The ends of two rods of different materials with their thermal conductivities, radii of cross-sections and lengths all are in the ratio $1:2$ are maintained at the same temperature difference. If the rate of flow of heat in the larger rod is $4\;cal/\sec $, that in the shorter rod in $cal/\sec $ will be
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(a) $\frac{{dQ}}{{dt}} = \frac{{K(\pi {r^2})d\theta }}{{dl}}$
==> $\frac{{{{\left( {\frac{{dQ}}{{dt}}} \right)}_s}}}{{{{\left( {\frac{{dQ}}{{dt}}} \right)}_l}}} = \frac{{{K_s} \times r_s^2 \times {l_l}}}{{{K_l} \times r_l^2 \times {l_s}}}$= $\frac{1}{2} \times \frac{1}{4} \times \frac{2}{1}$
==> ${\left( {\frac{{dQ}}{{dt}}} \right)_s} = \frac{{{{\left( {\frac{{dQ}}{{dt}}} \right)}_l}}}{4} = \frac{4}{4} = 1$
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