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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2012

MCQ (Single Correct Answer)
Helium gas goes through a cycle $ABCD$ (consisting of two isochoric and isobaric lines) as shown in figure efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas)
A
$15.4\%$
B
$9.1\%$
C
$10.5\%$
D
$12.5\%$

Explanation

Heat given to system $= {\left( {n{C_v}\Delta T} \right)_{A \to B}} + {\left( {n{C_p}\Delta T} \right)_{B \to C}}$
$= {\left[ {{3 \over 2}\left( {nR\Delta T} \right)} \right]_{A \to B}} + {\left[ {{5 \over 2}\left( {nR\Delta T} \right)} \right]_{B \to C}}$
$= {\left[ {{3 \over 2} \times {V_0}\Delta P} \right]_{A \to B}} + {\left[ {{5 \over 2} \times 2{P_0} \times {V_0}} \right]_{B \to C}}$
$= {{13} \over 2}{P_0}{V_0}$
and ${W_0} = {P_0}{V_0}$
$\eta = {{Work} \over {heat\,\,given}}$
$= {{{P_0}{V_0}} \over {{{13} \over 2}{P_0}{V_0}}} \times 100$
$= 15.4\%$
2

AIEEE 2012

MCQ (Single Correct Answer)
A Carnot engine, whose efficiency is $40\%$, takes in heat from a source maintained at a temperature of $500$ $K.$ It is desired to have an engine of efficiency $60\% .$ Then, the intake temperature for the same exhaust (sink) temperature must be :
A
efficiency of Carnot engine cannot be made larger than $50\%$
B
$1200$ $K$
C
$750$ $K$
D
$600$ $K$

Explanation

$0.4 = 1 - {{{T_2}} \over {500}}\,\,\,\,$ and $\,\,\,\,0.6 = 1 - {{{T_2}} \over {{T_1}}}$
on solving we get ${T_2} = 750\,K$
3

AIEEE 2011

MCQ (Single Correct Answer)
$100g$ 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

$\Delta U = \Delta Q = mc\Delta T$
$= 100 \times {10^{ - 3}} \times 4184\left( {50 - 30} \right) \approx 8.4\,kJ$
4

AIEEE 2011

MCQ (Single Correct Answer)
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
${P_1}{V_1} + {P_2}{V_2} + {P_3}{V_3} = PV$
${{{n_1}} \over {{N_A}}}R{T_1} + {{{n_2}} \over {{N_A}}}R{T_2} + {{{n_3}} \over {{N_A}}}R{T_3}$
$= {{{n_1} + {n_2} + {n_3}} \over {{N_A}}}R{T_{mix}}$
$\Rightarrow {T_{mix}} = {{{n_1}{T_1} + {n_2}{T_2} + {n_3}{T_3}} \over {{n_1} + {n_2} + {n_3}}}$

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