In the given figure, the resistance of the meter bridge wire AB is 4Ω. With the emf of the cell is, ε = 0.5 V and rheostat resistance R_h = 2Ω, the null point is obtained at some point J. When the cell is replaced by another one of emf ε = ε₂, we get the same null point J for R_h = 6 Ω. The emf ε₂ must be:

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0.3 V

0.5 V

0.6 V

0.4 V

answer is B.

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Detailed Solution

In circuit, when $R_{h} = 2 Ω$

$\begin{array}{l} i_{1} = \frac{6}{4 + 2} = 1 A, ε = \frac{1 \times 4}{AB} \times AJ \\ A J = \frac{A B}{4} \times e \dots \dots \end{array}$

when $R_{h} = 6 Ω$

$\begin{array}{l} i_{2} = \frac{6}{4 + 6} = 0.6 A \\ ε_{2} = \frac{0.6 \times 4}{A B} \times A J \\ A J = \frac{E_{2} \times A B}{0.6 \times 4} \dots \dots \dots \dots . . (2) \end{array}$

From (1) & (2)

$\begin{array}{l} \frac{ε_{2} \times AB}{0 .6 \times 4} = \frac{E \times AB}{4} \\ ε_{2} = 0 .6 \times ε \\ = 0 .6 \times 0 .5 = 0 .3 V \end{array}$

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