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(B) The thermal resistance of a plate is given by $R_{th} = \frac{\ell}{KA}$,where $\ell$ is the thickness,$K$ is the thermal conductivity,and $A$ is

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$\frac{2 K_1 K_2}{K_1 + K_2}$

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(B) The thermal resistance of a plate is given by $R_{th} = \frac{\ell}{KA}$,where $\ell$ is the thickness,$K$ is the thermal conductivity,and $A$ is the area of cross-section.Since the plates are joined in series,the total thermal resistance $R_{eq}$ is the sum of individual resistances: $R_{eq} = R_1 + R_2$.Let the thickness of each plate be $\ell$. The total thickness of the composite plate is $2\ell$.Substituting the values: $\frac{2\ell}{K_{eq} A} = \frac{\ell}{K_1 A} + \frac{\ell}{K_2 A}$.Canceling $\frac{\ell}{A}$ from both sides,we get: $\frac{2}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$.Simplifying the right side: $\frac{2}{K_{eq}} = \frac{K_1 + K_2}{K_1 K_2}$.Therefore,the equivalent thermal conductivity is $K_{eq} = \frac{2 K_1 K_2}{K_1 + K_2}$.

Two metal plates of equal thickness have thermal conductivities $K_1$ and $K_2$ respectively. They are joined together to form a single plate. The equivalent thermal conductivity of this composite plate is ..........

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