The resistivity of a potentiometer wire is $40 \times 10^{-8} \Omega \mathrm{m}$ and its area of cross - section is $8 \times 10^{-6} \mathrm{~m}^2$. If $0.2 \mathrm{~A}$ current is flowing through the wire, the potential gradient will be:

10^-2 m^-1

The resistance of 1m length wire will be $R = \frac{\left(40 \times 10^{-8}\right)}{\left(8 \times 10^{-6}\right)}=5 \times 10^{-2} \Omega$

Potential gradient,

$\frac{\Delta \mathrm{V}}{\Delta \mathrm{x}}=\frac{\mathrm{R} . \mathrm{I}}{l}$

$\frac{\Delta \mathrm{V}}{\Delta \mathrm{x}}=\left(5 \times 10^{-2}\right) \times 0.2=10^{-2} \mathrm{Vm}^{-1}$