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A potentiometer wire of length L and resistance 10Ω is connected in series with a battery of emf 2.5V and a resistance in its primary circuit. The null point corresponding to a cell of a emf 1V is obtained at a distance L/2. If the resistance in the primary circuit is doubled then the position of new null point will be

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0.6L

0.8L

0.4L

0.5L

answer is C.

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Detailed Solution

$P o t e n t i a l g r a d i e n t = (\frac{2.5}{10 + R}) \times \frac{10}{L} = \frac{25}{(10 + R)}$

$\frac{25}{(10 + R) L} \times \frac{L}{2} = 1 \Rightarrow 10 + R = 12.5$

$R = 2.5 Ω$

$\begin{array}{l} N e w p o t e n t i a l g r a d i e n t = (\frac{2.5}{10 + 2 R}) \times \frac{10}{L} \\ = \frac{25}{(10 + 5) L} = \frac{5}{3 L} \end{array}$

$\frac{5}{3 L} x = 1 \Rightarrow x = 0.6 L$

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