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 Omega`. If the balance point is obtained at I = 30 cm from the positive end, the e.m.f. of the battery is .
where i is the current in the potentiometer wire.

Options

A. `(30E)/100.5`
B. `(30E)/(100-0.5)`
C. `(30(E-0.5I))/100` where I is the current in the potentiometer wire
D. `(30E)/100`

Explanation

`V prop 1 therefore V/E=l/L`
where,=I balance point distance
L= length of potentiometer wire or `V=i/L E or V=(30 times E)/100=30/100E`