### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2012

A wooden wheel of radius $R$ is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area $S$ and length $L.$ $L$ is slightly less than $2\pi R.$ To fit the ring on the wheel, it is heated so that its temperature rises by $\Delta T$ and it just steps over the wheel. As it cools down to surrounding temperature, it process the semicircular parts together. If the coefficient of linear expansion of the metal is $\alpha$, and its Young's modulus is $Y,$ the force that one part of the wheel applies on the other part is :
A
$2\pi SY\alpha \Delta T$
B
$SY\alpha \Delta T$
C
$\pi SY\alpha \Delta T$
D
$2SY\alpha \Delta T$

## Explanation

$\gamma = {{F/S} \over {\Delta L/L}} \Rightarrow \Delta L = {{FL} \over {SY}}$
$\therefore$ $L\alpha \Delta T = {{FL} \over {SY}}$
$\left[ \, \right.$ as ${\Delta L = L\alpha \Delta T}$ $\left. \, \right]$
$\therefore$ $F = SY\alpha \Delta T$
$\therefore$ The ring is pressing the wheel from both sides,
$\therefore$ ${F_{net}} = 2F = 2YS\alpha \Delta T$
2

### 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.

3

### 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 ${W_0} = {P_0}{V_0}$
$\eta = {{Work} \over {heat\,\,given}}$
$= {{{P_0}{V_0}} \over {{{13} \over 2}{P_0}{V_0}}} \times 100$
$= 15.4\%$
4

### 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 ${T_2} = 750\,K$

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Class 12