A cell of emf 1.45 volt gives a balance print at 45 Cm length of a potentiometer wire For another cell , the balance point shifts to 90 Cm . Find the value of emf of second cell .

Explanation

`E_(1)=1.45`Volt
`l_(1)=45`Cm
`E_(2)=`?
`l_(2)=90Cm`
`(E_(1))/(E_(2))=(l_(1))/(l_(2))`
`(1.45)/(E_(2))=(45)/(90)`
`E_(2)=(90)/(45)xx1.45rArrE_(2)=(overset(1)cancel90)/(cancel45_(5))xx(overset(29)cancel145)/(100)`
`E_(2)=(29)/(10)`
`E_(2)=2.9`Volt .