The resistance of a wire of 0.02 cm radius is 8 $\Omega$. If the resistivity of the material of the wire is 40 $\times$10^-8 ohmmeter, find the length of the wire.

Given, Radius of the base of the wire = 0. 02 𝑐𝑚 

= 0. 02 × 10⁻² 𝑚 

The resistivity of the material of the wire, ρ = 40 × 10⁻⁸ Ω𝑚 

Resistance of wire, 𝑅 = 8 Ω

= ρ × 𝑙/𝐴

Where ‘𝑅’ is the resistance of wire 

‘𝑙’ is the length of wire 

‘ρ’ is the resistivity of the material of the wire 

‘𝐴’ is the area of cross-section of wire

$\Rightarrow \mathrm{R}=\frac{\mathrm{\rho }\times \mathrm{l}}{{\mathrm{\pi r}}^2}$

$\Rightarrow \mathrm{l}=\frac{{\mathrm{R\pi r}}^2}{\mathrm{\rho }}$

$\Rightarrow \mathrm{l}=\frac{8\times 3.14\times 0.02\times {10}^{-2}\times 0.02\times {10}^{-2}}{40\times {10}^{-8}}$

$\Rightarrow \mathrm{l}=\frac{1004.8\times {10}^{-4}}{40\times {10}^{-8}\times {10}^5}$

$\Rightarrow \mathrm{l}=\frac{25.12\times {10}^{-4}}{{10}^{-3}}$

$\Rightarrow \mathrm{l}=2.512 \mathrm{m}$

Hence, the length of the wire is 2. 512 𝑚.