 JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

Which of the following statements is correct for any thermodynamic system ?
A
The change in entropy can never be zero
B
Internal energy and entropy and state functions
C
The internal energy changes in all processes
D
The work done in an adiabatic process is always zero,

Explanation

Internal energy and entropy are state function, they do not depend upon path taken.
2

AIEEE 2004

Two thermally insulated vessels $1$ and $2$ are filled with air at temperatures $\left( {{T_1},{T_2}} \right),$ volume $\left( {{V_1},{V_2}} \right)$ and pressure $\left( {{P_1},{P_2}} \right)$ respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be
A
${T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)/\left( {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}} \right)$
B
$\left( {{T_1} + {T_2}} \right)/2$
C
${{T_1} + {T_2}}$
D
${T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)/\left( {{P_1}{V_1}{T_1} + {P_2}{V_2}{T_2}} \right)$

Explanation

Here $Q=0$ and $W=0.$ Therefore from first law of thermodynamics $\Delta U = Q + W = 0$
$\therefore$ Internal energy of the system with partition $=$ Internal energy of the system without partition.
${n_1}{C_v}\,{T_1} + {n_2}\,{C_v}{T_2} = \left( {{n_1} + {n_2}} \right){C_v}\,T$
$\therefore$ $T = {{{n_1}{T_1} + {n_2}T{}_2} \over {{n_1} + {n_2}}}$
But ${n_1} = {{{P_1}{V_1}} \over {R{T_1}}}$ and ${n_2} = {{{P_2}{V_2}} \over {R{T_2}}}$
$\therefore$ $T = {{{{{P_1}{V_1}} \over {R{T_1}}} \times {T_1} + {{{P_2}{V_2}} \over {R{T_2}}} \times {T_2}} \over {{{{P_1}{V_1}} \over {R{T_1}}} + {{{P_2}{V_2}} \over {R{T_2}}}}}$
$= {{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)} \over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$
3

AIEEE 2004

If the temperature of the sun were to increase from $T$ to $2T$ and its radius from $R$ to $2R$, then the ratio of the radiant energy received on earth to what it was previously will be
A
$32$
B
$16$
C
$4$
D
$64$

Explanation

$E = \sigma A{T^4};\,\,A \propto {R^2}$
$\therefore$ $E \propto {R^2}{T^4}$
$\therefore$ ${{{E_2}} \over {{E_1}}} = {{R_2^2T_2^4} \over {R_1^2T_1^4}}$
$\Rightarrow {{{E_2}} \over {{E_1}}}$
$= {{{{\left( {2R} \right)}^2}{{\left( {2T} \right)}^4}} \over {{R^2}{T^4}}}$
$= 64$
4

AIEEE 2004

One mole of ideal monatomic gas $\left( {\gamma = 5/3} \right)$ is mixed with one mole of diatomic gas $\left( {\gamma = 7/5} \right)$. What is $\gamma$ for the mixture? $\gamma$ Denotes the ratio of specific heat at constant pressure, to that at constant volume
A
$35/23$
B
$23/15$
C
$3/2$
D
$4/3$

Explanation

${{{n_1} + {n_2}} \over {\gamma - 1}} = {{{n_1}} \over {{\gamma _1} - 1}} + {{{n_2}} \over {{\gamma _2} - 1}}$
$\Rightarrow {{1 + 1} \over {\gamma - 1}}$
$= {1 \over {{5 \over 3} - 1}} + {1 \over {{7 \over 5} - 1}}$
$\Rightarrow \gamma = {3 \over 2}$