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JEE Mains Previous Years Questions with Solutions

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1

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)
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)

MCQ (Single Correct Answer)

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

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

Questions Asked from Heat and Thermodynamics

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