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MHT CET 2019 (3 May, Shift 1) · Potentiometer

Potentiometer question from MHT CET 2019 (3 May, Shift 1)

Q71MHT CET 2019 · 3 MAY · SHIFT 1Medium

Potentiometer

A potentiometer wire has length LL For given cell of emf EE, the balancing length is L3\frac{L}{3} from the positive end of the wire. If the length of potentiometer wire is increased by 50%50\%, then for the same cell, the balance point is obtained at length

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Answer: (a)

  1. k=VLk = \dfrac{V}{L}, so E=VL⋅L3=V3E = \dfrac{V}{L}\cdot\dfrac{L}{3} = \dfrac{V}{3}.
  2. New length 1.5L1.5L: V1.5L x=V3⇒x=L2\dfrac{V}{1.5L}\,x = \dfrac{V}{3} \Rightarrow x = \dfrac{L}{2}
  3. So the answer is (a) L2\dfrac{L}{2} from positive end.
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