The circuit shown here is used to compare the | vedclass

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(A) In a potentiometer,the balancing length $l$ is directly proportional to the emf $E$ of the cell,given by $E = kl$,where $k$ is the potential gradi

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to the left of $C$

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(A) In a potentiometer,the balancing length $l$ is directly proportional to the emf $E$ of the cell,given by $E = kl$,where $k$ is the potential gradient along the wire.For two cells,we have $E_1 = kl_1$ and $E_2 = kl_2$.Taking the ratio,we get $E_1 / E_2 = l_1 / l_2$.Given that $E_1 > E_2$,it follows that $l_1 > l_2$.Since the null point $C$ corresponds to length $l_1$ for cell $E_1$,the null point for cell $E_2$ will correspond to a shorter length $l_2$.Therefore,the null point for cell $E_2$ will shift to the left of $C$ towards the end $A$.

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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