A potentiometer wire of length 100cm has a resistance of 10Ω. It is connected in series

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

Options

(a) 760Ω
(b) 640Ω
(c) 790Ω
(d) 840Ω

Correct Answer:

790Ω

Explanation:

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  • Let the resistance be r and current be I .According to loop rule :
    E=V+Ir
    2=I(10+r)
    And also
    I*(40/100)=0.01V
    Here you get I=2.5*10^-3ampere
    Put this Vue in eqn 2=I(10+r) and get solution

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