The resistance of the meter bridge AB in given figure is 4 Ω. With a… | JEE Main 2019 (11 Jan, Shift 1)

Q72JEE Main 2019 · 11 JAN · SHIFT 1Difficult

Potentiometer

The resistance of the meter bridge AB in given figure is 4 $\Omega$ . With a cell of emf $\varepsilon$ = 0.5 V and rheostat resistance Rh = 2 $\Omega$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\varepsilon = \varepsilon$ 2 the same null point J is found for Rh = 6 $\Omega$ . The emf $\varepsilon$ 2 is:

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Answer: (a)

Primary current $I = \frac{6}{R_{AB} + R_h}$ ; at balance $\varepsilon = I \cdot R_{AJ}$ (same J both times).
Case 1: $I_1 = \frac{6}{4+2} = 1\ \text{A}$ , so $R_{AJ} = \frac{0.5}{1} = 0.5\ \Omega$ .
Case 2: $I_2 = \frac{6}{4+6} = 0.6\ \text{A}$ .
$\varepsilon_2 = I_2 R_{AJ} = 0.6 \times 0.5 = 0.3\ \text{V}$ .
So the answer is (a) 0.3 V.

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Potentiometer question from JEE Main 2019 (11 Jan, Shift 1)

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