In the potentiometer, when the cell in the secondary circuit is...

In a potentiometer, the internal resistance $r$ of a cell is given by the formula $r = R \left( \frac{l_0}{l} - 1 \right)$, where $l_0$ is the balancing length in open circuit, $l$ is the balancing length when shunted with resistance $R$.
This can be rewritten as $\frac{l_0}{l} = 1 + \frac{r}{R}$ or $l = \frac{l_0}{1 + r/R}$.
Case 1: $R_1 = 4 \Omega$ and $l_1 = 120$ cm.
$120 = \frac{l_0}{1 + r/4} = \frac{4l_0}{4 + r}$ ... (i)
Case 2: $R_2 = 12 \Omega$ and $l_2 = 180$ cm.
$180 = \frac{l_0}{1 + r/12} = \frac{12l_0}{12 + r}$ ... (ii)
Dividing equation (ii) by (i):
$\frac{180}{120} = \frac{12l_0}{12 + r} \times \frac{4 + r}{4l_0}$
$\frac{3}{2} = \frac{3(4 + r)}{12 + r}$
$\frac{1}{2} = \frac{4 + r}{12 + r}$
$12 + r = 8 + 2r$
$r = 4 \Omega$
The internal resistance of the cell is $4 \Omega$.