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MHT CET 2025 (21 Apr, Shift 1) · Potentiometer

Potentiometer question from MHT CET 2025 (21 Apr, Shift 1)

Q30MHT CET 2025 · 21 APR · SHIFT 1Medium

Potentiometer

The length of a potentiometer wire is ' LL '. A cell of e.m.f. EE is balanced at a length L5\frac{L}{5} from the positive end of the wire. If the length of the wire is increased by L2\frac{L}{2}, the same cell will give balance point at distance ' xx '. The value of ' xx ' is

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Answer: (c)

  1. k=VLk = \dfrac{V}{L}, so E=VL⋅L5=V5E = \dfrac{V}{L}\cdot\dfrac{L}{5} = \dfrac{V}{5}.
  2. New length 3L2\dfrac{3L}{2}: 2V3L x=V5\dfrac{2V}{3L}\,x = \dfrac{V}{5}
  3. x=3L10x = \dfrac{3L}{10}
  4. So the answer is (c) 3L10\dfrac{3L}{10}.
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