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(D) In a potentiometer,the balancing length $l$ is directly proportional to the e.m.f. of the cell,i.e.,$E = k \cdot l$,where $k$ is the potential gra

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$5 : 3$

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(D) In a potentiometer,the balancing length $l$ is directly proportional to the e.m.f. of the cell,i.e.,$E = k \cdot l$,where $k$ is the potential gradient.When cells are connected in series with the same polarity,the effective e.m.f. is $E_1 + E_2 = k \cdot l_1$.Given $l_1 = 80 \ cm$,so $E_1 + E_2 = 80k$ (Equation $1$).When the polarity of $E_2$ is reversed,the effective e.m.f. is $E_1 - E_2 = k \cdot l_2$.Given $l_2 = 20 \ cm$,so $E_1 - E_2 = 20k$ (Equation $2$).Dividing Equation $1$ by Equation $2$:$\frac{E_1 + E_2}{E_1 - E_2} = \frac{80k}{20k} = \frac{4}{1}$.Applying componendo and dividendo:$\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{4 + 1}{4 - 1}$.$\frac{2E_1}{2E_2} = \frac{5}{3}$.Therefore,$\frac{E_1}{E_2} = \frac{5}{3}$ or $5 : 3$.

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$ and the balancing length is $80 \ cm$. If the polarity of $E_2$ is reversed,the balancing length becomes $20 \ cm$. The ratio $E_1 / E_2$ is:

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