### JEE Mains Previous Years Questions with Solutions

4.5
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1

### 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

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

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

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

2

### AIEEE 2002

At what temperature is the $r.m.s$ velocity of a hydrogen molecule equal to that of an oxygen molecule at ${47^ \circ }C?$
A
$80K$
B
$-73$ $K$
C
$3$ $K$
D
$20$ $K$

## Explanation

${v_{rms}} =$$\sqrt {{{RT} \over M}}$

For ${v_{rms}}$ to be equal ${{{T_{{H_2}}}} \over {{M_{{H_2}}}}} = {{{T_{{O_2}}}} \over {{M_{{O_2}}}}}$

Here ${M_{{H_2}}} = 2;\,\,{M_{{O_2}}} = 32;$

${T_{{O_2}}} = 47 + 273 = 320K$

$\therefore$ ${{{T_{{H_2}}}} \over 2} = {{320} \over {32}}$

$\Rightarrow {T_{{H_2}}} = 20K$
3

### AIEEE 2002

Even Carnot engine cannot give $100\%$ efficiency because we cannot
A
B
find ideal sources
C
reach absolute zero temperature
D
eliminate friction.

## Explanation

$\eta = 1 - {{{T_2}} \over {{T_1}}}$
For $\eta = 1$ or $100\%$, ${T_2} = 0K.$
The temperature of $0$ $K$ (absolute zero) can not be obtained.
4

### AIEEE 2002

Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will
A
increase
B
decrease
C
remain same
D
decrease for some, while increase for others

## Explanation

Since pressure and volume are not changing, so temperature remains same.