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

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

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

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

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K_{1} + K_{2}

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Solution

The correct option is C $\frac{K_{1} + K_{2}}{2}$
At thermal equilibrium
$H = H_{1} + H_{2}$
Given that two slab are in parallel and have same dimensions.
$\frac{K \times A \times (T_{2} - T_{1})}{L} = \frac{K_{1} \times \frac{A}{2} \times (T_{2} - T_{1})}{L} + \frac{K_{2} \times \frac{A}{2} \times (T_{2} - T_{1})}{L}$ ,
where $K$ is the effective thermal conductivity, $T_{2} - T_{1}$ is the temperature difference between the ends of the slab, $A$ is area of the slab and $L$ is the thickness of the each layer of the slab
So, $\frac{K \times A}{L} = \frac{K_{1} \times \frac{A}{2}}{L} + \frac{K_{2} \times \frac{A}{2}}{L}$
or, $K = \frac{K_{1} + K_{2}}{2}$

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

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