(C) Since the heat flows along the length of the cylinders,the two cylinders are in parallel configuration.The equivalent thermal conductivity $K_{eq}

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(C) Since the heat flows along the length of the cylinders,the two cylinders are in parallel configuration.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:$K_{eq} = \frac{K_1(\pi R^2) + K_2(3\pi R^2)}{4\pi R^2} = \frac{K_1 + 3K_2}{4}$.

Class 11 Physics 10-2.Heat Transfer Mix Examples-Heat Transfer

Medium

$A$ cylinder of radius $R$ made of a material of thermal conductivity $K_1$ is surrounded by a cylindrical shell of inner radius $R$ and outer radius $2R$ made of material of thermal conductivity $K_2$. The two ends of the combined system are maintained at two different temperatures. There is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is

IIT 1988 All IIT

A
$K_1 + K_2$
B
$\frac{K_1 K_2}{K_1 + K_2}$
C
$\frac{K_1 + 3K_2}{4}$
D
$\frac{3K_1 + K_2}{4}$

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