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(A) For two wires connected in series,the thermal resistance $R$ is given by $R = \frac{l}{KA}$,where $l$ is the length,$K$ is the thermal conductivit

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

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(A) For two wires connected in series,the thermal resistance $R$ is given by $R = \frac{l}{KA}$,where $l$ is the length,$K$ is the thermal conductivity,and $A$ is the cross-sectional area.Since the wires are identical,they have the same length $l$ and cross-sectional area $A$.The total thermal resistance $R_{eff}$ of the series combination is the sum of individual resistances:$R_{eff} = R_{1} + R_{2} = \frac{l}{K_{1}A} + \frac{l}{K_{2}A}$For the equivalent wire of length $2l$ and thermal conductivity $K_{eq}$,the resistance is:$R_{eff} = \frac{2l}{K_{eq}A}$Equating the two expressions for $R_{eff}$:$\frac{2l}{K_{eq}A} = \frac{l}{K_{1}A} + \frac{l}{K_{2}A}$$\frac{2}{K_{eq}} = \frac{1}{K_{1}} + \frac{1}{K_{2}} = \frac{K_{1} + K_{2}}{K_{1}K_{2}}$$K_{eq} = \frac{2K_{1}K_{2}}{K_{1} + K_{2}}$

Two identical metal wires of thermal conductivities $K_{1}$ and $K_{2}$ respectively are connected in series. The effective thermal conductivity of the combination is

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