### JEE Mains Previous Years Questions with Solutions

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1

### JEE Main 2016 (Offline)

$'n'$ moles of an ideal gas undergoes a process $A$ $\to$ $B$ as shown in the figure. The maximum temperature of the gas during the process will be :
A
${{9{P_0}{V_0}} \over {2nR}}$
B
${{9{P_0}{V_0}} \over {nR}}$
C
${{9{P_0}{V_0}} \over {4nR}}$
D
${{3{P_0}{V_0}} \over {2nR}}$

## Explanation

The equation for the line is
$P = {{ - {P_0}} \over {{V_0}}}V + 3P$
[slope $=$ ${{ - {P_0}} \over {{V_0}}},\,\,c = 3{P_0}$ ]
$P{V_0} + {P_0}V = 3{P_0}{V_0}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$
But $\,\,\,\,\,\,\,\,\,\,\,\,pv = nRT$
$\therefore$ $p = {{nRT} \over v}\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$
From $(i)$ & $(ii)$ ${{nRT} \over v}{V_0} + {P_0}V = 3{P_0}{V_0}$
$\therefore$ $nRT{V_0} + {P_0}{V^2} = 3{P_0}{V_0}\,\,\,\,\,...\left( {iii} \right)$
For temperature to be maximum ${{dT} \over {dv}} = 0$
Differentiating e.q.$(iii)$ by $'v'$ we get
$nR{V_0}{{dT} \over {dv}} + {P_0}\left( {2v} \right) = 3{P_0}{V_0}$
$\therefore$ $nR{V_0}{{dT} \over {dv}} = 3{P_0}{V_0} - 2{P_0}V$
${{dT} \over {dv}} = {{3{P_0}{V_0} - 2{P_0}V} \over {nR{V_0}}} = 0$
$V = {{3{V_0}} \over 2}$
$\therefore$ $p = {{3{P_0}} \over 2}\,\,\,\,\,\,\,\,\,$ [From $(i)$ ]
$\therefore$ ${T_{\max }} = {{9{P_0}{V_0}} \over {4nR}}\,\,\,\,\,\,\,\,\,\,$ [ From $(ii)$ ]
2

### JEE Main 2016 (Offline)

A pendulum clock loses $12$ $s$ a day if the temperature is ${40^ \circ }C$ and gains $4$ $s$ a day if the temperature is ${20^ \circ }C.$ The temperature at which the clock will show correct time, and the co-efficient of linear expansion $\left( \alpha \right)$ of the metal of the pendulum shaft are respectively :
A
${30^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 3}}/{}^ \circ C$
B
${55^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 2}}/{}^ \circ C$
C
${25^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 5}}/{}^ \circ C$
D
${60^ \circ }C;\,\,\alpha = 1.85 \times {10^{ - 4}}/{}^ \circ C$

## Explanation

Time lost/gained per day $= {1 \over 2} \propto \Delta \theta \times 86400$ second
$12 = {1 \over 2}\alpha \left( {40 - \theta } \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$
$4 = {1 \over 2}\alpha \left( {\theta - 20} \right) \times 86400\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$
On dividing we get, $\,\,\,3 = {{40 - \theta } \over {\theta - 20}}$
$3\theta - 60 = 40 - \theta$
$4\theta = 100 \Rightarrow \theta = {25^ \circ }C$
3

### JEE Main 2016 (Offline)

An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity $C$ remains constant. If during this process the relation of pressure $P$ and volume $V$ is given by $P{V^n} =$ constant, then $n$ is given by (Here ${C_p}$ and ${C_v}$ are molar specific heat at constant pressure and constant volume, respectively:
A
$n = {{{C_p} - C} \over {C - {C_v}}}$
B
$n = {{C - {C_v}} \over {C - {C_p}}}$
C
$n = {{{C_p}} \over {{C_v}}}$
D
$n = {{C - {C_p}} \over {C - {C_v}}}$

## Explanation

For a polytropic process
$C = {C_v} + {R \over {1 - n}}$
$\therefore$ $C - {C_v} = {R \over {1 - n}}$
$\therefore$ $1 - n = {R \over {C - {C_v}}}$
$\therefore$ $1 - {R \over {C - {C_v}}} = n$
$\therefore$ $n = {{C - {C_v} - R} \over {C - {C_v}}}$
$= {{C - {C_v} - {C_p} + {C_v}} \over {C - {C_v}}}$
$= {{C - {C_p}} \over {C - {C_v}}}$
( as ${C_p} - {C_{v = R}}$ )
4

### JEE Main 2015 (Offline)

Consider a spherical shell of radius $R$ at temperature $T$. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume $u = {U \over V}\, \propto \,{T^4}$ and pressure $p = {1 \over 3}\left( {{U \over V}} \right)$ . If the shell now undergoes an adiabatic expansion the relation between $T$ and $R$ is:
A
$T\, \propto {1 \over R}$
B
$T\, \propto {1 \over {{R^3}}}$
C
$T\, \propto \,{e^{ - R}}$
D
$T\, \propto \,{e^{ - 3R}}$

## Explanation

As, $P = {1 \over 3}\left( {{U \over V}} \right)$
But $\,\,\,\,$ ${U \over V} = KT{}^4$
So, $\,\,\,\,\,P = {1 \over 3}K{T^4}$
or $\,\,\,\,{{uRT} \over V} = {1 \over 3}K{T^4}\,\,\,\,$
$\left[ \, \right.$ As $PV = uRT$ $\left. \, \right]$
${4 \over 3}\pi {R^3}{T^3} =$ $constant$
$\therefore$ $\,\,\,\,$ $T \propto {1 \over R}$