Question

# AB is a potentiometer wire of length 100 cm and its resistance is 10 Ω. It is connected in series with a resistance R=40 Ω and a battery of emf 2 V and negligible internal resistance. If a source of unknown emf E is balanced by 40 cm length of the potentiometer wire, then the value of E is

A
0.8 V
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B
1.6 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 D 0.16 VSuppose source of unknown emf is not connected, Then, I=210+40 A=250 A=125 A Resistance per unit length of wire=10100 Ω/m Now, voltage drop across 40 cm length of wire =125×10100×40=425 V=0.16 V We know that in balanced condition E=voltage drop across 40 cm length of wire=0.16 V

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