A cell of e.m.f.1.4 volt gives a balance point at 30.0 Cm length of a potentiometer wire for another cell , the balance points shifts to 45.0 C m . Find the e.m.f. of the second cell .

Explanation

`E_(1)=1.4` Volt
`l_(1)=30` Cm
`E_(2)=?`
`l_(2)=45Cm`
`(E_(1))/(E_(2))=(l_(1))/(l_(2))`
`(1.4)/(E_(2))=(30)/(45)`
`E_(2)=(45)/(30)xx1.4rArrE_(2)=(cancel45^(cancel9^(3)))/(cancel30_(cancel10_(5)))xx(cancel14^(7))/(cancel10_(2))`
`E_(2)=(21)/(10)`
`E_(2)=2.1`volt .