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Question

A potentiometer has a wire of 100 cm length and its resistance is 10 ohms. It is connected in series with a resistance of 40 ohms and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E connected in the secondary is balanced by 40 cm length of potentiometer wire, the value of E is:


A
0.2 V
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B
0.4 V
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C
0.08 V
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D
0.16 V
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Solution

The correct option is A 0.2 V
The potentiometer has 10Ω resistance at 100cm length . at 40cm length , it'll have resistance R1=10100×40=4Ω Drives cell has emf E1 connected to a secondary circuit having a cell of emf E2=2V and series resistance R2=40Ω At balance point,current through both primary and secondary circuits would be same.

I=E1R1=E2R2 (follows from ohm's law V=1R )

E14=E2×240E2=240×4=0.2V

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