### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio ${C_p}/{C_V}$ for the gas is
A
${4 \over 3}$
B
$2$
C
${5 \over 3}$
D
${3 \over 2}$

## Explanation

$P \propto {T^3} \Rightarrow P{T^{ - 3}} =$ constant ....$(i)$
But for an adiabatic process, the pressure temperature relationship is given by
${P^{1 - \gamma }}\,\,{T^\gamma } =$ constant $\Rightarrow P{T^{{\gamma \over {1 - \gamma }}}} =$ constant. ....$(ii)$
From $(i)$ and $(ii)$ ${\gamma \over {1 - \gamma }} = - 3 \Rightarrow \gamma = - 3 + 3\gamma \Rightarrow \gamma = {3 \over 2}$
2

### AIEEE 2003

''Heat cannot by itself flow from a body at lower temperature to a body at higher temperature'' is a statement or consequence of
A
second law of thermodynamics
B
conservation of momentum
C
conservation of mass
D
first law of thermodynamics

## Explanation

This is a statement of second law of thermodynamics
3

### AIEEE 2002

1 mole of a gas with $\gamma = 7/5$ is mixed with $1$ mole of a gas with $\gamma = 5/3,$ then the value of $\gamma$ for the resulting mixture is
A
$7/5$
B
$2/5$
C
$3/2$
D
$12/7$

## Explanation

If ${n_1}$ moles of adiabatic exponent ${\gamma _1}$ is mixed with ${n_2}$ moles of adiabatic exponent ${\gamma _2}$ then the adiabatic component of the resulting mixture is given by

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

$\therefore$ ${2 \over {\gamma - 1}} = {5 \over 2} + {3 \over 2} = 4$

$\therefore$ $2 = 4\gamma - 4 \Rightarrow \gamma = {6 \over 4} = {3 \over 2}$

4

### AIEEE 2002

Two spheres of the same material have radii $1$ $m$ and $4$ $m$ and temperatures $4000$ $K$ and $2000$ $K$ respectively. The ratio of the energy radiated per second by the first sphere to that by the second is
A
$1:1$
B
$16:1$
C
$4:1$
D
$1:9$

## Explanation

The energy radiated per second is given by $E = e\sigma {T^4}A$
For same material $e$ is same. $\sigma$ is stefan's constant
$\therefore$ ${{{E_1}} \over {{E_2}}} = {{T_1^4{A_1}} \over {T_2^4{A_2}}} = {{T_1^44\pi r_1^2} \over {T_2^44\pi r_2^2}}$
$= {{{{\left( {4000} \right)}^4} \times {1^2}} \over {{{\left( {2000} \right)}^4} \times {4^2}}} = {1 \over 1}$