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Question

# A potentiometer wire of length 100 cm has a resistance of 10 Ω. It is connected in series with a resistance and a cell of emf 2 V and of negligible internal resistance. A source of emf 10 mV is balanced against a length of 40 cm of the potentiometer wire. What is the value of external resistance?

A
790 Ω
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B
444 Ω
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C
512 Ω
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D
680 Ω
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Solution

## The correct option is A 790 ΩFigure shows the situation described in the question. As the source of emf, E′=10 mV=10×10−3 V is balanced by a length of 40 cm of the potentiometer wire, so we have IRcb=E′=10×10−3 V ⇒(2R+10)Rcb=10×10−3 V ....(1) Given, resistance of 100 cm isRab=10 Ω. So, the resistance of 40 cm of the potentiometer wire, ⇒Rcb=(40100)×10=4 Ω Substituting Rcb in (1), (2R+10)×4=10×10−3 V ⇒R=790 Ω Hence, (A) is the correct answer.

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