The length of a potentiometer wire is 200 cm and the e.m.f. of the battery connected to it is E. It is used to measure the e.m.f. of a cell whose internal resistance is `0.5 Omega` . If the balance point is obtained at l=40 cm from the positive end, then the e.m.f. of the cell is

Options

A. `(40 E)/(200)`
B. `(40 E)/(200+0.5)`
C. `(40(E-0.5))/(200)`
D. `(40 E)/(200-0.5)`

Answer

A

Explanation

`(V)/(E )=(l)/(L) therefore V=(El )/(L)=(40)/(200) E`