Two wires made of German silver are taken such that the length and area of the cross-section of the second wire are twice and thrice, respectively, those of the first wire. If the resistance of the second wire is 12 Ω, find the resistance of the first wire.

1st case

l₁ = length of German silver wire

a₁ = area of cross-section

R₁ = resistance

${\mathrm{R}}_1=\mathrm{\rho }\frac{{\mathrm{l}}_1}{{\mathrm{a}}_1}$

2nd case

l₂ = length of wire = 2l₁

a₂ = area of cross-section of wire = 3a₁

R₂ = resistance of wire = 12 Ω

$$\begin{gathered} {\mathrm{R}}_2=\mathrm{\rho }\frac{{\mathrm{\ell }}_2}{{\mathrm{a}}_2} \\ \therefore 12=\mathrm{\rho }\frac{2{\mathrm{l}}_1}{3{\mathrm{a}}_1} \\ \mathrm{dividing} \left(1\right) \mathrm{by} \left(2\right) \\ \frac{{\mathrm{R}}_1}{12}=\frac{\mathrm{\rho \ell }}{{\mathrm{a}}_1}\times \frac{3{\mathrm{a}}_1}{\mathrm{\rho }\left(2{\mathrm{\ell }}_1\right)}=\frac{3}{2} \\ \therefore {\mathrm{R}}_1=18\mathrm{\Omega } \end{gathered}$$