Two slabs A and B of different materials but of the same thickness are joined as shown in the figure. The thermal conductivities of A and B are $k_{1}$ and $k_{2}$ respectively. The thermal conductivity of the composite slab will be

$(k_{1} + k_{2})$

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$\frac{1}{2} (k_{1} + k_{2})$

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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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Solution

The correct option is D
$\frac{2 k_{1} k_{2}}{(k_{1} + k_{2})}$

The thermal resistance of a slab of length l, area A and thermal conductivity k is given by $R = \frac{l}{k A}$
For slab A: $R_{1} = \frac{\frac{l}{2}}{k_{1} A}$
For slab B: $R_{2} = \frac{\frac{l}{2}}{k_{2} A}$
Since the slabs are joined in series, the thermal resistance of the composite slab is $R_{c} = R_{1} + R_{2} \Rightarrow \frac{l}{k_{c} A} = \frac{\frac{l}{2}}{k_{1} A} + \frac{\frac{l}{2}}{k_{2} A} \Rightarrow \frac{1}{k_{c}} = \frac{1}{2} (\frac{1}{k_{1}} + \frac{1}{k_{2}}) \Rightarrow k_{c} = \frac{2 k_{1} k_{2}}{(k_{1} + k_{2})}$
Hence, the correct choice is (d).

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