An ideal battery of 4V and resistance R are connected in series in the primary circuit of a potentiometer of length 1m and resistance 5Ω. The value of R, to given a potential difference of 5mVacross 10cm of potentiometer wire is:
A
490Ω
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B
480Ω
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C
395Ω
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D
495Ω
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Solution
The correct option is C395Ω Let current flowing in the wire is i. ∴i=(4R+5)A If resistance of 10m length of wire is x then x=0.5Ω=5×0.11Ω ∴ΔV=P.d on wire =i.x 5×10−3=(4R+5)(0.5) ∴4R+5=10−2 ∴R=395Ω