What happens if the galvanometer and cell are interchanged at the balance point of the bridge ? Would the galvanometer show any current ?

Explanation

After interchanging x and y when equilibrium condition is obtained again,
`(y)/(l) = (x)/(100 - l)`
(Where l = balancing length = distance of null point from end A (positive terminal of battery ))
`therefore (12.5)/(l) = (8.16)/(100 - l)`
`therefore 1250 - 12.5 l = 8.16 l`
`therefore ` 1250 = 20.66 l
`therefore l = (1250)/(20.66) = 60.5 ` cm