What equal length of an iron wire and a copper-nickel alloy wire, each of 2 mm diameter connected parallel to give an equivalent resistance of 3Ω?

(Given, resistivities of iron and copper-nickel alloy wire are 12 µΩ cm and 51 µΩ cm respectively)

Let the resistance of iron wire be R₁ and that of copper nickel alloy wire be R₂

                  $\begin{array}{l}{\mathrm{r}}_1={\mathrm{r}}_2=1\mathrm{mm}={10}^{-3}\mathrm{m}\\ {\mathrm{\rho }}_1=12\mathrm{\mu \Omega cm}\\ =12\times {10}^{-6}\mathrm{\Omega Cm}\\ =12\times {10}^{-8}\mathrm{\Omega m}\\ {\mathrm{\rho }}_2=51\mathrm{\mu \Omega cm}\\ =51\times {10}^{-6}\mathrm{\Omega Cm}\\ =51\times {10}^{-8}\mathrm{\Omega m}\end{array}$

For parallel combination,

$\begin{array}{l} {\mathrm{R}}_{\mathrm{eq}}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}\\ 3=\frac{\frac{{\mathrm{\rho }}_1\mathrm{I}}{{\mathrm{\pi r}}_1^2}\frac{{\mathrm{\rho }}_2\mathrm{I}}{{\mathrm{\pi r}}_2^2}}{\frac{{\mathrm{\rho }}_1\mathrm{I}}{{\mathrm{\pi r}}_1^2}+\frac{{\mathrm{\rho }}_2\mathrm{I}}{{\mathrm{\pi r}}_2^2}}\\ \Rightarrow 3=\frac{\frac{\left(12\times {10}^{-8}\right)\mathrm{I}}{\mathrm{\pi }\times {10}^{-6}}\times \frac{51\times {10}^{-8}\mathrm{I}}{\mathrm{\pi }\times {10}^{-6}}}{\frac{12\times {10}^{-8}\mathrm{I}}{\mathrm{\pi }\times {10}^{-6}}+\frac{51\times {10}^{-8}\mathrm{I}}{\mathrm{\pi }\times {10}^{-6}}}\\ \mathrm{On} \mathrm{solving}, \mathrm{I}=97\mathrm{m}\end{array}$