Circuit diagram of metre bridge is shown in figure. The null point is found at a distance of 40cm from A. it now a resistance of `12Omega` is connected in parallel with S, the null point occurs at 64cm.

Explanation

Fro m the first balance point `(R)/(S)=(40)/(100-40)=(40)/(60)`
If `12Omega` is connected in parallel with S then `S_(eq)=(12S)/(12+S)`
so from new balance conditions `(R)/(S_(eq))=(64)/(36)`
`rArr (R(S+12))/(12S)=(64)/(36)rArr((40)/(36))((S+12)/(12))=(64)/(36)rArrS+12=32rArrS=20Omega`
So `R=((2)/(3))(20)=(40)/(3)Omega`