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The resistance of a 100 cm long potentiometer wire is 10 . It is connected to a 2 V battery and a resistance R in series. A source of 10 mV gives a null point at 40 cm length. The external resistance R is

1. 490
2. 790
3. 590
4. 990

Correct answer: 790

Solution

The balance length 40 cm for 10 mV gives the potential gradient as 10/40 = 0.25 mV/cm. So the drop across 100 cm wire is 25 mV, implying current through the 10 wire is 0.025/10 = 0.0025 A. With a 2 V battery, total circuit resistance is 2/0.0025 = 800 , so the external resistance is 800 - 10 = 790 .

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