A cell of emf ‘E’ and internal resistance ‘r’ connected in the secondary gets balanced against length ‘l’ of potentiometer wire. If a resistance ‘R’ is connected in parallel with the cell, then the new balancing length for the cell will be

Given length l balances cell of EMF E and internal resistance r.

$\Rightarrow$Potential gradient is $k = \frac{E}{l}$

When a resistance is added in parallel to the cell :

Current in secondary circuit is

$\mathrm{i} = \frac{\mathrm{E}}{\mathrm{r}+\mathrm{R}}$

Potential Difference is 

$\mathrm{V} = \mathrm{iR} = \frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}\mathrm{R}$

Therefore, new balance point will be at $\frac{\mathrm{V}}{\mathrm{k}}$

$\frac{\mathrm{V}}{\mathrm{k}}=\frac{{\displaystyle \frac{ER}{r+R}}}{{\displaystyle \frac{E}{l}}}=\left(\frac{R}{r+R}\right)l$