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In potentiometer experiment, cells of e.m.f. E and E 2 are connected in series (E $_{1}>\mathrm{E}_{2}$ ), the balancing length is $64 \mathrm{~cm}$ of the wire. If the polarity of $\mathrm{E}_{2}$ is reversed, the balancing length becomes $32 \mathrm{~cm}$. The ratio $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}$ is
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$3: 1$
$\begin{aligned} \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} &=\frac{\ell_{1}+\ell_{2}}{\ell_{1}-\ell_{2}} \\ &=\frac{490+90}{490-90}=\frac{580}{400}=1.45 \end{aligned}$
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