The figure shows experimental set up of a meter bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of `10 Omega` is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the `10 Omega` resistance is instead connected in series with resistance 'Y'. Determine the value of the resistance X and Y

Explanation

At the null point at D
`(X)/(Y) = (AD)/(DC)`
`(X)/(Y) = (40)/(100 - 40) rArr (X)/(Y) = (40)/(60)`
`(X)/(Y) = (2)/(3) rArr X = (2)/(3) Y`...(i)
When a resistance of `10 Omega` is connected in series with X
`:.` At the new null point
`(X + 30)/(Y) = (50)/(100 -50)`
`rArr (X + 30)/(Y) = (50)/(50) rArr (X + 30)/(Y) = 1`
`:. X + 30 = Y`..(ii)
Putting `X = (2)/(3)Y` in equation (ii)
`(2)/(3) Y + 30 = Y rArr 30 = Y - (2)/(3) Y`
`30 = (1)/(3) Y rArr Y = 90 Omega`
Putting the value of Y in (i), we get
`X = (2)/(3) xx 90 = 60 Omega`
`:. X = 60 Omega, Y = 90 Omega`
When the `10 Omega` resistance is connected in series with Y then
`(X)/(Y + 30) = (60)/(90 + 30) = (60)/(120) = (60)/(100 + 20)`
Then the null point is shifted 20cm towards right.