 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2013 (Offline) The above $p$-$v$ diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is

A
${p_0}{v_0}$
B
$\left( {{{13} \over 2}} \right){p_0}{v_0}$
C
$\left( {{{11} \over 2}} \right){p_0}{v_0}$
D
$4{p_0}{v_0}$

Explanation

Along path DA, volume is constant.

Hence, $\Delta$QDA = nCv$\Delta$T = nCv(TA – TD)

$\therefore$ $\Delta$QDA = $n\left( {{3 \over 2}R} \right)\left[ {{{2{p_0}{v_0}} \over {nR}} - {{{p_0}{v_0}} \over {nR}}} \right] = {3 \over 2}{p_0}{v_0}$

Along the path AB, pressure is constant.

Hence $\Delta$QAB = nCp$\Delta$T = nCp(TB – TA)

$\therefore$ $\Delta$QAB = $n\left( {{5 \over 2}R} \right)\left[ {{{2{p_0}2{v_0}} \over {nR}} - {{2{p_0}{v_0}} \over {nR}}} \right] = {{10} \over 2}{p_0}{v_0}$

$\therefore$ The amount of heat extracted from the source in a single cycle is

$\Delta$Q = $\Delta$QDA + $\Delta$QAB

$= {3 \over 2}{p_0}{v_0} + {{10} \over 2}{p_0}{v_0}$ = $\left( {{{13} \over 2}} \right){p_0}{v_0}$
2

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,
$\therefore$ ${F_{net}} = 2F = 2YS\alpha \Delta T$
3

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.

4

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