In potentiometer experiment in which we find internal resistance of a cell, let E be the emf of the cell and V the terminal potential difference, then

$\frac{E}{V}=\frac{l_1}{l_2}$

Where $l_1$ and $l_2$ are lengths of potentiometer wire with and without short circuited through a resistance.

Since,

$\frac{E}{V}=\frac{R+r}{R} \left[\because E=I\left(R+r\right)\mathrm{a}\mathrm{n}\mathrm{d} V=IR\right]$

$\therefore \frac{R+r}{R}=\frac{l_1}{l_2}$

$\mathrm{o}\mathrm{r} 1+\frac{r}{R}=\frac{110}{100} \mathrm{o}\mathrm{r} \frac{r}{R}=\frac{10}{100}$

$\mathrm{o}\mathrm{r} r=\frac{1}{10}\times 10=\mathrm{1 }\mathrm{\Omega }$