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NEET2016PhysicsCurrent ElectricityActual

A potentiometer wire is 100   cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50   cm and 10   cm from the positive end of the wire in the two cases. The ratio of EMF's is :

Options

  1. A5 : 1
  2. B5 : 4
  3. C3 : 4
  4. D3 : 2

Correct answer

D. 3 : 2

Step-by-step solution

as E ∝ L E 1 + E 2 E 1 − E 2 = λ 50 λ 10 E 1 + E 2 = 5 E 1 - 5 E 2 6 E 2 = 4 E 1 3 2 = E 1 E 2

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