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The length of the potentiometer wire is 600 c m and a current of 40 m A is flowing in it. When a cell of emf 2 V and internal resistance 10 Ω is balanced on this potentiometer the balance length is found to be 500 c m . The resistance of potentiometer wire will be

Options

  1. A20 Ω
  2. B40 Ω
  3. C60 Ω
  4. D80 Ω

Correct answer

C. 60 Ω

Step-by-step solution

EMF   o f   e x ternal   c i r c u i t = P o t e n t i a l   g r a d i e n t   ×   B a l a n c e   l e n g t h 2 = I R L × l 0 R = 2 L I × l 0 = 2 × 600 40 × 1 0 - 3 × 500 = 12 200 × 1 0 - 3 = 12 × 1 0 3 200 =   60   Ω .

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