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in Physics by (44.7k points)

A 10m wire potentiometer is connected to an accumulator of steady voltage. A 7.8 m length of it balances the emf of a cell on 'open-circuit'. When cell delivers current through a conductor of resistance 10 Ω it is balanced against 7.0 m of the same potentiometer. What is the internal resistance of the cell ? 

(A) 1.24Ω 

(B) 1.36Ω 

(C) 1.14Ω 

(D) 1Ω

1 Answer

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by (69.7k points)
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Best answer

The correct option is  (C) 1.14Ω.

Explanation:

Internal resistance of cell = r = R[(ℓ1 – ℓ2) / ℓ2

1 = length when in open circuit = 7.8m 

2 = length when key is closed = 7m 

R = 10Ω 

∴ r = 10[(7.8 – 7) / 7] = (8/7)Ω = 1.14Ω

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