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Question

A potentiometer wire of length 10 m and resistance 30 ohm is connected in series with a battery of emf 2.5 V, internal resistance 5 ohm and external resistance R. If the fall of potential along the potentiometer wire is 50 mV/m the value of R in Ω is

A
115
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B
115.0
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C
115.00
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Solution


Potential gradient Vl=50×103 V/m

The P.D across the wire is,

VAB=vl×10=50×103×10=0.5 V

I=VABRw=0.530=160 A

E=I[Rw+R+r]

2.5=160[30+R+5]

2.5×60=R+35

R=15035

R=115 Ω

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