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The ratio of radii of two cylindrical rods of same material is 2:1 and ratio of their lengths is 2:3. Their ends are maintained at same temperature difference. If rate of flow of heat in the longer rod is $2 c a l s^{‑ 1}$ , then that in the shorter rod will be

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$12 c a l s^{‑ 1}$

$1 c a l s^{‑ 1}$

$4 c a l s^{‑ 1}$

$8 c a l s^{‑ 1}$

answer is B.

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Detailed Solution

$\dot{Q} = \frac{Q}{t} \propto \frac{r^{2}}{l}$

$\Rightarrow \frac{\dot{Q}}{2} = \frac{r_{1}^{2}}{r_{2}^{2}} \times \frac{l_{2}}{l_{1}} = {(\frac{2}{1})}^{2} \times (\frac{3}{2})$

$\Rightarrow \dot{Q} = 12 cal s^{- 1}$

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