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(D) Let $E_1$ and $E_2$ be the $EMF$ of the two cells,and $x$ be the potential gradient of the potentiometer wire.When the cells assist each other,the

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$2 : 1$

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(D) Let $E_1$ and $E_2$ be the $EMF$ of the two cells,and $x$ be the potential gradient of the potentiometer wire.When the cells assist each other,the total $EMF$ is $E_1 + E_2 = x \cdot l_1$,where $l_1 = 6 \ m$.So,$E_1 + E_2 = 6x$ ... $(i)$When the cells oppose each other,the total $EMF$ is $E_1 - E_2 = x \cdot l_2$,where $l_2 = 2 \ m$.So,$E_1 - E_2 = 2x$ ... $(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{E_1 + E_2}{E_1 - E_2} = \frac{6x}{2x} = \frac{3}{1}$Applying componendo and dividendo:$\frac{(E_1 + E_2) + (E_1 - E_2)}{(E_1 + E_2) - (E_1 - E_2)} = \frac{3 + 1}{3 - 1}$$\frac{2E_1}{2E_2} = \frac{4}{2}$$\frac{E_1}{E_2} = 2$Thus,the ratio $E_1 : E_2 = 2 : 1$.

When two cells are connected in series in a potentiometer circuit to assist each other,the balancing length is $6 \ m$. When they are connected in series to oppose each other,the balancing length is $2 \ m$. What is the ratio of the $EMF$ of the two cells?

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