The length of a wire of a potentiometer is $100 \mathrm{~cm}$, and the e.m.f. of its standard cell is $\mathrm{E}$ volt. It is employed to measure the e.m.f of a battery whose internal resistance is $0.5 \Omega$. If the balance point is obtained at $\mathrm{l}=30$ $\mathrm{cm}$ from the positive end, the e.m.f. of the battery is

• A

  $\frac{30 \mathrm{E}}{100.5}$
• B

  $\frac{30 \mathrm{E}}{(100-0.5)}$
• C

  $\frac{30(\mathrm{E}-0.5 \mathrm{i})}{100}$, where $\mathrm{i}$ is the current in the potentiometer wire
• Correct answer
  D

  $\frac{30 \mathrm{E}}{100}$

Hence, the correct answer is option: 'D'.