The potential difference in open circuit for a cell is 2.2 volts. When a 4 ohm resistor is connected between its two electrodes the potential difference becomes 2 volts. The internal resistance of the cell will be

The potential difference in open circuit for the cell is 2.2 volts, it means the emf of the battery should be 2.2 volts.

Now a 4 ohm resistor is connected between its two electrodes, then the potential difference across the battery becomes 2 volts.

$\begin{array}{l}\mathrm{V}=\mathrm{\epsilon }-\mathrm{ir}\Rightarrow 2=2.2-\mathrm{ir}\\ \Rightarrow \mathrm{ir}=0.2\mathrm{V}....\left(\mathrm{i}\right)\\ \mathrm{But}\mathrm{i}=\frac{\mathrm{\epsilon }}{\mathrm{r}+\mathrm{R}}=\frac{2.2}{\mathrm{r}+4}.....\left(\mathrm{ii}\right)\end{array}$

Putting the value of i in (i), we get r = 0.4 ohm.