the particle to go from x = 4 cm to x = – 3 cm can be

(A*) 10 sec (B*) 15 sec (C*) 5 sec (D*) 25 sec[Sol. B to C 5 sec

A to D 15 secB to B to C 25 sec

Q.9 One mole of an ideal gas at pressure P0 and temperature T0 volume V0 is expanded isothermallyto twice its volume and then compressed at constant pressure to (V0/2) and the gas is broughtto original state by a process in which P V (Pressure is directly proportional to volume).The correct representation of process is

(A) (B) ( C * ) ( D )

H E A T N T H E R M O

[Sol. For isothermal process Vf = 2V0 Pf = P0/2

For isobaric process Vf = V0/2, Tf = 00

0 T•V22

V

=

4

T0

For P V processP–V must be straight line

T V2 V–T must be parabolicP2 T P–T must be parabolic

Q.10 A particle executes SHM about a point other than x = 0 as shown in the graph. Find itsamplitude, equilibrium position and angular frequency. [ 1+ 1 + 3 ]

[Sol. From the graph

Amplitude = 2

)2(6 = 4 m

t = 1 =

3

2 =

2

T·

3

2 =

3

T

T = 3 secEquilibrium position; x = 2

Angular frequency = 3

2

Q.11 What is the temperature of the steel-copper junction in the steady state of the system shown inthe figure. Length of the steel rod = 25 cm, length of the copper rod = 50 cm, temperature ofthe furnace = 300 °C, temperature of the other end = 0°C. The area of cross section of thesteel rod is twice that of the copper rod. (Thermal conductivity of steel = 50 J s–1 m–1 K–1

and of copper = 400 J s–1 m–1 K–1)

H E A T N T H E R M O

[Ans. 100°C]

[Sol.1

111

L

)TT(Ak =

2

222

L

)TT(Ak

300 – T =

2

1

L

L

1

2

k

k

1

2

A

A(T – 0)

300 – T = 2TT = 100°C ]

Q.12 A source of frequency 700 Hz is placed between a man and a wall at the same height of person'sear. Find the velocity v (assume v << c) of the source with which it should approach the wallsuch that person will detect 3 beats per second. [Take velocity of sound = 350 m/s]

[Ans. (3/4) m/s]

[Sol.

vc

c

vc

c700 = 3

700 × 2vc = 3c2 – 3v2

assuming v << c1400 vc = 3c2

v = 1400

c3 =

1400

3503 =

4

3m/s ]

Q.13 One end of a long string of linear mass density 10–2 kg m–1 is connected to an electricallydriven tuning fork of frequency 150 Hz. The other end passes over a pulley and is tied to apan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so thatreflected waves from this end have negligible amplitude. At t = 0, the left end (fork end) ofthe string is at x = 0 has a transverse displacement of 2.5 cm and is moving along positive y-direction. The amplitude of the wave is 5 cm. Write down the transverse displacement y (incm) as function of x (in m) and t (in sec) that describes the wave on the string.

[Ans. y = =

6

xt300sin5 ]

[Sol. = /6

v = /T = 01.0

900 = 300 m/s

frequency = 150 Hz = (1/2) mThen equation will be

y =

x

T

t2sinA =

62

xt1502sin5 =

6

xt300sin5

H E A T N T H E R M O

Q.14 A conical pendulum is composed of a mass 100 gm and a massless string of length 50 cm. Itswings with constant angular velocity 5 rad/s as shown in the figure.

(a) What is the angle, , between the string and the vertical?(b) What is the angular momentum vector with respect to the pivot at the instant shown in the

diagram?(c) What is the torque vector with respect to the pivot at the instant shown in the diagram?

[Ans. (a) 37°, (b) 1.5 × 10–3 )j3i4( N-m/s, (c) j3.0 N-m ]

[Sol.(a) Tcos – mg = 0

Tsin = rm 20

tan = g

r20 =

g

sin20 l

cos = 20

g

l = cos–1

20

g

l = 37°

(b) )kcos–isin(r ll

& v = jsin 0l

vmrL

= )icossinlksinl(m 222o = )ksini(cossinlm o

2

= 1.5 × 10–3 )j3i4( N-m/s

(c) )k(mgr = )kmg()kcosisin( ll = jsinmg l = j3.0 N-m]

Q.15 One mole of an ideal monoatomic gas is enclosed in a chamber at 300 K. The gas undergoes aprocess in which pressure is proportional to the volume. At the end of the process, thevolume of the gas is doubled. The change in the internal energy of the gas is [R is gasconstant]

(A) 450 R (B) 700 R (C*) 1350 R (D) data insufficient[Sol. P V (given)

PV = nRT gives

V2 T 1

21

T

V =

2

22

T

V

T2 = 21

122

V

TV = 2

12

V

T)v2( = 4T1

T = T2 – T1 = 3T1 = 3 × 300 = 900 K

and U = nCvT = 1 × 2

3R × 900 = 1350 R ]

H E A T N T H E R M O

Q.16 A heat engine uses an ideal gas ( = 1.40) that undergoes the reversible cycle shown in figure.Obtain the thermodynamic percentage efficiency of the engine.

[Ans. 24%]

[Sol. v = 1.4 CV = 2

R5

4.0

R

PROCESS U W Q

1-2 0T92

R5n

0 0nRT2

45

2-3 0T902

R5n

90PoVo 315 PoVo

3-4 0T902

R5n

0 – 405 PoVo

4-1 0T92

R5n

– 9 Po Vo ooVP

2

63

Total 0 81 PoVouse PoVo = nRTo

efficiency = 24.05.337

81

VP5.22VP315

VP81

oooo

oo

%24 ]

H E A T N T H E R M O

Q.17 Two conducting movable smooth pistons are kept inside a nonconducting, adiabatic container with initial positions as shown. Gas ispresent in the three parts A, B & C having initial pressures as shown.Now the pistons are released and are allowed to attain equilibriumposition slowly. Then the final equilibrium position length of part Awill be

(A) L/8 (B*) L/4 (C) L/6 (D) L/5

[Sol. n = RT

PV

Final pressure & temperature is same hence volume will be in ratio of number of moles. ]

Q.18 Which of the following statements is/are true in case of specific heat of an ideal gas?(A*) specific heat of an ideal gas is zero when it undergoes an adiabatic process(B*) specific heat of an ideal gas is infinite when it undergoes an isothermal process(C*) the difference between specific heats at constant pressure and constant volume is the

same for all ideal gases(D) the ratio of specific heats at constant pressure and constant volume is the same for all ideal

gases

Q.19 Monoatomic, diatomic and triatomic gases whose initial volume and pressure are same, eachis compressed till their pressure becomes twice the initial pressure then.

(A*) If the compression is isothermal then their final volumes will be same(B*) If the compression is adiabatic, then their final volume will be different(C) If the compression is adiabatic, then triatomic gas will have maximum final volume(D*) If the compression is adiabatic then monoatomic gas will have maximum final volume

[Sol. m > d > tIsothermal process, PV = constant, which is independent of . So vfinal is same (A)Adiabatic process, PV = constant

2111 V)P2(VP

/11

22

VV If is more than V2 will be more. So, B,D.

Q.20 Which of the following statements is/are correct?(A*) A real gas approaches perfect gas behaviour at high temperature and low pressure(B) For ideal gas if gas pressure increases then its temperature must also increase.(C*) An ideal gas would never condense into the liquid state(D*) The average translational kinetic energy per molecule at any given temperature is

independent of the type of ideal gas

Q.21 A certain amount of perfect gas undergoes changes in pressure andvolume

(Process PVN = C) N and C are constant as shown in figure. Duringthis change from A to B.

(A) no heat is absorbed by the gas from outside(B*) positive work is done by the gas(C*) the temperature of the gas is constant(D*) the internal energy of the gas remains constant

H E A T N T H E R M O

Q.22 Entries in column I consists of diagrams of thermal conductors. The type of conductor &direction of heat flows are listed below. Entries in column II consists of the magnitude of rateof heat flow belonging to any of the entries in column I. If temperature difference in all thecases is (T1 – T2), then match column

Column I Column II

(A) (P) )TT(Rk6 210

(B) (Q) )TT(2n3

Rk21

0

l

(C) (R) k0R(T1 – T2)

(D) (S) )TT(2n

Rk421

0 l

[Ans. (A) R; (B) P ; (C) S; (D) Q ]

H E A T N T H E R M O

A calorimeter of mass m contains an equal mass of water in it. The temperature of the water andcalorimeter is t2. A block of ice of mass m and temperature t3 < 0° C is gently dropped intothe calorimeter. Let C1, C2 and C3 be the specific heats of calorimeter, water and icerespectively and L be the latent heat of fusion of ice.

Q.23 The whole mixture in the calorimeter becomes ice if(A) C1t2 + C2t2 + L + C3t3 > 0 (B*) C1t2 + C2t2 + L + C3t3 < 0(C) C1t2 + C2t2 – L + C3t3 > 0 (D) C1t2 + C2t2 – L – C3t3 < 0

[Sol. For ice tf < 0°C, say tf = – t0Heat lost = Heat gain

)tt(mC)t0(mCmL)0t(mC)0t(mC 303032221

03332221 tC2tCLtCtC

or 0tCLtCtC 332221 ]

Q.24 The whole mixture in the calorimeter becomes water if(A) (C1 + C2)t2 – C3t3 + L > 0 (B) (C1 + C2)t2 + C3t3 + L > 0(C) (C1 + C2)t2 – C3t3 – L > 0 (D*) (C1 + C2)t2 + C3t3 – L > 0

[Sol. For water tf > 0°C say tf = t0Heat lost = Heat gain

)0t(mCmL)t0(mC)tt(mC)tt(mC 0233022021

02133221 t)C2C(LtCt)CC(

or 0LtCt)CC( 33221 ]

Q.25 Water equivalent of calorimeter is:

(A) mC1 (B*) 2

1

C

mC(C)

1

2

C

mC(D) none

[Sol. Let water equivalent be m0

m0C2 = mC1 2

10 C

mCm

Q.26 O ne m ole of an ideal m onoatom ic gas is taken from state A to state B through the process P =

2/1T2

3. It is found that its temperature increases by 100 K in this process. Now it is taken

from state B to C through a process for which internal energy is related to volume as U =

2/1V2

1. Find the total work performed by the gas (in Joule), if it is given that volume at B

is 100 m3 and at C it is 1600 m3.[Use R = 8.3 J/mol-K]

[Ans. 435][Sol. Process A B

H E A T N T H E R M O

WAB = dvP = dvT2

3 2/1 = dTRT

3

1T

2

3 2/12/1

On solving, WAB = 50 R = 50 × 8.3 = 415 JProcess B C

U = 2/1V

2

1

RT2

3 =

2/1V2

1 3PV1/2 = 1

P = V3

1

Now WBC = dvP = 1600

100

dvV3

1 = V

3

2 = ]1040[

3

2 = 30

3

2 = 20 J

Total W = 415 + 20 = 435 ]

Q.27 n moles of an ideal gas undergoes a process A to B as shown.Maximum temperature of the gas during the process is

(A) nR

VP3 00 (B*) nR

VP4 00

(C) nR

VP6 00 (D) nR

VP9 00

[Sol: From graph P = oo

o P4VV

P

PV = nRT => T = nR

PV

T is maximum when PV is maximum.

=> 0P4V

VP2VP4V

V

P

dV

d)PV(

dV

do

o

oo

2

o

o

=> V = 2Vo

There for Tmax = nR

PV (at V = 2Vo) =

nR

VP4 oo (B) ]

Q.28 The order of magnitude of the number of nitrogen molecules in an air bubble of diameter 2 mmunder ordinary conditions (pressure = 1 atm; temperature = 27°C) is:

(A) 105 (B) 109 (C) 1013 (D*) 1017

[Sol: PV = (No/N)RT No= No of molecules.Solving No = 1017]

H E A T N T H E R M O

Q.29 A rod of length l with thermally insulated lateral surface andcross-section area A consists of a material whose thermal

conductivity varies with temperature as K = bTa

K0

where

K0, a and b are constants. T1 and T2 (< T1) are the temperatureof two ends of rod. The rate of flow of heat across the rod is

(A)

2

10

bTa

bTa

b

AK

l (B)

2

10

bTa

bTa

b

AK(C*)

2

10

bTa

bTaln

b

AK

l (D)

2

10

aTb

aTbln

b

AK

l

[Sol: Q = dx

dT

bTa

AK

dx

dTKA o

0o

T

T

dxAK

Q

bTa

dT2

1

2

1o

bTa

bTan

b

AKQ

(C) ]

Q.30 We would like to increase the length of a 15 cm long copper rod of cross-section 4 mm2 by1mm. The energy absorbed by the rod if it is heated is E1. The energy absorbed by the rod ifit is stretched slowly is E2. Then find E1/E2.

[Various parameters of Copper are: Density = 9 × 103 kg/m3, Thermal coefficient of linearexpansion = 16 × 10–6 K–1, Young's modulus = 135 × 109 Pa, Specific heat = 400 J/kg-K]

[Ans. 500][Sol. Temp. is increased by then

l = l

= l

l

E1 = (Al)S = Al Sl

l

when stretched, Stress = Yl

l

E2 =

2

1

l

lY

l

l × AAl =

l

l

2

A)(Y 2

So, 2

1

E

E =

A)Y(

2SA2ll

lll

= Y)(

S2

l

l

= 500 ]

H E A T N T H E R M O

Q.31 A uniform rod of cross-section area A = 0.20 cm2 is bent to form a ring.The length of rod L = 60 cm. Three points on this ring P1, P2 & P3 aremaintained at temperature 800C, 800C & 1100C. Here P1P2 = P2P3 =P3P1. The thermal

conductivity of material of rod K = 385 W/m0C. Neglect the heat lossthrough lateral side. Find the rate of heat flow (in watt) in branch P3 to P1.

[Ans: 1.16 watt]

[Sol:R

TTQ 13 ;

KA3

L

KA3L

R

L

)TT(KA3Q 13

60

30x10x2.0x385x3Q

4

= 1.16 watt ]

Q.32 An ideal gas has molar heat capacity at constant pressure Cp = 5R/2. The gas is kept in acylindrical vessel fitted with a piston which is free to move. Side walls of the container andthe piston are non conducting. Mass of the frictionless piston is 9 kg. Initial volume of thegas is 0.0027 m3 and cross-section area of the piston is 0.09 m2. The initial temperature ofthe gas is 300 K. Atmospheric pressure P0 = 1.05 × 105 N/m2. An amount of 2.5 × 104 J ofheat energy is supplied to the gas by a heating coil, then

(A*) Initial pressure of the gas is 1.06 × 105 N/m2

(B) Final temperature of the gas is 1000 K(C*) Final pressure of the gas is 1.06 × 105 N/m2

(D) Work done by gas is 9.94 × 103 J

[Sol: P iA = P0A + mg Pi = P0 + A

mg

Pi = 1.05× 105 + 09.0

109

Pi = 1.06× 105 N/m2

Outside pressure = P0 + A

mg remians constant always

Hence the final pressure, Pf = P0 + A

mg = Pi always

Hence, the process is isobaric

Q = nCpT, n = i

ii

RT

VP =

300R

0027.01006.1 5 = R10

06.19

CP = 2

R5, Q = 2.5 × 104 J

2.5 × 104 = R10

05.19 ×

2

5R T

T = 06.19

105

K

H E A T N T H E R M O

W = PV = nRT

W = R10

05.19R

06.19

105

= 104J

HEnce answer is (A) & (C) ]

Q.33 The time period of a particle executing SHM is T. There is a point P, at a distance x fromthe mean position O. When the particle passes P towards OP (moving away from meanposition O), it has speed V then find the time in which it return to P again

(A) T (B*)

x2

VTtan

T 1(C) Tsin–1

x

VT(D)

x2

VTcot

2

T 1

Q.34 The spring is compressed by a distance a and released. The block again comes to rest when thespring is elongated by a distance b. During this process

(A) work done by the spring on the block = 2

1k(a2 + b2)

(B*) work done by the spring on the block = 2

1k(a2 – b2)

(C*) coefficient of friction = mg2

)ba(k

(D) coefficient of friction = mg2

)ba(k

[Sol: Energy equation

2

1 ka2 = mmg(a + b) +

2

1kb2 m =

mg2

bak

From A 0, wspring > 0

wspring = 2

1 ka2

From 0 B, wspring < 0

H E A T N T H E R M O

Q.35 One mole of an ideal monoatomic gas undergoes the cyclicprocess as

shown. Find out efficiency (in percent) of the cycle.[Take ln 2 = 0.7] [10]

[Sol. P0 = 0

0

V2

RT [from 1 – 2], CV =

2

R3, CP =

2

R5

1 – 2, )1(Q = nCPT = –2

5RTT0

2 – 3, )2(Q = W = –RTT0ln2

3 – 4, )3(Q = nCPT = 2

5RTT0

4 – 1, )4(Q = W = R(2T0)ln2 = 2RTT0 ln2

Qinput = Q(3) + Q(4) = RT0

2ln2

2

5

Qtotal = RT0ln2

= %100Q

Q

Input

Total

= 2ln2

2ln

25 ×100% =

4.15.2

7.0

× 100%

= 39

700% = 17.95% ]

Q.36 One mole of an ideal gas requires 207 J heat to raise the temperature by 10 K when heated atconstant pressure. If the same gas is heated at constant volume to raise the temperature by 5K, then the heat required may be

(A*) 62.1 J (B) 124 J (C) 12.4 J (D) 6.2 J[Sol: CV = Cp – R

required heat = CVT = (Cp – R) T

As CP = 10

207 J/mol-K

Required heat 62.1 JQ.37 Which of the following graphs correctly shows the change in internal energy of an ideal gas

with pressure for the isobaric, isochoric and isothermal processes.

(A*) (B) (C) (D)

[Sol: Isobaric P = const. & hence the line normal to pressure U axis.Similarly, isothermal will be normal to x axisIsochoric V = const.

P T & hence the straight line passing through origin. ]

H E A T N T H E R M O

Q.38 A black body radiates maximum energy around the wavelength 0. If the temperature of theblack body is changed so that it radiates maximum energy around the wavelength 20, theratio of final to initial power radiated by it will be

(A) 1/2 (B) 2 (C*) 1/16 (D) 16[Sol: From wein’s displacement law

0T1 = 20T2

T2 = 21T

as P T4

P2 = 1

4

2

1P

= 16

1P]

Q.39 The figure shows a system of spherical shell of inner radii r1 and outer radii r2and kept at temperatures T1 and T2, respectively. The radial rate of flowof heat in a substance is proportional to

(A*) 12

21

rr

rr

(B) (r2 – r1)

(C)

12

12

rr

rr (D)

1

2lnr

r

[Sol: effective thermal resistance = 2

1

24

1r

r r

dr

K = 21

12

4 rKr

rr

Heat current resistancethermal

1]

Q.40 n1 mole of a monatomic gas is mixed with n2 mole of diatomic gas such that mixture = 1.5. Then(A) n1 = 3n2 (B) n2 = 2n1 (C*) n1 = n2 (D) none of these

[Sol: mix = 1 + mixVC

R

as mix = 1.5, mixVC = 2R

Hence, 2R = 21

21 21

nn

CnCn VV

=

21

25

223

1

nn

RnRn

4n1 + 4n2 = 3n1 + 5n2 n1 = n2 ]

Q.41 A ideal gas is heated isobarically from temperature T1 to T2 as shown in theV–T diagram. Which of the following statement must be correct.(A) Heat must be rejected from gas.(B) work must be done on the gas by surrounding(C) First law of thermodynamics must hold for process a to b(D*) Gas must leak from vessel

H E A T N T H E R M O

Q.42 If there are no heat loss, the heat released by the condensation of x gram of steam at 1000Cinto water at 1000C can be used to convert y gram of ice at 00C into water at 1000C. Thenthe ratio of y : x is nearly [Given Lf = 80 cal/gm and Lv = 540 cal/gm]

(A) 1 : 1 (B) 2 : 1 (C*) 3 : 1 (D) 2.5 : 1[Sol: Heat released from condensation of x gm of steam at 1000C to water at 1000C

= xLVHeat required for y gm of ice at 00C to be converted to water at 1000C

= yLf + (100)ySwAs, xLV = yLf + (100)ySwwe get, y/x = 3 ]

Q.43 Three identical metal rods A, B and C are placed end to end and a temperature difference ismaintained between the free ends of A and C. If the thermal conductivity of B (KB) is twicethat of C (KC) and half that of A (KA), then the effective thermal conductivity of the systemwill be

(A) KA/7 (B*) 6KB/7 (C) 6KA/7 (D) none[Sol: Req = R1 + R2 + R3

= AK

L

AK

L

AK

L

CBA

=

CBA K

L

K

L

K

L

A

L

=

BKA

L

2

7[as KA = 2KB; KC = KB/2]

But, Req = AK

L

eq

3[as length of combined rod = 3L]

Hence, Keq = 7

6 BK]

Q.44 A wall has two layer A and B each made of different material, both the layers have the samethickness. The thermal conductivity of the material A is twice that of B. Under steady statethe temperature difference across the wall B is 360C. The temperature difference across thewall A is

(A) 60C (B) 120C (C*) 180C (D) 240C[Sol: temperature difference across wall A

= Br

TT 02 =

Ar

TT 12

= 12 TT = 02 TTr

r

B

A

= 362

1 = 180C ]

H E A T N T H E R M O

Q.45 Two isochoric cooling process AB and CD are shown in P-V diagram for the same gas. Inwhich case heat lost will be more.

(A) AB (B) same in both cases (C*) CD (D) can't say[Sol: Q = u + w

As 'w' (work done) is zero [Volume is constant] in both cases, Q would would depend on uu T (for a given gas)

and T VVPAs, P is same in both cases but V (volume) is more in CD, more heat would be lost in process

CD.]

Q.46 PV versus T graph of equal masses of H2, He and CO2 is shown in figure. Choose the correctalternative

(A*) 3 corresponds to H2, 2 to He and 1 to CO2(B) 1 corresponds to He, 2 to H2 and 3 to CO2(C) 1 corresponds to He, 3 to H2 and 2 to CO2(D) 1 corresponds to CO2, 2 to H2 and 3 to He

[Sol: The slope i.e. MT

PV 1 (M = Molar mass)

as22 HHeCO MMM

Slope of graph for H2 > slope of graph for He > slope of graph for CO2Q.47 A beaker contains 200 g of water. The water equivalent of beaker is 20 gm. The initial

temperature of water in the beaker is 20°C. If 440 g of hot water at 92°C is poured in, thenthe final temperature, will be nearest to: (neglect heat loss)

(A) 58°C (B*) 68°C (C) 73°C (D) none[Sol: Loss of heat by 440 gm of hot water as it comes down from 920C to some temperature,

say T = 440 [92 – T]Gain of heat by (water + beaker) = 220 (T – 200) equating loss of heat to gain of heat, we getT = 680C ]

Q.48 Two different isotherms representing the relationship between pressureP and volume V at a same temperature of the same ideal gas areshown for masses m1

and m2 of the gas respectively in the figure given, then:

(A) m1 > m2 (B) m1 = m2 (C*) m1< m2 (D) All of the above are possible

[Sol: PV = nRT2

2

1

1

n

P

n

P

But P1 < P2, hence n1 < n2 or m1 < m2(C) ]

H E A T N T H E R M O

Q.49 The molar heat capacity C for an ideal gas going through a given process is given by C = a/T,where ‘a’ is a constant. If = Cp / Cv, the work done by one mole of gas during heating fromT0 to T0 through the given process will be

(A) a

1ln (B*) a ln–

1

1RTT0

(C) a ln– (–1) RT0 (D) None of these

[Sol:ndT

dw

ndT

dV

ndT

dQ

C = Cv + ndT

dw

T

a – Cv =

ndT

dw ndTC

T

adw v

w = an lnT – 0

0

2T

TvnC (n = 1)

= aln – Cv (h – 1) T0

= aln – 1

1

r

R TT0 (B) ]

Q.50 Two moles of O2

5

7at temperature TT0 and 3 moles of CO2

3

4at temperature 2T0

are allowed to mix together in a closed adiabatic vessel. The resulting mixture finally comesin thermal equilibrium. Then,

(A*) final temperature of the mixture is 14

T23 0

(B) final temperature of the mixture is 19

T31 0

(C) adiabatic exponent () of the mixture formed is5

14

(D*) adiabatic exponent () of the mixture formed is 14

19

[Sol: dW and dQ = 0 hence dV = 0

11 vCn (T – TT0) = 22 vCn (2T0 – T) 1 = 7/5

2 11 R

(T – TT0) = 3 12 R

(2T0 – T) 2 = 4/3

Final temperature T = 14

23 0T

121

nn

= 11

1

n

+ 12

2

n

H E A T N T H E R M O

solving = 14

19(A,D) ]

Q.51 An ideal gas can be expanded from an initial state to a certain volume through two differentprocesses : (a) PV2 = K and (b) P = KV2, where K is a positive constant. Then, choose thecorrect option from the following.

(A) Final temperature in (a) will be greater than in (b).(B*) Final temperature in (b) will be greater than in (a).(C) Work done by the gas in both the processes would be equal.(D) Total heat given to the gas in (a) is greater than in (b).

[Sol: P = v

RT

v

RT · v2 = K RTv = K for (a)

v

RT = Kv2 RT = Kv3 for (b)

Hence B]

Q.52 The potential energy of a particle of mass 0.1 kg, moving along the x-axis, is given by U = 5x(x – 4)J, where x is in metres. It can be concluded that :

(A) the particle is acted upon by a constant force.(B*) the speed of the particle is maximum at x = 2 m.(C*) the particle executes simple harmonic motion(D*) the period of oscillation of the particle is /5 s.

[Sol: u = 5x2 – 20 x

F = dx

dv = – (10x – 20)

F = 20 – 10 x a = 200 – 100xSHM (since F x)

F = 0 at mean position i.e. x = 2 T = 2

2 is coefficient of x

i.e. 100 T = 10

2 =

5

s ]

Q.53 In an adiabatic expansion of air ( = 7/5) the volume increases by 5%. Then, the percentagechange in pressure is approximately

(A) + 7% (B) + 3% (C) – 5% (D*) – 7%

[Sol: 05.0V

dV

PVr = C

P · r Vr – 1 + Vr · dV

dP = 0

dV

dP =

V

rP

P

dP = –r

V

dV = 1.4 × 0.05 = – 0.07 = –7 % (D)