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Q.

A potentiometer has a wire of 100cm length and its resistance is 10ohms. It is connected in series with a resistance of 40ohms and a battery of emf 2V and negligible internal resistance. If a source of unknown emf  E connected in the secondary is balanced by 40cm length of potentiometer wire, the value of E is

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a

0.8V

b

0.16V

c

0.4V

d

0.08V

answer is D.

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Detailed Solution

Length L=100cm

Resistance R=10Ω

Series resistance RS=40Ω

Battery of  emf E=2V

Balancing length l=40cm

emf of the cell in secondary circuit,

E1=(iρ)l

E1=(ER+RS)RLl

     =(210+40)10100×40

    =2×10×4050×100

    =850

E1=0.16V

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