The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the emf of a battery whose internal resistance is 0.5 Ω. If the balance point is obtained at l = 30 cm from the positive end, the emf of the battery is
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a
30E100
b
30E100.5
c
30E(100−0.5)
d
30(E−0.5i)100
answer is A.
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Detailed Solution
Using the principle of potentiometer, V ∞ l. SoVE=lLor V=lLE=30100E=30E100