 JEE Mains Previous Years Questions with Solutions

4.5
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1

AIEEE 2008

The speed of sound in oxygen $\left( {{O_2}} \right)$ at a certain temperature is $460\,\,m{s^{ - 1}}.$ The speed of sound in helium $(He)$ at the same temperature will be (assume both gases to be ideal)
A
$1421\,\,m{s^{ - 1}}$
B
$500\,\,m{s^{ - 1}}$
C
$650\,\,m{s^{ - 1}}$
D
$300\,\,m{s^{ - 1}}$

Explanation

The speed of sound in a gas is given by $v = \sqrt {{{\gamma RT} \over M}}$
$\therefore$ ${{{v_{{O_2}}}} \over {{v_{He}}}} = \sqrt {{{{\gamma _{{O_2}}}} \over {{M_{{O_2}}}}} \times {{{M_{He}}} \over {{\gamma _{He}}}}}$
$= \sqrt {{{1.4} \over {32}} \times {4 \over {1.67}}} = 0.3237$
$\therefore$ ${v_{He}} = {{{v_{{O_2}}}} \over {0.3237}}$
$= {{460} \over {0.3237}}$
$= 1421\,m/s$
2

AIEEE 2007

One end of a thermally insulated rod is kept at a temperature ${T_1}$ and the other at ${T_2}$. The rod is composed of two sections of length ${L_1}$ and ${L_2}$ and thermal conductivities ${K_1}$ and ${K_2}$ respectively. The temperature at the interface of the two section is A
${{\left( {{K_1}{L_1}{T_1} + {K_2}{L_2}{T_2}} \right)} \over {\left( {{K_1}{L_1} + {K_2}{L_2}} \right)}}$
B
${{\left( {{K_2}{L_2}{T_1} + {K_1}{L_1}{T_2}} \right)} \over {\left( {{K_1}{L_1} + {K_2}{L_2}} \right)}}$
C
${{\left( {{K_2}{L_1}{T_1} + {K_1}{L_2}{T_2}} \right)} \over {\left( {{K_2}{L_1} + {K_1}{L_2}} \right)}}$
D
${{\left( {{K_1}{L_2}{T_1} + {K_2}{L_1}{T_2}} \right)} \over {\left( {{K_1}{L_2} + {K_2}{L_1}} \right)}}$

Explanation ${{{K_1}A\left( {{T_1} - T} \right)} \over {{\ell _1}}} = {{{K_2}A\left( {T - {T_2}} \right)} \over {{\ell _2}}}$
$\therefore$ $T = {{{K_1}{T_1}{\ell _2} + {K_2}{T_2}{\ell _1}} \over {{K_2}{\ell _1} + {K_1}{\ell _2}}}$
$= {{{K_1}{\ell _2}{T_1} + {K_2}{\ell _1}{T_2}} \over {{K_1}{\ell _2} + {K_2}{\ell _1}}}$
3

AIEEE 2007

When a system is taken from state $i$ to state $f$ along the path iaf, it is found that $Q=50$ cal and $W=20$ $cal$. Along the path $ibf$ $Q=36$ $cal.$ $W$ along the path $ibf$ is A
$14$ $cal$
B
$6$ $cal$
C
$16$ $cal$
D
$66$ $cal$

Explanation

For path iaf, $\Delta U = Q - W = 50 - 20 = 30\,cal.$
For path ibf, $W = Q - \Delta U = 36 - 30 = 6\,cal.$
4

AIEEE 2007

If ${C_p}$ and ${C_v}$ denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then
A
${C_p} - {C_v} = 28R$
B
${C_p} - {C_v} = R/28$
C
${C_p} - {C_v} = R/14$
D
${C_p} - {C_v} = R$

Explanation

According to Mayer's relationship ${C_p} - {C_v} = R$
$\therefore$ ${{{C_p}} \over M} - {{{C_v}} \over M} = {R \over M}$ $\,\,\,\,\,\,$ Here $M=28.$