 JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

Three rods of Copper, Brass and Steel are welded together to form a $Y$ shaped structure. Area of cross - section of each rod $= 4c{m^2}.$ End of copper rod is maintained at ${100^ \circ }C$ where as ends of brass and steel are kept at ${0^ \circ }C$. Lengths of the copper, brass and steel rods are $46,$ $13$ and $12$ $cms$ respectively. The rods are thermally insulated from surroundings excepts at ends. Thermal conductivities of copper, brass and steel are $0.92, 0.26$ and $0.12$ $CGS$ units respectively. Rate of heat flow through copper rod is:
A
$1.2$ $cal/s$
B
$2.4$ $cal/s$
C
$4.8$ $cal/s$
D
$6.0$ $cal/s$

Explanation

Rate of heat flow is given by,
$Q = {{KA\left( {{\theta _1} - {\theta _2}} \right)} \over l}$
Where, $K=$ coefficient of thermal conductivity $l=$ length of rod and $A=$ Area of cross-section of rod If the junction temperature is $T,$ then
${Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}$
${{0.92 \times 4\left( {100 - T} \right)} \over {46}} = {{0.26 \times 4 \times \left( {T - 0} \right)} \over {13}} + {{0.12 \times 4 \times \left( {T - 0} \right)} \over {12}}$
$\Rightarrow 200 - 2T = 2T + T$
$\Rightarrow T = {40^ \circ }C$
$\therefore$ ${Q_{Copper}} = {{0.92 \times 4 \times 60} \over {46}} = 4.8\,cal/s$

2

JEE Main 2014 (Offline)

The pressure that has to be applied to the ends of a steel wire of length $10$ $cm$ to keep its length constant when its temperature is raised by ${100^ \circ }C$ is: (For steel Young's modulus is $2 \times {10^{11}}\,\,N{m^{ - 2}}$ and coefficient of thermal expansion is $1.1 \times {10^{ - 5}}\,{K^{ - 1}}$ )
A
$2.2 \times {10^8}\,\,Pa$
B
$2.2 \times {10^9}\,\,Pa$
C
$2.2 \times {10^7}\,\,Pa$
D
$2.2 \times {10^6}\,\,Pa$

Explanation

Young's modulus $Y = {{stress} \over {strain}}$
$stress = Y \times strain$
$Stress$ in steel wire $=$ Applied $pressure$
$Pressure$ $=$ $stress$ $=$ $Y \times \,strain$
$Strain = {{\Delta L} \over L} = \alpha \Delta T$ (As length is constant)
$= 2 \times {10^{11}} \times 1.1 \times {10^{ - 5}} \times 100$
$= 2.2 \times {10^8}Pa$
3

JEE Main 2013 (Offline)

If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature ${\theta _0},$ the graph between the temperature $T$ of the metal and time $t$ will be closest to
A B C D Explanation

According to Newton's law of cooling, the temperature goes on decreasing with time non-linearly.
4

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

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