Two cells of e.m.f.s $E_1$ and $E_2\left(E_1>E_2\right)$ are connected as shown in figure.
When the potentiometer is connected between A and B , the balancing length of the potentiometer wire is 3.60 m . On connecting the potentiometer between A and C , the balancing length is 0.90 m . The ratio $E_1 / E_2$ is

Let $k$ be the potential gradient of the potentiometer wire. According to the potentiometer principle, the potential difference across a wire length is proportional to that length: $V = kL$.

When connected between points A and B, the balanced length $L_1 = 3.60\text{ m}$ measures the e.m.f. of cell $E_1$, giving:

$E_1 = k \cdot 3.60$

When connected between A and C, the configuration measures the net e.m.f. of $E_1$ and $E_2$ connected in opposition. Given $E_1 > E_2$, the effective e.m.f. is $E_1 - E_2$, balanced at length $L_2 = 0.90\text{ m}$, so:

$E_1 - E_2 = k \cdot 0.90$

Dividing the first equation by the second:

$\frac{E_1}{E_1 - E_2} = \frac{k \cdot 3.60}{k \cdot 0.90} = 4$

Solving for the ratio:

$E_1 = 4(E_1 - E_2)$

$E_1 = 4E_1 - 4E_2$

$4E_2 = 3E_1$

$\frac{E_1}{E_2} = \frac{4}{3}$

The ratio is 4:3, corresponding to option B.