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(B) Let the length of each wire be $\ell$ and the cross-sectional area be $A$.For the equivalent wire,the total length is $2\ell$ and the cross-sectio

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

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(B) Let the length of each wire be $\ell$ and the cross-sectional area be $A$.For the equivalent wire,the total length is $2\ell$ and the cross-sectional area is $A$.The resistance of a wire is given by $R = \frac{\ell}{\sigma A}$,where $\sigma$ is the conductivity.Since the wires are connected in series,the equivalent resistance $R_{eq}$ is the sum of individual resistances:$R_{eq} = R_{1} + R_{2}$Substituting the formula for resistance:$\frac{2\ell}{\sigma_{eq} A} = \frac{\ell}{\sigma_{1} A} + \frac{\ell}{\sigma_{2} A}$Dividing both sides by $\frac{\ell}{A}$:$\frac{2}{\sigma_{eq}} = \frac{1}{\sigma_{1}} + \frac{1}{\sigma_{2}}$$\frac{2}{\sigma_{eq}} = \frac{\sigma_{1} + \sigma_{2}}{\sigma_{1} \sigma_{2}}$Therefore,the effective conductivity is $\sigma_{eq} = \frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1} + \sigma_{2}}$.

Two metallic wires of identical dimensions are connected in series. If $\sigma_{1}$ and $\sigma_{2}$ are the conductivities of these wires respectively,the effective conductivity of the combination is:

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