### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2012

A liquid in a beaker has temperature $\theta \left( t \right)$ at time $t$ and ${\theta _0}$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between ${\log _e}\left( {\theta - {\theta _0}} \right)$ and $t$ is:
A
B
C
D

## Explanation

Newton's law of cooling
${{d\theta } \over {dt}} = - k\left( {\theta - {\theta _0}} \right)$
$\Rightarrow {{d\theta } \over {\left( {\theta - {\theta _0}} \right)}} = - kdt$

Intergrating
$\Rightarrow \log \left( {\theta - {\theta _0}} \right) = - kt + c$
Which represents an equation of straight line.
Thus the option $(a)$ is correct.

2

### AIEEE 2012

Helium gas goes through a cycle $ABCD$ (consisting of two isochoric and isobaric lines) as shown in figure efficiency of this cycle is nearly : (Assume the gas to be close to ideal gas)
A
$15.4\%$
B
$9.1\%$
C
$10.5\%$
D
$12.5\%$

## Explanation

Heat given to system $= {\left( {n{C_v}\Delta T} \right)_{A \to B}} + {\left( {n{C_p}\Delta T} \right)_{B \to C}}$
$= {\left[ {{3 \over 2}\left( {nR\Delta T} \right)} \right]_{A \to B}} + {\left[ {{5 \over 2}\left( {nR\Delta T} \right)} \right]_{B \to C}}$
$= {\left[ {{3 \over 2} \times {V_0}\Delta P} \right]_{A \to B}} + {\left[ {{5 \over 2} \times 2{P_0} \times {V_0}} \right]_{B \to C}}$
$= {{13} \over 2}{P_0}{V_0}$
and ${W_0} = {P_0}{V_0}$
$\eta = {{Work} \over {heat\,\,given}}$
$= {{{P_0}{V_0}} \over {{{13} \over 2}{P_0}{V_0}}} \times 100$
$= 15.4\%$
3

### AIEEE 2012

A Carnot engine, whose efficiency is $40\%$, takes in heat from a source maintained at a temperature of $500$ $K.$ It is desired to have an engine of efficiency $60\% .$ Then, the intake temperature for the same exhaust (sink) temperature must be :
A
efficiency of Carnot engine cannot be made larger than $50\%$
B
$1200$ $K$
C
$750$ $K$
D
$600$ $K$

## Explanation

$0.4 = 1 - {{{T_2}} \over {500}}\,\,\,\,$ and $\,\,\,\,0.6 = 1 - {{{T_2}} \over {{T_1}}}$
on solving we get ${T_2} = 750\,K$
4

### AIEEE 2011

$100g$ of water is heated from ${30^ \circ }C$ to ${50^ \circ }C$. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is $4184$ $J/kg/K$):
A
$8.4$ $kJ$
B
$84$ $kJ$
C
$2.1$ $kJ$
D
$4.2$ $kJ$

## Explanation

$\Delta U = \Delta Q = mc\Delta T$
$= 100 \times {10^{ - 3}} \times 4184\left( {50 - 30} \right) \approx 8.4\,kJ$