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