A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity K2 . 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

Both the cylinders are in parallel, for the heat flow from one end as shown.Hence ; where A1 = Area of cross-section of inner cylinder = R2 and A2 = Area of cross-section of cylindrical shell

A cylinder of radius R made of a material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity K2 . 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

Both the cylinders are in parallel, for the heat flow from one end as shown.Hence ; where A1 = Area of cross-section of inner cylinder = R2 and A2 = Area of cross-section of cylindrical shell

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A cylinder of radius R made of a material of thermal conductivity K₁ is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of material of thermal conductivity K₂ . 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

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$K_1+K_2$

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

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

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

answer is C.

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Detailed Solution

Both the cylinders are in parallel, for the heat flow from one end as shown.
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Hence ${K_{eq}} = \frac{{{K_1}{A_1} + {K_2}{A_2}}}{{{A_1} + {A_2}}}$ ; where A₁ = Area of cross-section of inner cylinder = $\pi$ R² and
A₂ = Area of cross-section of cylindrical shell $= \pi \{ {(2R)^2} - {(R)^2}\} = 3\pi {R^2}$
$\Rightarrow {K_{eq}} = \frac{{{K_1}(\pi {R^2}) + {K_2}(3\pi {R^2})}}{{\pi {R^2} + 3\pi {R^2}}} = \frac{{{K_1} + 3{K_2}}}{4}$