A cell can be balanced against $$110\,cm$$  and $$100\,cm$$  of potentiometer wire, respectively with and without being short circuited through a resistance of $$10\,\Omega .$$  Its internal resistance is

Solution :
In potentiometer experiment in which we find internal resistance of a cell. Let $$E$$ be the emf of the cell and $$V$$ be 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 being short circuited through a resistance.
Since, $$\frac{E}{V} = \frac{{R + r}}{R}\,\,\left[ {\because E = I\left( {R + r} \right)\,{\text{and}}\,V = IR} \right]$$
$$\eqalign{ & \therefore \frac{{R + r}}{R} = \frac{{{l_1}}}{{{l_2}}} \cr & {\text{or}}\,\,1 + \frac{r}{R} = \frac{{110}}{{100}} \cr & {\text{or}}\,\,\frac{r}{R} = \frac{{110}}{{100}} \cr & {\text{or}}\,\,r = \frac{1}{{10}} \times 10 \cr & = 1\,\Omega \cr} $$