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(D) The wall consists of blocks arranged in parallel with respect to the direction of heat flow (from left to right). Each block has the same length $

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The wall consists of blocks arranged in parallel with respect to the direction of heat flow (from left to right). Each block has the same length $d$ and the same cross-sectional area $A$.For blocks connected in parallel,the equivalent thermal conductivity $K_{eq}$ is given by the weighted average of the individual conductivities based on their cross-sectional areas:$K_{eq} = \frac{\sum K_i A_i}{\sum A_i}$In this arrangement,there are three blocks of conductivity $K_1$ and three blocks of conductivity $K_2$,each with area $A$. The total area is $6A$.$K_{eq} = \frac{K_1 A + K_2 A + K_1 A + K_2 A + K_1 A + K_2 A}{A + A + A + A + A + A}$$K_{eq} = \frac{3K_1 A + 3K_2 A}{6A} = \frac{3(K_1 + K_2)A}{6A} = \frac{K_1 + K_2}{2}$Thus,the equivalent coefficient of thermal conductivity is $\frac{K_1 + K_2}{2}$.

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