A slab consists of portions of different materials of same thickness and having the conductivities $K_{1}$ and $K_{2}$ . The equivalent thermal conductivity of the slab is

K_{1} + K_{2}

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\sqrt{K_{1} + K_{2}}

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

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

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Solution

The correct option is D $\frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$
For the first conductor, thermal rate is given as:
$\frac{H_{1}}{t} = \frac{K_{1} A (T - T_{1})}{L}$

For the second conductor, thermal rate is given as:
$\frac{H_{2}}{t} = \frac{K_{2} A (T_{2} - T)}{L}$

Since they are connected in series, $H_{1} = H_{2}$
$⟹ K_{1} (T - T_{1}) = K_{2} (T_{2} - T)$
Let the equivalent conductance be $K$ .

Hence, $\frac{K A (T_{2} - T_{1})}{2 L} = \frac{K_{1} A (T - T_{1})}{L} + \frac{K_{2} A (T_{2} - T)}{L}$

$⟹ K = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

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A slab consists of portions of different materials of same thickness and having the conductivities K1 and K2. The equivalent thermal conductivity of the slab is

A slab consists of portions of different materials of same thickness and having the conductivities $K_{1}$ and $K_{2}$ . The equivalent thermal conductivity of the slab is