The length of a potentiometer wire is 1200 cm and it carries a current of 60 mA. For a cell of emf 5 V and internal resistance of $20 \mathrm{\Omega },$ the null point on it is found to be at 1000 cm. The resistance of whole wire is:

Let R be the resistance of the whole wire potential gradient of the potentiometer wire AB

$\begin{array}{l}-\frac{\mathrm{dV}}{\mathrm{d\ell }}=\frac{\mathrm{I}\times \mathrm{R}}{\mathrm{\ell }}=\left[\frac{60\times \mathrm{R}}{{\mathrm{\ell }}_{\mathrm{AB}}}\right] mV/\mathrm{m}\\ {\mathrm{V}}_{\mathrm{AP}}=\left(\frac{\mathrm{dV}}{{\mathrm{d\ell }}_{\mathrm{AB}}}\right){\mathrm{\ell }}_{\mathrm{AP}}=\frac{60\times \mathrm{R}}{1200}\times 1000 \mathrm{mV}\end{array}$

$\Rightarrow {\mathrm{V}}_{\mathrm{AP}}=50R \mathrm{mV}$

Also, AP  V_AP = 5 V (for balance point at P)

$\therefore \mathrm{R}=\frac{{\mathrm{V}}_{\mathrm{AP}}}{50\times {10}^{-3}}=\frac{5}{50\times {10}^{-3}}=100 \mathrm{\Omega }$