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A potentiometer has a wire 100 cm long, and its standard cell has an e.m.f. of E volt. It is used to determine the e.m.f. of a battery whose internal resistance is 0.5 Ω. If the null point is found at 30 cm from the positive end, the battery’s e.m.f. is

1. 30E/100.5
2. 30E/(100 - 0.5)
3. 30(E - 0.5i)/100
4. 30E/100

Correct answer: 30E/100

Solution

Potential gradient = E/100 per cm; at null point (no current through the cell) internal resistance has no effect. EMF = (E/100)*30 = 30E/100.

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