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JEE Main 2020 (2 Sep, Shift 2) · Potentiometer

Potentiometer question from JEE Main 2020 (2 Sep, Shift 2)

Q67JEE Main 2020 · 2 SEP · SHIFT 2Medium

Potentiometer

A potentiometer wire PQ of 1 m length is connected to a standard cell E1. Another cell E2 of emf 1.02 V is connected with a resistance ‘r’ and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49 cm from Q. The potential gradient in the potentiometer wire is:

Figure

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Answer: (c)

  1. At null point no current flows through E2E_2 and rr, so the full emf E2E_2 is balanced.
  2. Null point is 49 cm from Q, so PJ=100−49=51PJ = 100-49 = 51 cm from P (where E2E_2 is connected).
  3. k=E2PJ=1.0251=0.02k = \dfrac{E_2}{PJ} = \dfrac{1.02}{51} = 0.02 V/cm
  4. So the answer is (c) 0.02 V/cm.
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