Two cells of emf $e 1$ and $e 2$ are joined in series and the balancing length of the potentiometer. wire is $625 c m$ . if the terminals of e1 are reversed the balancing length obtained is $125 c m$ . given $e 2 > e 1$ the ratio $e 1 : e 2$ will be?

1 : 2

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2 : 3

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1 : 4

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4 : 1

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Solution

The correct option is B $2 : 3$

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