The length of a wire of a potentiometer is 100 cm, and the e.m.f. of its standard cell is E volt. It is employed to measure the e.m.f. 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 e.m.f. of the battery is
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a
30E100
b
30E100.5
c
30E100−0.5
d
30E−0.5i100, where i is the current in the potentiometer
answer is A.
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Detailed Solution
From the principle of potentiometet, V ∝ l⇒ VE = lL ; where V = emf of battery E = emf of standard cell, L = Length of potentiometer wireV= ElL = 30E100