Two cells of emf $E_1$ and $E_2\left(E_1 > E_2\right)$ are connected individually to a potentiometer and their corresponding balancing length are $625 \,cm$ and $500 \,cm$, then the ratio $\frac{E_1}{E_2}$ is ...........
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(a)
$E_1=k l_1$ $\quad k$ is potential gradient
$E_2=k l_2$
$\frac{E_1}{E_2}=\frac{l_1}{l_2}=\frac{625}{500}=\frac{25}{20}=\frac{5}{4}$
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