Two cells of emf E₁ and E₂are joined in series and the balancing length of the potentiometer wire is 600 cm. If the terminals of E₂ are reversed $(E_{2} < E_{1}),$ the balancing length obtained is 150 cm. The ratio of $E_{1} : E_{2} will be$

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5:3

3:2

1:4

4:3

answer is A.

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Detailed Solution

$E_{1} + E_{2} = K (600) & E_{1} - E_{2} = K (150) \frac{(E_{1} + E_{2})}{E_{1} - E_{2}} = 4 \begin{aligned} E_{1} + E_{2} = 4 E_{1} - 4 E_{2} \\ 5 E_{2} = 3 E_{1} \\ E_{1} : E_{2} = 5 : 3 \end{aligned}$

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