The ends of two rods of different materials, having thermal conductivities, radii of cross-section and lengths in the ratio 1 : 2, are maintained at the same temperature difference. If the rate of flow of heat in the larger rod is ${4 cal s}^{- 1}$ , then that in shorter rod is( in ${cal s}^{- 1}$ ) will be

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Solution

Since we know that the rate of heat conduction
$\frac{Q}{t} = \frac{k A Δ T}{l}$ But since $Δ T$ for both the rods is same,
$\frac{{(\frac{Q}{t})}_{1}}{4} = \frac{k_{1} r_{1}^{2} l_{2}}{k_{2} r_{2}^{2} l_{1}}$
${(\frac{Q}{t})}_{1} = {\frac{1}{2} (\frac{1}{4}) 2} 4$
$\Rightarrow (\frac{Q}{t}) = {1 cal s}^{- 1}$

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