A cell of e.m.f. 1 V gives a balance point at 40 cm length of a potentioeter wire. For another cell, the balance point shifts to 60 cm. Find the e.m.f. of the second cell.

Explanation

`E_(1)= 1.0` volt `l_(1)= 40cm`
`E_(2)`= ?
`l_(2)=60cm`
We know `(E_(1))/(E_(2))= (l_(1))/(l_(2))= (40)/(60)= (2)/(3)`
`rArr (1)/(E_(2))= (2)/(3)`
`rArr E_(2)= (3)/(2) = 1.5V`