### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2007

When a system is taken from state $i$ to state $f$ along the path iaf, it is found that $Q=50$ cal and $W=20$ $cal$. Along the path $ibf$ $Q=36$ $cal.$ $W$ along the path $ibf$ is
A
$14$ $cal$
B
$6$ $cal$
C
$16$ $cal$
D
$66$ $cal$

## Explanation

For path iaf, $\Delta U = Q - W = 50 - 20 = 30\,cal.$
For path ibf, $W = Q - \Delta U = 36 - 30 = 6\,cal.$
2

### AIEEE 2007

If ${C_p}$ and ${C_v}$ denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then
A
${C_p} - {C_v} = 28R$
B
${C_p} - {C_v} = R/28$
C
${C_p} - {C_v} = R/14$
D
${C_p} - {C_v} = R$

## Explanation

According to Mayer's relationship ${C_p} - {C_v} = R$
$\therefore$ ${{{C_p}} \over M} - {{{C_v}} \over M} = {R \over M}$ $\,\,\,\,\,\,$ Here $M=28.$
3

### AIEEE 2007

A Carnot engine, having an efficiency of $\eta = 1/10$ as heat engine, is used as a refrigerator . If the work done on the system is $10$ $J$, the amount of energy absorbed from the reservoir at lower temperature is
A
$100$ $J$
B
$99$ $J$
C
$90$ $J$
D
$1$ $J$

## Explanation

The efficiency $\left( \eta \right)$ of a Carnot engine and the coefficient of performance $\left( \beta \right)$ of a refrigerator are related as
$\beta = {{1 - \eta } \over \eta }$
$\,\,\,\,\,\,$ Here, $\eta = {1 \over {10}}$
$\,\,\,\,\,\,$ $\therefore$ $\beta = {{1 - {1 \over {10}}} \over {\left( {{1 \over {10}}} \right)}} = 9.$

Also, Coefficient of performance $\left( \beta \right)$ is given by $\beta = {{{Q_2}} \over W},$
where ${Q_2}$ is the energy absorbed from the reservoir.
or, $9 = {{{Q_2}} \over {10}}$
$\therefore$ ${Q_2} = 90\,J.$

4

### AIEEE 2006

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature ${T_0},$ while Box contains one mole of helium at temperature $\left( {{7 \over 3}} \right){T_0}.$ The boxes are then put into thermal contact with each other, and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases, ${T_f}$ in terms of ${T_0}$ is
A
${T_f} = {3 \over 7}{T_0}$
B
${T_f} = {7 \over 3}{T_0}$
C
${T_f} = {3 \over 2}{T_0}$
D
${T_f} = {5 \over 2}{T_0}$

## Explanation

Heat lost by He $=$ Heat gained by ${N_2}$
${n_1}C{v_1}\Delta {T_1} = {n_2}C{v_2}\Delta {T_2}$
${3 \over 2}R\left[ {{7 \over 3}{T_0} - {T_f}} \right]$
$= {5 \over 2}R\left[ {{T_f} - {T_0}} \right] \Rightarrow {T_f}$
$= {3 \over 2}{T_0}$