JEE MAIN - Physics (2014 (Offline) - No. 22)
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:
$$1.2$$ $$cal/s$$
$$2.4$$ $$cal/s$$
$$4.8$$ $$cal/s$$
$$6.0$$ $$cal/s$$
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
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$$Q = {{KA\left( {{\theta _1} - {\theta _2}} \right)} \over l}$$
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
$${Q_{Copper}}\,\, = \,\,{Q_{Brass}}\,\, + \,\,{Q_{Steel}}$$
$${{0.92 \times 4\left( {100 - T} \right)} \over {46}} = {{0.26 \times 4 \times \left( {T - 0} \right)} \over {13}} + {{0.12 \times 4 \times \left( {T - 0} \right)} \over {12}}$$
$$ \Rightarrow 200 - 2T = 2T + T$$
$$ \Rightarrow T = {40^ \circ }C$$
$$\therefore$$ $${Q_{Copper}} = {{0.92 \times 4 \times 60} \over {46}} = 4.8\,cal/s$$
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