The temperature of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T₂ and T₁ (T₂ > T₁). The rate of heat transfer through the slab, in a steady state is $(\frac{A (T_{2} - T_{1}) K}{x}) f$ , with f equal to

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$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{3}$

answer is D.

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Detailed Solution

Equation of thermal conductivity of the given combination

$K_{e q} = \frac{l_{1} + l_{2}}{\frac{l_{1}}{K_{1}} + \frac{l_{2}}{K_{2}}} = \frac{x + 4 x}{\frac{x}{K} + \frac{4 x}{2 K}} = \frac{5}{3} K$

Hence rate of flow of heat through the given combination is

$\frac{Q}{t} = \frac{K_{e q} . A (T_{2} - T_{1})}{(x + 4 x)} = \frac{\frac{5}{3} K A (T_{2} - T_{1})}{5 x}$

$= \frac{\frac{1}{3} K A (T_{2} - T_{1})}{x}$

On comparing it with given equation we get $f = \frac{1}{3}$

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