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Q.

A potentiometer has a uniform wire of length 5m. A battery of emf 10V and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at 200cm.The emf of secondary cell.

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a

6V

b

12V

c

8V

d

4V

answer is C.

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Detailed Solution

$\begin{array}{l} l_{1} = 5 m, l_{2} = 200 c m = 2 m \\ ε_{1} = 10 v, ε_{2} = ? \\ \frac{ε_{1}}{ε_{2}} = \frac{l_{1}}{l_{2}} = \frac{ε_{2}}{ε_{1}} = \frac{l_{2}}{l_{1}} \Rightarrow ε_{2} = \frac{l_{2}}{l_{1}} \times \frac{ε_{2}}{ε_{1}} \\ = \frac{2}{5} \times 10 = 4 V \end{array}$

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