### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2011

$100g$ of water is heated from ${30^ \circ }C$ to ${50^ \circ }C$. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is $4184$ $J/kg/K$):
A
$8.4$ $kJ$
B
$84$ $kJ$
C
$2.1$ $kJ$
D
$4.2$ $kJ$

## Explanation

$\Delta U = \Delta Q = mc\Delta T$
$= 100 \times {10^{ - 3}} \times 4184\left( {50 - 30} \right) \approx 8.4\,kJ$
2

### AIEEE 2011

A thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $\gamma .$ It is moving with speed $v$ and it's suddenly brought to rest. Assuming no heat is lost to the surroundings, Its temperature increases by:
A
${{\left( {\gamma - 1} \right)} \over {2\gamma R}}M{v^2}K$
B
${{\gamma {M^2}v} \over {2R}}K$
C
${{\left( {\gamma - 1} \right)} \over {2R}}M{v^2}K$
D
${{\left( {\gamma - 1} \right)} \over {2\left( {\gamma + 1} \right)R}}M{v^2}K$

## Explanation

Here, work done is zero.
So, loss in kinetic energy $=$ change in internal energy of gas
${1 \over 2}m{v^2} = n{C_v}\Delta T = n{R \over {\gamma - 1}}\Delta T$
${1 \over 2}m{v^2} = {m \over M}{R \over {\gamma - 1}}\Delta T$
$\therefore$ $\Delta T = {{M{v^2}\left( {\gamma - 1} \right)} \over {2R}}K$
3

### AIEEE 2011

A Carnot engine operating between temperatures ${{T_1}}$ and ${{T_2}}$ has efficiency ${1 \over 6}$. When ${T_2}$ is lowered by $62$ $K$ its efficiency increases to ${1 \over 3}$. Then ${T_1}$ and ${T_2}$ are, respectively:
A
$372$ $K$ and $330$ $K$
B
$330$ $K$ and $268$ $K$
C
$310$ $K$ and $248$ $K$
D
$372$ $K$ and $310$ $K$

## Explanation

Efficiency of engine
${1 \over 6} = 1 - {{{T_2}} \over {{T_1}}}$ and ${\eta _2} = 1 - {{{T_2} - 62} \over {{T_1}}} = {1 \over 3}$
$\therefore$ ${T_1} = 372\,K$ and ${T_2} = {5 \over 6} \times 372 = 310\,K$
4

### AIEEE 2011

Three perfect gases at absolute temperatures ${T_1},\,{T_2}$ and ${T_3}$ are mixed. The masses of molecules are ${m_1},{m_2}$ and ${m_3}$ and the number of molecules are ${n_1},$ ${n_2}$ and ${n_3}$ respectively. Assuming no loss of energy, the final temperature of the mixture is:
A
${{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}$
B
${{{n_1}T_1^2 + {n_2}T_2^2 + {n_3}T_3^2} \over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}$
C
${{n_1^2T_1^2 + n_2^2T_2^2 + n_3^2T_3^2} \over {{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}}}$
D
${{\left( {{T_1} + {T_2} + {T_3}} \right)} \over 3}$

## Explanation

Number of moles of first gas $= {{{n_1}} \over {{N_A}}}$
Number of moles of second gas $= {{{n_2}} \over {{N_A}}}$
Number of moles of third gas $= {{{n_3}} \over {{N_A}}}$
If there is no loss of energy then
${P_1}{V_1} + {P_2}{V_2} + {P_3}{V_3} = PV$
${{{n_1}} \over {{N_A}}}R{T_1} + {{{n_2}} \over {{N_A}}}R{T_2} + {{{n_3}} \over {{N_A}}}R{T_3}$
$= {{{n_1} + {n_2} + {n_3}} \over {{N_A}}}R{T_{mix}}$
$\Rightarrow {T_{mix}} = {{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}$