### JEE Mains Previous Years Questions with Solutions

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1

### 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.
2

### 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 $BC,$ $\Delta T = - 200\,K$
$\therefore$ $\Delta {U_{BC}} = - 500R$

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)

Three rods of Copper, Brass and Steel are welded together to form a $Y$ shaped structure. Area of cross - section of each rod $= 4c{m^2}.$ End of copper rod is maintained at ${100^ \circ }C$ where as ends of brass and steel are kept at ${0^ \circ }C$. Lengths of the copper, brass and steel rods are $46,$ $13$ and $12$ $cms$ respectively. The rods are thermally insulated from surroundings excepts 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:
A
$1.2$ $cal/s$
B
$2.4$ $cal/s$
C
$4.8$ $cal/s$
D
$6.0$ $cal/s$

## Explanation

Rate of heat flow is given by,
$Q = {{KA\left( {{\theta _1} - {\theta _2}} \right)} \over l}$
Where, $K=$ coefficient of thermal conductivity $l=$ length of rod and $A=$ Area of cross-section of rod

If the junction temperature is $T,$ then
${Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}$
${{0.92 \times 4\left( {100 - T} \right)} \over {46}} = {{0.26 \times 4 \times \left( {T - 0} \right)} \over {13}} + {{0.12 \times 4 \times \left( {T - 0} \right)} \over {12}}$
$\Rightarrow 200 - 2T = 2T + T$
$\Rightarrow T = {40^ \circ }C$
$\therefore$ ${Q_{Copper}} = {{0.92 \times 4 \times 60} \over {46}} = 4.8\,cal/s$