### JEE Mains Previous Years Questions with Solutions

4.5
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1

### AIEEE 2009

Two moles of helium gas are taken over the cycle $ABCD,$ as shown in the $P$-$T$ diagram.

The work done on the gas in taking it from $D$ to $A$ is :

A
$+414$ $R$
B
$-690$ $R$
C
$+690$ $R$
D
$-414$ $R$

## Explanation

Work done by the system in the isothermal process
$DA$ is $W = 2.303nRT\,{\log _{10}}{{{P_D}} \over {{P_A}}}$
$= 2.303 \times 2R \times 300{\log _{10}}{{1 \times {{10}^5}} \over {2 \times {{10}^5}}} = - 414R.$
Therefore work done on the gas is $+ \,414\,R.$
2

### AIEEE 2009

Two moles of helium gas are taken over the cycle $ABCD,$ as shown in the $P$-$T$ diagram.

Assuming the gas to be ideal the work done on the gas in taking it from $A$ to $B$ is :

A
$300$ $R$
B
$400$ $R$
C
$500$ $R$
D
$200$ $R$

## Explanation

$A$ to $B$ is an isobaric process. The work done
$W = nR\left( {{T_2} - {T_1}} \right) = 2R\left( {500 - 300} \right) = 400R$
3

### AIEEE 2009

A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature $\theta$ along the length $x$ of the bar from its hot end is best described by which of the following figures?
A
B
C
D

## Explanation

The heat flow rate is given by

${{dQ} \over {dt}} = {{kA\left( {{\theta _1} - \theta } \right)} \over x}$

$\Rightarrow {\theta _1} - \theta$ $= {x \over {kA}}{{dQ} \over {dt}}$

$\Rightarrow \theta$ $= {\theta _1} - {x \over {kA}}{{dQ} \over {dt}}$

where ${\theta _1}$ is the temperature of hot end and $\theta$ is temperature at a distance $x$ from hot end.
The above equation can be graphically represented by option $(a).$
4

### AIEEE 2009

One $kg$ of a diatomic gas is at a pressure of $8 \times {10^4}\,N/{m^2}.$ The density of the gas is $4kg/{m^3}$. What is the energy of the gas due to its thermal motion ?
A
$5 \times {10^4}\,J$
B
$6 \times {10^4}\,J$
C
$7 \times {10^4}\,J$
D
$3 \times {10^4}\,J$

## Explanation

$Volume\,\, = \,\,{{mass} \over {density}} = {1 \over 4}{m^3}$
$K.E = {5 \over 2}PV$
$= {5 \over 2} \times 8 \times {10^4} \times {1 \over 4}$
$= 5 \times {10^4}J$