Two cells of $e.m.f.$ $E_1$ and $E_2$ are joined in series and the balancing length of the potentiometer wire is $625$ $cm$. If the terminals of $E_1$ are reversed, the balancing length obtained is $125 \,cm$. Given $E_2 > E_1$, the ratio $E_1: E_2$ will be
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(a)
$\frac{E_1+E_2}{E_1-E_2}=\frac{625}{125}=5$
$\frac{E_1}{E_2}=\frac{3}{2}$
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