A cell of emf 1.5 volt given a balance point 30 Cm length of potentiometer wire .For another cell , the balance point , shifts to 50 Cm . Find the value of emf of the cell .

Explanation

`E_(1)=1.5`Volt
`l_(1)=30Cm`
`E_(2)=?`
`l_(2)=50Cm`
`(E_(1))/(E_(2))=(l_(1)/(l_(2)))`
`(1.5)/(E_(2))=(30)/(50)rArrE_(2)=(1.5xx50)/(30)`
`E_(2)=(cancel50)/(cancel30)xx(cancel15^(5))/(cancel10_(2))`
`E_(2)=(5)/(2)`
`E_(2)=2.5`Volt .