### JEE Mains Previous Years Questions with Solutions

4.5
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1

### 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 ${Q_3} = 0$
$\eta = 1 - {{{Q_2}} \over {{Q_1}}} = 1 - {{{T_0}{S_0}} \over {{3 \over 2}{T_0}{S_0}}} = {1 \over 3}$
2

### AIEEE 2005

The figure shows a system of two concentric spheres of radii ${r_1}$ and ${r_2}$ are kept at temperatures ${T_1}$ and ${T_2}$, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to
A
$In\left( {{{{r_2}} \over {{r_1}}}} \right)$
B
${{\left( {{r_2} - {r_1}} \right)} \over {\left( {{r_1}{r_2}} \right)}}$
C
${\left( {{r_2} - {r_1}} \right)}$
D
${{{r_1}{r_2}} \over {\left( {{r_2} - {r_1}} \right)}}$

## Explanation

Consider a shell of thickness $(dr)$ and of radius $(r)$ and the temperature of inner and outer surfaces of this shell be $T,(T-dT)$
$H = {{KA\left[ {\left( {T - dT} \right) - T} \right]} \over {dr}} = {{ - KAdT} \over {dr}}$
$H = - 4\pi K{r^2}{{dT} \over {dr}}$ ( as $A = 4\pi {r^2}$ )
Then, $\left( H \right)\int\limits_{{r_1}}^{{r^2}} {{1 \over {{r^2}}}} dr = - 4\pi K\int\limits_{{T_1}}^{{T_2}} {dT}$
$H\left[ {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right] = - 4\pi K\left[ {{T_2} - {T_1}} \right]$
or $H = {{ - 4\pi K{r_1}{r_2}\left( {{T_2} - {T_1}} \right)} \over {\left( {{r_2} - {r_1}} \right)}}$
3

### AIEEE 2005

A system goes from $A$ to $B$ via two processes $I$ and $II$ as shown in figure. If $\Delta {U_1}$ and $\Delta {U_2}$ are the changes in internal energies in the processes $I$ and $II$ respectively, then
A
relation between $\Delta {U_1}$ and $\Delta {U_2}$ can not be determined
B
$\Delta {U_1} = \Delta {U_2}$
C
$\Delta {U_2} < \Delta {U_1}$
D
$\Delta {U_2} > \Delta {U_1}$

## Explanation

Change in internal energy do not depend upon the path followed by the process. It only depends on initial and final states $i.e.,$ $\Delta U{}_1 = \Delta {U_2}$
4

### AIEEE 2005

A gaseous mixture consists of $16$ $g$ of helium and $16$ $g$ of oxygen. The ratio ${{Cp} \over {{C_v}}}$ of the mixture is
A
$1.62$
B
$1.59$
C
$1.54$
D
$1.4$

## Explanation

${{{n_1} + {n_2}} \over {r - 1}} = {{{n_1}} \over {{r_1} - 1}} + {{{n_2}} \over {{r_2} - 1}}$
${{{{16} \over 4} + {{16} \over {32}}} \over {r - 1}} = {{16/4} \over {{5 \over 3} - 1}} + {{16/32} \over {1.4 - 1}}$
$\therefore$ $\gamma = 1.62$