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 Cal s^-1, then that in the shorter rod will be

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4 Cal s^-1

1 Cal s^-1

8 Cal s^-1

12 Cal s^-1

answer is B.

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Detailed Solution

$\frac{r_{1}}{r_{2}} = \frac{2}{1}, \frac{l_{1}}{l_{2}} = \frac{2}{3}, {(\frac{Q}{t})}_{2} = 2 cal / \sec \frac{Q}{t} \propto \frac{r^{2}}{l}, i . e ., \frac{{(\frac{Q}{t})}_{1}}{{(\frac{Q}{t})}_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} \frac{l_{2}}{l_{1}} \frac{{(\frac{Q}{t})}_{1}}{2} = {(\frac{2}{1})}^{2} (\frac{3}{2}) \Rightarrow {(\frac{Q}{t})}_{1} = 12 cal / \sec$