b
As both metal wires are of identical dimensions, so their length and area of cross-section will be same. Let them be \(l\) and \(A\) respectively. Then the resistance of the first wire is

\(R_{1}=\frac{l}{\sigma_{1} A}\)         ...\((i)\)

and that of the second wire is

\(R_{2}=\frac{l}{\sigma_{2} A}\)        ....\((ii)\)

As they are connected in series, so their effective

resistance is

\(R_{s} =R_{1}+R_{2} \)

\(=\frac{l}{\sigma_{1} A}+\frac{l}{\sigma_{2} A}\)       \( \quad(\text { using }(\mathrm{i}) \text { and (ii) })\)

\(=\frac{l}{A}\left(\frac{1}{\sigma_{1}}+\frac{1}{\sigma_{2}}\right)\)       ....\((iii)\)

If \(\sigma_{\mathrm{eff}}\) is the effective conductivity of the combination, then

\(R_{s}=\frac{2 l}{\sigma_{\mathrm{eff}} A}\)        ....\((iv)\)

Equating eqns. \((iii)\) and \((iv),\) we get

\({\frac{2 l}{\sigma_{\mathrm{eff}} A}=\frac{l}{A}\left(\frac{1}{\sigma_{1}}+\frac{1}{\sigma_{2}}\right)} \)

\({\frac{2}{\sigma_{\mathrm{eff}}}=\frac{\sigma_{2}+\sigma_{1}}{\sigma_{1} \sigma_{2}} \text { or } \sigma_{\mathrm{eff}}=\frac{2 \sigma_{1} \sigma_{2}}{\sigma_{1}+\sigma_{2}}}\)