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(C) Since the heat flows from one end to the other,the inner cylinder and the outer cylindrical shell act as thermal conductors in parallel.The equiva

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$\frac{K_1 + 3K_2}{4}$

Vedclass · Answer

(C) Since the heat flows from one end to the other,the inner cylinder and the outer cylindrical shell act as thermal conductors in parallel.The equivalent thermal conductivity $K_{eq}$ for parallel combination is given by $K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$.Here,$A_1$ is the cross-sectional area of the inner cylinder: $A_1 = \pi R^2$.$A_2$ is the cross-sectional area of the outer cylindrical shell: $A_2 = \pi(2R)^2 - \pi R^2 = 4\pi R^2 - \pi R^2 = 3\pi R^2$.The total area $A = A_1 + A_2 = \pi R^2 + 3\pi R^2 = 4\pi R^2$.Substituting these values into the formula:$K_{eq} = \frac{K_1(\pi R^2) + K_2(3\pi R^2)}{4\pi R^2} = \frac{K_1 + 3K_2}{4}$.

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