### JEE Mains Previous Years Questions with Solutions

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1

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

### AIEEE 2006

Assuming the Sun to be a spherical body of radius $R$ at a temperature of $TK$, evaluate the total radiant powered incident of Earth at a distance $r$ from the Sun

Where r0 is the radius of the Earth and $\sigma$ is Stefan's constant.

A
$4\pi r_0^2{R^2}\sigma {{{T^4}} \over {{r^2}}}$
B
$\pi r_0^2{R^2}\sigma {{{T^4}} \over {{r^2}}}$
C
$r_0^2{R^2}\sigma {{{T^4}} \over {4\pi {r^2}}}$
D
${R^2}\sigma {{{T^4}} \over {{r^2}}}$

## Explanation

Total power radiated by Sun $= \sigma {T^4} \times 4\pi {R^2}$

The intensity of power at earth's surface $= {{\sigma {T^4} \times 4\pi {R^2}} \over {4\pi {r^2}}}$

Total power received by Earth $= {{\sigma {T^4}{R^2}} \over {{r^2}}}\left( {\pi r_0^2} \right)$
3

### AIEEE 2006

The work of $146$ $kJ$ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by ${7^ \circ }C.$ The gas is $\left( {R = 8.3J\,\,mo{l^{ - 1}}\,{K^{ - 1}}} \right)$
A
diatomic
B
triatomic
C
a mixture of monoatomic and diatomic
D
monoatomic

## Explanation

$W = {{nR\Delta T} \over {1 - \gamma }} \Rightarrow - 146000$
$= {{1000 \times 8.3 \times 7} \over {1 - \gamma }}$
or $1 - \gamma = - {{58.1} \over {146}} \Rightarrow \gamma$
$= 1 + {{58.1} \over {146}} = 1.4$
Hence the gas is diatomic.
4

### AIEEE 2005

The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
A
${1 \over 4}$
B
${1 \over 2}$
C
${2 \over 3}$
D
${1 \over 3}$

## Explanation

${Q_1} = {T_0}{S_0} + {1 \over 2}{T_0}{S_0} = {3 \over 2}{T_0}{S_0}$
${Q_2} = {T_0}\left( {2{S_0} - {S_0}} \right)$ $= {T_0}{S_0}$
and ${Q_3} = 0$
$\eta = 1 - {{{Q_2}} \over {{Q_1}}} = 1 - {{{T_0}{S_0}} \over {{3 \over 2}{T_0}{S_0}}} = {1 \over 3}$