Two cells of e.m.f. E_1 and E_2 (E_1 E_2) a | vedclass

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

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$4 : 3$

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(C) In a potentiometer, the balancing length $l$ is directly proportional to the e.m.f. $E$ of the cell, i.e.,$E = kl$, where $k$ is the potential gradient of the potentiometer wire.When the potentiometer is connected between points $A$ and $B$, the e.m.f. measured is $E_1$. Given the balancing length $l_1 = 412 	ext{ cm}$, we have:$E_1 = k 	imes 412$  --- $(1)$When the potentiometer is connected between points $A$ and $C$, the total e.m.f. measured is the sum of the two cells connected in series, which is $E_1 + E_2$. Given the balancing length $l_2 = 103 	ext{ cm}$, we have:$E_1 + E_2 = k 	imes 103$  --- $(2)$Wait, looking at the circuit diagram, the cells are connected in opposition (positive terminal of $E_1$ faces positive terminal of $E_2$). Thus, the effective e.m.f. between $A$ and $C$ is $E_1 - E_2$.$E_1 - E_2 = k 	imes 103$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{E_1}{E_1 - E_2} = \frac{412}{103}$$\frac{E_1}{E_1 - E_2} = 4$$E_1 = 4(E_1 - E_2)$$E_1 = 4E_1 - 4E_2$$3E_1 = 4E_2$$\frac{E_1}{E_2} = \frac{4}{3}$Therefore, the ratio $E_1 : E_2$ is $4 : 3$.

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