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JEE Main 2019 (11 Jan, Shift 1) · Potentiometer

Potentiometer question from 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 Ω\Omega. With a cell of emf ε\varepsilon = 0.5 V and rheostat resistance Rh = 2 Ω\Omega the null point is obtained at some point J. When the cell is replaced by another one of emf ε=ε\varepsilon = \varepsilon2 the same null point J is found for Rh = 6 Ω\Omega. The emf ε\varepsilon2 is:

Figure

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Answer: (a)

  1. Primary current I=6RAB+RhI = \frac{6}{R_{AB} + R_h}; at balance ε=I⋅RAJ\varepsilon = I \cdot R_{AJ} (same J both times).
  2. Case 1: I1=64+2=1 AI_1 = \frac{6}{4+2} = 1\ \text{A}, so RAJ=0.51=0.5 ΩR_{AJ} = \frac{0.5}{1} = 0.5\ \Omega.
  3. Case 2: I2=64+6=0.6 AI_2 = \frac{6}{4+6} = 0.6\ \text{A}.
  4. ε2=I2RAJ=0.6×0.5=0.3 V\varepsilon_2 = I_2 R_{AJ} = 0.6 \times 0.5 = 0.3\ \text{V}.
  5. So the answer is (a) 0.3 V.
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