Resistivity of potentiometer wire is $${10^{ - 7}}\Omega - m$$    and its area of cross-section is $${10^{ - 6}}{m^2}.$$  When a current $$i = 0.1\,A$$  flows through the wire, its potential gradient is

Solution :
Potential gradient = Potential fall per unit length $$ = \frac{V}{l}$$
$$\eqalign{ & = {\text{current}}\, \times {\text{resistance per unit length}} \cr & = i \times \frac{R}{l} \cr} $$
but $$R = \frac{{\rho l}}{A} \Rightarrow \frac{R}{l} = \frac{\rho }{A}$$
where, symbols have their usual meanings.
∴ Potential gradient $$ = i \times \frac{\rho }{A}$$
Here, $$\rho = {10^{ - 7}}\Omega {\text{ - }}m,\,i = 0.1\,A$$
and $$A = {10^{ - 6}}{m^2}$$
Hence, potential gradient $$ = 0.1 \times \frac{{{{10}^{ - 7}}}}{{{{10}^{ - 6}}}} = \frac{{0.1}}{{10}}$$
$$ = 0.01 = {10^{ - 2}}\,V/m$$