A cell in secondary circuit gives null deflection for 2.5 m length of potentiometer having 10m length of wire. If the length of the potentiometer wire is increased 1 m without changing the cell in the primary, the position of the null point now is

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3 m

2.75 m

2.0 m

3.5 m

answer is C.

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Detailed Solution

$\begin{array}{l} E^{'} = \frac{E \times R_{P} \times l}{(r + R_{s} + R_{p}) L}; \\ ℓ \propto L, \frac{ℓ_{1}}{ℓ_{2}} = \frac{L_{1}}{L_{2}} \\ \frac{2 .5}{ℓ_{2}} = \frac{10}{11} \Rightarrow ℓ_{2} = 2 .75 m \end{array}$

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