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In this page we have *Multiple Choice questions on heat and thermodynamics for JEE Main and Advanced* . Hope you like them and do not forget to like , social shar
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**Question 1 **The coefficient of linear expansion of an in homogenous rod changes linearly from α

_{1}to α

_{2}from one end to the other end of the rod. The effective coefficient of linear expansion of the rod is

a α

_{1}+α

_{2}
b 1/2(α

_{1}+α

_{2})

c. √α

_{1}α

_{2}
d. (α

_{1}-α

_{2})

Solution
Consider a small element of length dx from one end of the rod. Let L be the length of the rod. Now the increase in the coefficient of linear expansion by unit length of the rod is (α_{2}-α_{1})/L. Therefore the value of α of the element located at x is

α_{x}=α_{1} +x(α_{2}-α_{1})/L.

Therefore increase in length of the element is α_{x}dxΔT where ΔT is the rise in temperature. Therefore the increase in the length of the rod is

L=∫α_{x}dxΔT

integrating between the limits 0 and L = 1/2(α_{1}+α_{2})LΔT

**Question 2 **An Aluminum Rod of length L

_{0 }rest on a smooth horizontal base if the temperature is increase by ΔT °C. What will be the longitudinal strain developed

a. αΔT

b. Zero

c. -αΔT

d. None of the above

Solution
Zero Since no tensile stress is there, so strain will be zero

**Question 3 **The ratio of adiabatic bulk modulus and isothermal bulk modulus of a gas (

)

a)

b) 1

c)

d)

Solution
Bulk Modulus is defined as

$B=-\frac {V \Delta P){\Delta V}$

For adiabatic process

$PV^{\gamma }=Constant$

Differentiating partially we get

$\Delta P V^{\gamma } + P \gamma V^{\gamma -1} =0$

or $- \frac {V \Delta P}{\Delta V} =\gamma P$

For isothermal Process

PV=constant

or $ - \frac {V \Delta P}{\Delta V}=P$

So ratio is equal to $\gamma $

**Question 4 **An animal of 70 kg is running with a speed of 6 m/s. If all the KE of animal can be used in increasing water from 18 °C to 32 °C, how much water can be heated with this energy

Solution
KE of animal = 1/2 × 70 × (6)^{2}

= 12.60J

Q = 1260 J is the amount of heat being given to heat the water

since, Q = ms Δθ

m = Q/ sΔθ

= 12605 / 4186 × (32-18)

= 0.0215 kg = 21.5 gm

**Question 5** A copper cube of mass 300g slides down on a rough inclined plane of inclination 40 ° at a constant speed. Assume that any loss in mechanical energy goes into copper block as thermal energy find the increase in thermal energy of the block as it slides down through 50 cm. specific heat capacity of copper = 420J/kg-k.

Solution
work done in sliding 50 cm=Change in Potential energy=mgh

=mg(dsin40)

W = = .95 J.

If this mechanical work W produces same temperature change as heat Q then

W = J Q

where J = mechanical equivalent of heat and J = 1 if same unit of W and Q are used

Q= .95 J

since, Q= msΔθ

change in temperature Δθ of block

Δθ= Q/ ms

= .95 / .3 x 420

= 7.53 x 10^{-3 °C.}

**Question 6** It is known that curves A,B,C are Isobaric, Isothermal, Adiabatic process then when one is correct

a, A - Adiabatic , B - Isothermal,, C - Isobaric

b, A- Isothermal,, B - Adiabatic, C - Isobaric

c, A - Isobaric, B - Isothermal C - Adiabatic

d, None of these

Solution
. Isobaric Pressure is constant

So Curve C is Isobaric

Adiabatic slope is more steep the Isothermal

So Adiabatic - A

And Isothermal - B

so (a) is correct

**Question 7 A container has a mixture of 1 mole of oxygen and 2 moles of nitrogen at 330K.The ratio of average rotational kinetic energy per O**_{2} molecule to that per H_{2} molecule is
a) 2:1

b) 1:2

c) 1:1

d) None of these

Solution
Rotational kinetic energy depends only on degree of freedom associated with which is same for both diatomic molecule ie 1:1

So c is correct

**Question 8 **The sprinkling of water reduces the temperature of the closed room

a) The water has large latent heat of vaporization

b) Water is bad conductor of heat

c) Specific heat of water is high

d) the temperature of water is less than that of room

Solution
Sprinkled water vapourises by taking the heat from the room.The latent heat of vaporization is very high so it takes large heat to vaporize and room becomes cool

So a is correct

**Question 9** Two boxes A and B containing different ideal gases are placed on table

Box A contain one mole of gas m where (C

_{v}=5R/2) at Temperature T

_{0}
Box B contains one mole of gas n where (C

_{v}=3R/2) at temperature (7/3) T

_{0}
The boxes are then put into thermal contact with each other and heat flows between until the gases reach a common final temperature T

_{f}
Which of the following relation is correct?

a) 2T

_{f}-3T

_{0}=0

b) 2T

_{f}-7T

_{0}=0

c) 2T

_{f}-5T

_{0}=0

d) T

_{f}-3T

_{0}=0

Solution
Change in the internal energy of the system is zero i.e increase in internal of one gas is equal to decrease in internal energy of other

So

$\Delta U_A= 1x \frac {5R}{2} (T_f - T_0)$

$\Delta U_B= 1x \frac {3R}{2} (T_f - \frac {7}{3} T_0)$

$ \Delta U_A + \Delta U_B =0$

Solving we get

2T_{f}-3T_{0}=0

So a is correct

**Question 10 **During an adiabatic process the square of the pressure of a gas is proportional to the fifth power of its absolute temperature. The ratio of specific heat C

_{p} / C

_{v} for that gas is

a. 3/5

b. 4/3

c. 5/3

d. 3/2

Solution
. P^{2}= k T^{5}

P^{2} =k(PV/nR)^{5}

P^{2} = k^{'}P^{5}V^{5} where k^{'}=k/(nR)^{5} which is a constant

P^{3}V^{5} = constant

PV^{5/3} = constant

so C_{p} / C_{v} = 5/3

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