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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2016 (Offline)

MCQ (Single Correct Answer)
A pendulum clock loses $$12$$ $$s$$ a day if the temperature is $${40^ \circ }C$$ and gains $$4$$ $$s$$ a day if the temperature is $${20^ \circ }C.$$ The temperature at which the clock will show correct time, and the co-efficient of linear expansion $$\left( \alpha \right)$$ of the metal of the pendulum shaft are respectively :
A
$${30^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 3}}/{}^ \circ C$$
B
$${55^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 2}}/{}^ \circ C$$
C
$${25^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 5}}/{}^ \circ C$$
D
$${60^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 4}}/{}^ \circ C$$

Explanation

Time lost/gained per day $$ = {1 \over 2} \propto \Delta \theta \times 86400$$ second
$$12 = {1 \over 2}\alpha \left( {40 - \theta } \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
$$4 = {1 \over 2}\alpha \left( {\theta - 20} \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
On dividing we get, $$\,\,\,3 = {{40 - \theta } \over {\theta - 20}}$$
$$3\theta - 60 = 40 - \theta $$
$$4\theta = 100 \Rightarrow \theta = {25^ \circ }C$$
2

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
Consider a spherical shell of radius $$R$$ at temperature $$T$$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $$u = {U \over V}\, \propto \,{T^4}$$ and pressure $$p = {1 \over 3}\left( {{U \over V}} \right)$$ . If the shell now undergoes an adiabatic expansion the relation between $$T$$ and $$R$$ is:
A
$$T\, \propto {1 \over R}$$
B
$$T\, \propto {1 \over {{R^3}}}$$
C
$$T\, \propto \,{e^{ - R}}$$
D
$$T\, \propto \,{e^{ - 3R}}$$

Explanation

As, $$P = {1 \over 3}\left( {{U \over V}} \right)$$
But $$\,\,\,\,$$ $${U \over V} = KT{}^4$$
So, $$\,\,\,\,\,P = {1 \over 3}K{T^4}$$
or $$\,\,\,\,{{uRT} \over V} = {1 \over 3}K{T^4}\,\,\,\,$$
$$\left[ \, \right.$$ As $$PV = uRT$$ $$\left. \, \right]$$
$${4 \over 3}\pi {R^3}{T^3} = $$ $$constant$$
$$\therefore$$ $$\,\,\,\,$$ $$T \propto {1 \over R}$$
3

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $${V^q},$$ where $$V$$ is the volume of the gas. The value of $$q$$ is: $$\left( {\gamma = {{{C_p}} \over {{C_v}}}} \right)$$
A
$${{\gamma + 1} \over 2}$$
B
$${{\gamma - 1} \over 2}$$
C
$${{3\gamma + 5} \over 6}$$
D
$${{3\gamma - 5} \over 6}$$

Explanation

$$\tau = {1 \over {\sqrt 2 \pi {d^2}\left( {{N \over V}} \right)\sqrt {{{3RT} \over M}} }}$$
$$\tau \propto {V \over {\sqrt T }}$$
As, $$\,\,\,\,T{V^{\gamma - 1}} = K$$
So, $$\,\,\,\,\tau \propto {V^{\gamma + 1/2}}$$
Therefore, $$q = {{\gamma + 1} \over 2}$$
4

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)
A solid body of constant heat capacity $$1$$ $$J/{}^ \circ C$$ is being heated by keeping it in contact with reservoirs in two ways:
$$(i)$$ Sequentially keeping in contact with $$2$$ reservoirs such that each reservoir
$$\,\,\,\,\,\,\,\,$$supplies same amount of heat.
$$(ii)$$ Sequentially keeping in contact with $$8$$ reservoirs such that each reservoir
$$\,\,\,\,\,\,\,\,\,\,$$supplies same amount of heat.
In both the cases body is brought from initial temperature $${100^ \circ }C$$ to final temperature $${200^ \circ }C$$. Entropy change of the body in the two cases respectively is :
A
$$ln2, 2ln2$$
B
$$2ln2, 8ln2$$
C
$$ln2, 4ln2$$
D
$$ln2, ln2$$

Explanation

The entropy change of the body in the two cases is same as entropy is a state function.

Questions Asked from Heat and Thermodynamics

On those following papers in MCQ (Single Correct Answer)
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