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 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

(1) $\frac{30E}{100}$

(2) $\frac{30E}{100.5}$

(3) $\frac{30E}{\left(100-0.5\right)}$

(4) $\frac{30\left(E-0.5i\right)}{100}$, where i is the current in the potentiometer

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