A potentiometer having a wire `10 m` long stretched on it is connected to a battery having a steady voltage. A length of potentiometer wire is increase by `100 cm`, find the new position of null point.

Answer

` 825 cm`

Explanation

Let `x` be the emf of battery used in potentiometer wire `= 10 m = 1000 cm `
emf of Lenlache cell `= (epsilon)/(1000) xx 750 V`
New length of potentimeter wire
`= 1000 + 100 = 1100 cm `
Let `l_(1)` be the balancing length of potentiometer wire be the emf of lachacles cell .Then
`(epsilon)/(1000) xx 750 = (epsilon)/(1100) l_(1)`
or `l_(1) = (1100 xx 750)/(1000) = 825 cm`