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Question

A potentiometer wire of length 1 m and resistance 10Ω is connected in series with a cell of e.m.f 2 volt with internal resistance 1Ω and a resistance box including a resistance R. If the potential difference between the ends of the wire is 1 mV, then the value of R will be

A
9989Ω
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B
10000Ω
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C
19989Ω
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D
20000Ω
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Solution

The correct option is C 19989Ω
i=V2R1=1×10310=104AE=i[R+r+RL]2=104[R+1+10]20000=R+11R=2000011=19989Ω

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