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)= 1.08` volt `l_(1)= 40cm`
`E_(2)`= ? `l_(2)= 50cm`
We know `(E_(1))/(E_(2))= (l_(1))/(l_(2))`
`rArr (1.08)/(E_(2))= (40)/(50)`
`rArr E_(2)= (5)/(4) xx 1.08 = 5 xx 0.27`
`rArr E_(2)= 1.35` volt