The length of a potentiometer wire is L. A ce | vedclass

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(B) Let $E_0$ be the potential difference across the potentiometer wire.In the first case,the potential gradient is $k_1 = \frac{E_0}{L}$.The balance

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$\frac{L}{2}$

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(B) Let $E_0$ be the potential difference across the potentiometer wire.In the first case,the potential gradient is $k_1 = \frac{E_0}{L}$.The balance length is $l_1 = \frac{L}{3}$.So,the e.m.f. of the cell is $E = k_1 l_1 = \left(\frac{E_0}{L}\right) \left(\frac{L}{3}\right) = \frac{E_0}{3} \quad \dots (1)$In the second case,the new length of the wire is $L' = L + \frac{L}{2} = \frac{3L}{2}$.The new potential gradient is $k_2 = \frac{E_0}{L'} = \frac{E_0}{3L/2} = \frac{2E_0}{3L}$.Let $x$ be the new balance length for the same cell $E$.Then,$E = k_2 x = \left(\frac{2E_0}{3L}\right) x \quad \dots (2)$Equating $(1)$ and $(2)$,we get:$\frac{E_0}{3} = \left(\frac{2E_0}{3L}\right) x$$\Rightarrow x = \frac{L}{2}$.

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{3}$ from the positive end of the wire. If the length of the wire is increased by $\frac{L}{2}$,at what distance will the same cell give a balance point?

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