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.

Answer

`64%`

Explanation

`COLM = - ((1.6v) - v_(A))/(0 - v) v_(A) = 0.6V` ….(i)
`COLM rArr m_(A)v = m_(A) xx (0.6V) + m_(B) (1.6 v)`
`(m_(A))/(m_(B)) = 4` ….(ii)
`DeltaE_(B) = (1)/(2)m_(B) (1.6 v^(2)) = 0 = 2m_(B) (0.8 v)^(2)`
`K_(A) = (1)/(2) (4m_(B))v^(2) = 2m_(B)v^(2) rArr (DeltaK_(B))/(K_(A)) = 0.64 = 64%`