Two metal wires of same length, same area of crosssection have conductivities of their material $\sigma_1$…
Two metal wires of same length, same area of crosssection have conductivities of their material $\sigma_1$ and $\sigma_2$. If they are connected in series, the effective conductivity is
$\frac{\sigma_1 \sigma_2}{\sigma_1+\sigma_2}$
$\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}$
$\frac{\sigma_1 \sigma_2}{\sigma_1-\sigma_2}$
$\frac{\sigma_1 \sigma_2}{\sigma_1-\sigma_2}$
Solution
If they are connected in series, then
$\begin{aligned}
& R_{e q}=R_1+R_2 \\
& \frac{L+L}{\sigma_{e q} A}=\frac{L}{\sigma_1 A}+\frac{L}{\sigma_2 A} \\
& \frac{2}{\sigma_{e q}}=\frac{1}{\sigma_1}+\frac{1}{\sigma_2} \Rightarrow \sigma_{e q}=\frac{2 \sigma_1 \sigma_2}{\sigma_1+\sigma_2}
\end{aligned}$