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(D) The system consists of two parallel paths for heat flow: the inner solid cylinder and the outer cylindrical shell.Since the system is in a steady

Vedclass · Accepted Answer

$\frac{K_{1}+3 K_{2}}{4}$

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(D) The system consists of two parallel paths for heat flow: the inner solid cylinder and the outer cylindrical shell.Since the system is in a steady state and there is no heat loss across the cylindrical surface,the total heat flow $Q_{	ext{total}}$ is the sum of heat flows through the two parts: $Q_{	ext{total}} = Q_{1} + Q_{2}$.The rate of heat flow is given by $Q = \frac{KA \Delta 	heta}{L}$.For the inner cylinder,area $A_{1} = \pi R^{2}$.For the outer shell,area $A_{2} = \pi (2R)^{2} - \pi R^{2} = 3\pi R^{2}$.The total area is $A = A_{1} + A_{2} = 4\pi R^{2}$.Equating the heat flow: $\frac{K(4\pi R^{2}) \Delta 	heta}{L} = \frac{K_{1}(\pi R^{2}) \Delta 	heta}{L} + \frac{K_{2}(3\pi R^{2}) \Delta 	heta}{L}$.Canceling common terms $\frac{\pi R^{2} \Delta 	heta}{L}$ from both sides:$4K = K_{1} + 3K_{2}$.Therefore,the effective thermal conductivity is $K = \frac{K_{1} + 3K_{2}}{4}$.

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