 JEE Mains Previous Years Questions with Solutions

4.5
star star star star star
1

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

${{{n_1} + {n_2}} \over {r - 1}} = {{{n_1}} \over {{r_1} - 1}} + {{{n_2}} \over {{r_2} - 1}}$
${{{{16} \over 4} + {{16} \over {32}}} \over {r - 1}} = {{16/4} \over {{5 \over 3} - 1}} + {{16/32} \over {1.4 - 1}}$
$\therefore$ $\gamma = 1.62$
2

AIEEE 2004

The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2K$ and thickness $x$ and $4x,$ respectively, are ${T_2}$ and ${T_1}\left( {{T_2} > {T_1}} \right).$ The rate of heat transfer through the slab, in a steady state is $\left( {{{A\left( {{T_2} - {T_1}} \right)K} \over x}} \right)f,$ with $f$ equal to A
${2 \over 3}$
B
${1 \over 2}$
C
$1$
D
${1 \over 3}$

Explanation

The thermal resistance
${x \over {KA}} + {{4x} \over {2KA}} = {{3x} \over {KA}}$
$\therefore$ ${{dQ} \over {dt}} = {{\Delta T} \over {{{3x} \over {KA}}}} = {{\left( {{T_2} - {T_1}} \right)KA} \over {3x}}$
$= {1 \over 3}\left\{ {{{A\left( {{T_2} - {T_1}} \right)K} \over x}} \right\}$
$\therefore$ $f = {1 \over 3}$
3

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.
4

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 ${n_2} = {{{P_2}{V_2}} \over {R{T_2}}}$
$\therefore$ $T = {{{{{P_1}{V_1}} \over {R{T_1}}} \times {T_1} + {{{P_2}{V_2}} \over {R{T_2}}} \times {T_2}} \over {{{{P_1}{V_1}} \over {R{T_1}}} + {{{P_2}{V_2}} \over {R{T_2}}}}}$
$= {{{T_1}{T_2}\left( {{P_1}{V_1} + {P_2}{V_2}} \right)} \over {{P_1}{V_1}{T_2} + {P_2}{V_2}{T_1}}}$