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 Multiple Choice QuestionsMultiple Choice Questions

1.

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn=constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively):

  • straight n space equals straight C subscript straight p over straight C subscript straight v
  • straight n equals space fraction numerator straight C minus space straight C subscript straight p over denominator straight C minus straight C subscript straight v end fraction
  • straight n equals fraction numerator straight C subscript straight p minus straight C over denominator straight C minus straight C subscript straight v end fraction
  • straight n space equals space fraction numerator straight C minus straight C subscript straight v over denominator straight C minus straight C subscript straight p end fraction

B.

straight n equals space fraction numerator straight C minus space straight C subscript straight p over denominator straight C minus straight C subscript straight v end fraction

For the polytropic process, specific heat for an ideal gas.
straight C space equals space fraction numerator straight R over denominator 1 minus straight n end fraction space plus space straight C subscript straight v
therefore comma space fraction numerator straight R over denominator 1 minus straight n end fraction space plus straight C subscript straight v space equals space straight C
rightwards double arrow space fraction numerator straight R over denominator 1 minus straight n end fraction space equals space straight C minus straight C subscript straight v
rightwards double arrow fraction numerator straight R over denominator straight C minus straight C subscript straight v end fraction equals space 1 minus straight n space left parenthesis where comma space straight R thin space equals space straight C subscript straight p minus straight C subscript straight v right parenthesis
rightwards double arrow space fraction numerator straight C subscript straight p minus straight C subscript straight v over denominator straight C minus straight C subscript straight v end fraction space equals space 1 minus straight n
straight n space equals space 1 minus space fraction numerator straight C subscript straight p minus straight C subscript straight v over denominator straight C minus straight C subscript straight v end fraction space rightwards double arrow space straight n space equals space fraction numerator straight C minus straight C subscript straight p over denominator straight C minus straight C subscript straight v end fraction
Thus comma space number space of space moles space straight n space is space given space by space
straight n space equals space fraction numerator straight C minus straight C subscript straight p over denominator straight C minus straight C subscript straight v end fraction

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2.

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, the density of liquid is ρ and L is its latent heat of vaporisation. 

  • ρL/T

  • square root of straight T divided by ρL end root
  • T/ρL

  • 2T/ρL


D.

2T/ρL

Decrease in surface energy = Heat required in vapourisation.... (i)
Decrease in surface energy = T x ΔA

 = 2T x 4πrdr  [ For two surface]
heat required in vapourisation = Latent heat
= ML = V x ρL
= (4πr2dr) ρL
Now from eq (i)
2T x 4πrdr = 4πr2ρL
r = 2T/ρL

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3.

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heat γ.It is moving with speed v and it suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by 

  • fraction numerator left parenthesis straight gamma minus 1 right parenthesis over denominator 2 γR end fraction Mv squared straight K
  • fraction numerator γMv squared over denominator 2 straight R end fraction minus straight K
  • fraction numerator left parenthesis straight gamma minus 1 right parenthesis over denominator 2 γR end fraction Mv squared straight K
  • fraction numerator left parenthesis straight gamma minus 1 right parenthesis over denominator 2 left parenthesis straight gamma plus 1 right parenthesis straight R end fraction Mv squared straight K

C.

fraction numerator left parenthesis straight gamma minus 1 right parenthesis over denominator 2 γR end fraction Mv squared straight K

As no heat is lost, therefore,
Loss of kinetic energy = Gain of internal energy of gas
i.e., 
1 half mv squared space equals space nC subscript straight v increment straight T
rightwards double arrow space 1 half mv squared space equals space straight m over straight M. fraction numerator straight R over denominator straight gamma minus 1 end fraction increment straight T
rightwards double arrow increment straight T space equals space fraction numerator Mv squared space left parenthesis straight gamma minus 1 right parenthesis over denominator 2 straight R end fraction straight K

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4.

‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:

  • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction
  • fraction numerator 3 straight p subscript straight o straight V subscript straight o over denominator 2 nR end fraction
  • 9 over 2 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction
  • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator nR end fraction

A.

fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction

As, T will be maximum temperature where product of pV is maximum


Equation of line AB, we have

straight y minus straight y subscript 1 space equals space fraction numerator straight y subscript 2 minus straight y subscript 1 over denominator straight x subscript 2 minus straight x subscript 1 end fraction space left parenthesis straight x minus straight x subscript 1 right parenthesis
rightwards double arrow space straight p minus straight p subscript straight o space equals fraction numerator 2 straight p subscript straight o minus straight p subscript straight o over denominator straight V subscript straight o minus 2 straight V subscript straight o end fraction space left parenthesis straight V minus 2 straight V subscript straight o right parenthesis
straight p minus straight p subscript straight o space equals space fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction left parenthesis straight V minus 2 straight V subscript straight o right parenthesis
straight p space equals space fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction straight V squared space plus 3 straight p subscript straight o straight V
nRT space equals space fraction numerator negative straight p subscript straight o over denominator straight v subscript straight o end fraction straight V squared space plus space 3 straight p subscript straight o straight V
straight T space equals space 1 over nR open parentheses fraction numerator negative straight p subscript straight o straight V squared over denominator straight v subscript straight o end fraction space plus 3 straight p subscript straight o straight V close parentheses
For space maximum space temperature
fraction numerator partial differential straight T over denominator partial differential straight V end fraction space equals space 0
fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction left parenthesis 2 straight V right parenthesis space equals space minus space 3 straight p subscript straight o
straight V space equals space 3 divided by 2 straight V subscript straight o
left parenthesis condition space for space maximum space temperature right parenthesis
Thus comma space the space maximum space temperature space of space the space gas space during space the space process space will space be
straight T subscript max space equals space 1 over nR open parentheses fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction space straight x 9 over 4 straight V subscript 0 superscript 2 space plus space 3 straight p subscript straight o space straight x space 3 over 2 straight V subscript straight o close parentheses
equals space 1 over nR open parentheses negative 9 over 4 straight p subscript straight o straight V subscript straight o space plus space 9 over 2 straight p subscript straight o straight V subscript straight o close parentheses space equals space 9 over 4 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction

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5.

If a piece of metal is heated to temperature θ and then allowed to cool in a room which is at temperature θ0 the graph between the temperature T of the metal and time t will be closest to


C.

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6.

Three rods of Copper, Brass and Steel are welded together to from a Y –shaped structure. Area of cross – section of each rod = 4 cm2. End of the copper rod is maintained at 100oC where as ends of brass and steel are kept at 0oC. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is:

  • 1.2 cal/s

  • 2.4 cal/s

  • 4.8 cal/s

  • 6.0 cal/s


C.

4.8 cal/s


dQ subscript 1 over dt space equals space dQ subscript 2 over dt space equals space dQ subscript 3 over dt
rightwards double arrow fraction numerator 0.92 left parenthesis 100 minus straight T right parenthesis over denominator 46 end fraction space equals space fraction numerator 0.26 space left parenthesis straight T minus 0 right parenthesis over denominator 13 end fraction plus fraction numerator 0.12 space left parenthesis straight T minus 0 right parenthesis over denominator 12 end fraction
rightwards double arrow straight T space equals space 40 to the power of straight o straight C
dQ subscript 1 over dt space equals space fraction numerator 0.92 space straight x space 4 space left parenthesis 100 minus 40 right parenthesis over denominator 40 end fraction space equals space 4.8 space cal divided by straight s
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7.

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm 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/gm°C

  • 1250°C

  • 825°C

  • 800°C

  • 885° C


D.

885° C

Heat given = Heat taken
(100) (0.1)(T – 75) = (100)(0.1)(45) + (170)(1)(45)
10(T – 75) = 450 + 7650 = 8100
T – 75 = 810
T = 885 °C

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8.

Two conductors have the same resistance a 0°C but their temperature coefficients o resistance are α1 and α2. The respective temperature coefficients of their series parallel combinations are nearly

  • fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction comma space fraction numerator straight alpha subscript 1 space plus space straight alpha subscript 2 over denominator 2 end fraction
  • fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction comma space straight alpha subscript 1 space plus straight alpha subscript 2
  • space straight alpha subscript 1 space plus straight alpha subscript 2 comma space fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction
  • straight alpha subscript 1 space plus straight alpha subscript 2 space comma space fraction numerator straight alpha subscript 1 straight alpha subscript 2 over denominator straight alpha subscript 1 space plus straight alpha subscript 2 end fraction

A.

fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction comma space fraction numerator straight alpha subscript 1 space plus space straight alpha subscript 2 over denominator 2 end fraction

Re = Ro +Ro
2R(1 + αsΔT)
= R( 1 +α1ΔT) + R(1+α2ΔT)
= αs = α12 / 2
Rp = R x R / R+R
straight R over 2 left parenthesis space 1 space plus space straight alpha subscript straight p increment straight T right parenthesis space equals space fraction numerator straight R space left parenthesis 1 space plus straight alpha subscript 1 increment straight T right parenthesis space straight x space straight R left parenthesis space 1 space plus space straight alpha subscript 2 increment straight T right parenthesis over denominator straight R space left parenthesis 1 space plus straight alpha subscript 1 increment straight T right parenthesis space plus space straight R left parenthesis space 1 space plus space straight alpha subscript 2 increment straight T right parenthesis end fraction
fraction numerator begin display style 1 space plus space straight alpha subscript straight p increment straight T end style over denominator 2 end fraction space equals space open parentheses 1 space plus space straight alpha subscript 1 increment straight T close parentheses left parenthesis 1 space plus space straight alpha subscript 2 increment straight T right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket space 1 space plus space left parenthesis straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis increment straight T right square bracket to the power of negative 1 end exponent
space equals space left square bracket 1 space plus space left parenthesis straight alpha subscript 1 space plus straight alpha subscript 2 right parenthesis increment straight T right square bracket space left square bracket 2 space plus space open parentheses straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis increment straight T close parentheses to the power of negative 1 end exponent
space equals space 1 half space left square bracket space 1 space plus space left parenthesis straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis space increment straight T right square bracket space open square brackets 1 minus fraction numerator left parenthesis straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis over denominator 2 end fraction increment straight T close square brackets
fraction numerator 1 space plus space straight alpha subscript straight p increment straight T over denominator 2 end fraction space equals space open square brackets 1 plus fraction numerator begin display style left parenthesis straight alpha subscript 1 space plus space straight alpha subscript 2 right parenthesis end style over denominator begin display style 2 end style end fraction increment straight T close square brackets
straight alpha subscript straight p space equals space fraction numerator straight alpha subscript 1 space plus straight alpha subscript 2 over denominator 2 end fraction

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9.

The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the numbers of molecules in the room before and after heating, then nf – ni will be

  • 2.5 x 1025

  • -2.5 x 1025

  • -1.61 x 1023

  • 1.38 x 1023


B.

-2.5 x 1025

Using ideal gas equation
PV = nRT
(N is number of moles)
P0V0 = niR × 290 ...... (1)
[Ti = 273 + 17 = 290 K]
After heating
P0V0 = NfR × 300 ...... (2)
[Tf = 273 + 27 = 300 K]
from equation (1) and (2)
straight N subscript straight f minus straight N subscript straight i space equals space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 300 end fraction space minus space fraction numerator straight P subscript straight o straight V subscript 0 over denominator straight R space straight x 290 end fraction
difference space in space number space of space moles space minus fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 300 end fraction space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 290 end fraction
difference space in space number space of space moles space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R end fraction space open square brackets fraction numerator 10 over denominator 290 space straight x space 300 end fraction close square brackets
Hence comma space straight n subscript straight f space minus straight n subscript straight i space is
equals space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R end fraction space open square brackets fraction numerator 10 over denominator 290 space straight x space 300 end fraction close square brackets straight x space 6.023 space straight x space 10 to the power of 23

putting P0 = 105 PA and V0 = 30 m3
Number of molecules nf – ni = – 2.5 × 1025

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10.

100g of water is heated from 30°C to 50°C ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184 J/Kg/K)

  • 8.4 kJ

  • 84 kJ

  • 2.1 kJ

  • 4.2 kJ


A.

8.4 kJ

ΔQ = M,S,ΔT
= 100 × 10-3 × 4.184 × 20 = 8.4 × 103
ΔQ = 84 kJ, ΔW = 0
ΔQ = ΔV + ΔW
ΔV = 8.4 kJ

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