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(C) In a potentiometer,the balancing length $l$ is directly proportional to the emf $\varepsilon$ of the cell,i.e.,$\varepsilon = kl$,where $k$ is the

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$\frac{5}{3}$

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(C) In a potentiometer,the balancing length $l$ is directly proportional to the emf $\varepsilon$ of the cell,i.e.,$\varepsilon = kl$,where $k$ is the potential gradient.When the galvanometer is connected to point $(1)$,only cell $\varepsilon_{1}$ is in the circuit:$\varepsilon_{1} = k l_{1} = k(250) \ldots (i)$When the galvanometer is connected to point $(2)$,both cells $\varepsilon_{1}$ and $\varepsilon_{2}$ are in the circuit in series:$\varepsilon_{1} + \varepsilon_{2} = k l_{2} = k(400) \ldots (ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{\varepsilon_{1}}{\varepsilon_{1} + \varepsilon_{2}} = \frac{250}{400} = \frac{5}{8}$Cross-multiplying gives:$8 \varepsilon_{1} = 5 \varepsilon_{1} + 5 \varepsilon_{2}$$3 \varepsilon_{1} = 5 \varepsilon_{2}$Therefore,the ratio is:$\frac{\varepsilon_{1}}{\varepsilon_{2}} = \frac{5}{3}$

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