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(C) Let $k$ be the potential gradient of the potentiometer wire.When the cells are in series and in conjunction,the total $e.m.f.$ is $(E_1 + E_2)$. T

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$3$

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(C) Let $k$ be the potential gradient of the potentiometer wire.When the cells are in series and in conjunction,the total $e.m.f.$ is $(E_1 + E_2)$. The balancing length is $l_1 = 58 \ cm$.So,$(E_1 + E_2) = k \cdot l_1 = 58k$  --- $(1)$When the polarity of $E_2$ is reversed,the total $e.m.f.$ is $(E_1 - E_2)$. The balancing length is $l_2 = 29 \ cm$.So,$(E_1 - E_2) = k \cdot l_2 = 29k$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{E_1 + E_2}{E_1 - E_2} = \frac{58k}{29k} = 2$$E_1 + E_2 = 2(E_1 - E_2)$$E_1 + E_2 = 2E_1 - 2E_2$$3E_2 = E_1$Therefore,$\frac{E_1}{E_2} = \frac{3}{1}$.

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