### JEE Mains Previous Years Questions with Solutions

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1

### 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, $q = {{\gamma + 1} \over 2}$
2

### 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.
3

### JEE Main 2014 (Offline)

There is a circular tube in a vertical plane. Two liquids which do not mix and of densities ${d_1}$ and ${d_2}$ are filled in the tube. Each liquid subtends ${90^ \circ }$ angle at center. Radius joining their interface makes an angle $\alpha$ with vertical. Radio ${{{d_1}} \over {{d_2}}}$ is :
A
${{1 + \sin \,\alpha } \over {1 - \sin \,\alpha }}$
B
${{1 + \cos \,\alpha } \over {1 - \cos \,\alpha }}$
C
${{1 + \tan \,\alpha } \over {1 - \tan \,\alpha }}$
D
${{1 + \sin \,\alpha } \over {1 - \cos \,\alpha }}$

## Explanation

Pressure at interface A must be same from both the sides to be in equilibrium.

$\therefore$ $\left( {R\cos \alpha + R\sin \alpha } \right){d_2}g = \left( {R\cos \alpha - R\sin \alpha } \right){d_1}g$
$\Rightarrow {{{d_1}} \over {{d_2}}} = {{\cos \alpha + \sin \alpha } \over {\cos \alpha - \sin \alpha }} = {{1 + \tan \alpha } \over {1 - \tan \alpha }}$
4

### JEE Main 2014 (Offline)

One mole of a diatomic ideal gas undergoes a cyclic process $ABC$ as shown in figure. The process $BC$ is adiabatic. The temperatures at $A, B$ and $C$ are $400$ $K$, $800$ $K$ and $600$ $K$ respectively. Choose the correct statement :
A
The change in internal energy in whole cyclic process is $250$ $R.$
B
The change in internal energy in the process $CA$ is $700$ $R$.
C
The change in internal energy in the process $AB$ is - $350$ $R.$
D
The change in internal energy in the process $BC$ is - $500$ $R.$

## Explanation

In cyclic process, change in total internal energy is zero.
$\Delta {U_{cyclic}} = 0$
$\Delta {U_{BC}} = n{C_v}\Delta T = 1 \times {{5R} \over 2}\Delta T$

where, ${C_v} =$ molar specific heat at constant volume.
For $BC,$ $\Delta T = - 200\,K$
$\therefore$ $\Delta {U_{BC}} = - 500R$