A cells of e.m.f. 1.08 volt , gives a balance point of 40 Cm length of a potentioneter wire .For another cell the balance point shifts to 50 Cm . Find the value of e.m.f. of the second cell .

Explanation

`E_(1)=1.08` Volt
`l_(1)=40Cm`
`E_(2)=` ?
`l_(2)=50Cm`
`(E_(1))/(E_(2))=(l_(1))/(l_(2))`
`(1.08)/(E_(2))=(40)/(50)`
`E_(2)=(5)/(4)xx1.08rArrE_(2)=(cancel5^(1))/(cancel4_(1))xx(cancel108^(27))/(cancel100_(20))`
`E_(2)=(27)/(20)`
`E_(2)=1.35` Volt