The circuit shown here is used to compare the emf of two cells $E_1$ and $E_2 (E_1 > E_2)$. The null point is at $C$ when the galvanometer is connected to $E_1$. When the galvanometer is connected to $E_2$, the null point will be
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In case of potentiometer, $\mathrm{E}_{1} / \mathrm{E}_{2}=l_{1} / l_{2} .$ As given that $\mathrm{E}_{1}>\mathrm{E}_{2},$ therefore $l_{1}>l_{2} .$ Therefore, the null point for the cell of $\mathrm{emf}$ $\mathrm{E}_{2}$ must be at shortest length than that of cell $\mathrm{E}_{1} .$ Thus, the null point on potentiometer wire should shift towards left of $\mathrm{C}$
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