$\therefore E=K l$

Here $\mathrm{K}$ is potential gradient.

If the cell is short circuited by resistance $'R'$

Let balancing length obtained be $l'$ then

${V=k l^{\prime}}$

${r=\left(\frac{E-V}{V}\right) R}$

$\Rightarrow \quad V=E-V$  $[\because r=R \text { given }]$

$\Rightarrow \quad 2 V=E$

or, $\quad 2 K l^{\prime}=K l$

$\therefore \quad r=\frac{l}{2}$