The length of the wire of a potentiometer is 100 cm and the emf of a standard cell connected to it is E volt. While measuring the emf of a battery having internal resistance `0.5Omega` the null point is obtained at a length of 30 cm . Determine the emf of the battery .

Explanation

When a null ponit is obtained , no current flows through the battery . So there is no internal potential drop . In this case internal resistance of the battery has no influence . So if E' be the emf of the battery , then
`(E)/(100)=(E')/(30)" or, "(E)/(E')=(10)/(3)" or,"E'=0.3E`