### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

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)

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)

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)

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.