Physics - Heat & Thermo Question with Solution | TestHub

Let length of wire be $l$,
Area of wire as $A$
For equivalent wire length $=2l$ & area will be $A$

As the wires are connected in series combination, the thermal resistance

$R_{eq}=R_1+R_2$

$\Rightarrow \frac{2l}{{\sigma }_{\mathrm{eq}} A}=\frac{l}{{\sigma }_1 A}+\frac{l}{{\sigma }_1 A}$

$$\begin{gathered} \Rightarrow \frac{2}{{\sigma }_{\mathrm{eq} }}=\frac{1}{{\sigma }_1}+\frac{1}{{\sigma }_2} \\ \Rightarrow {\sigma }_{\mathrm{eq} }=\frac{2{\sigma }_1{\sigma }_2}{{\sigma }_1+{\sigma }_2} \end{gathered}$$