When cell of e.m.f. ' $E_1{ }^{\prime}$ is connected to potentiometer wire, the balancing length is ' $\ell_1{ }^{\prime}$. Another cell of e.m.f. ' $E_2{ }^{\prime}\left(E_1>E_2\right)$ is connected so that two cells oppose each other, the balancing length is ' $\ell_2{ }^{\prime}$. The ratio $\mathrm{E}_1: \mathrm{E}_2$ is

A potentiometer's potential drop across any wire length is proportional to that length when current is constant and cross-section uniform. Let $k$ represent the potential gradient.

For cell $E_1$ with balancing length $\ell_1$, the e.m.f. equals the potential drop: $E_1 = k \ell_1$.

When $E_2$ opposes $E_1$, the effective e.m.f. becomes $E_1 - E_2$ and balances at length $\ell_2$: $E_1 - E_2 = k \ell_2$.

Dividing these equations eliminates $k$:

$\frac{E_1}{E_1 - E_2} = \frac{k \ell_1}{k \ell_2} = \frac{\ell_1}{\ell_2}$

Cross-multiplying yields $E_1 \ell_2 = (E_1 - E_2) \ell_1$, which rearranges to $E_2 \ell_1 = E_1 (\ell_1 - \ell_2)$.

The ratio is therefore

$\frac{E_1}{E_2} = \frac{\ell_1}{\ell_1 - \ell_2}$

This corresponds to option B.