The temperature of the two outer surfaces of a composite slab consisting of two materials having coeffieient of thermal conductivity $K$ and $2 K$ and thickness $x$ and $4 x$ respectively are $T_{2}$ and $T_{1}$ ( $T_{2} > T_{1}$ ). 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

1

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\frac{1}{2}

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\frac{2}{3}

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\frac{1}{3}

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Solution

The correct option is D $\frac{1}{3}$
Given that slab are in series,hence there equivalent will be,
$R_{e q u i v a l e n t} = R_{1} + R_{2} = \frac{X}{K A} + \frac{4 X}{2 K A} = \frac{3 X}{K A}$
we know that at steady state,rate of heat transfer through slab is given by,
$\frac{d Q}{d t} = \frac{T_{2} - T_{1}}{R_{e q u i v a l e n t}} = \frac{K A T_{2} - T_{1}}{3 X}$ --------(1)
and it is given that $\frac{d Q}{d t} = (\frac{K A (T_{2} - T_{1})}{X}) f$ -------(2)
now comparing both equations (1) and (2),we find
$f = \frac{1}{3}$

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The temperatures 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 $T_{2}$ and $T_{1}$ ( $T_{2} > T_{1}$ ). 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