The length of a wire of a potentiometer is 100 cm, and the emf of its stand and cell is E volt. It is employed to measure the emf of a battery whose internal resistance is $0.5 \Omega$. If the balance point is obtained at $l = 30\; cm$ from the positive end, the emf of the battery is :

Question

( A ) $\frac{30E}{100-0.5}$
( B ) $\frac{30E}{100.5}$
( C ) $\frac{30(E -0.5i)}{100}$, where i is the current in the potentiometer wire
( D ) $\frac{30E}{100}$