Two unknown resistance X and Y are connected to left and right gaps of a meter bridge and the balancing point is obtained at 80 cm from left. When a `10Omega` resisance is connected in parallel to `x`, balance point is 50 cm from left. The values of X and Y respectively are

Options

A. `4 Omega , 9 Omega`
B. `30 Omega , 7.5 Omega`
C. `20 Omega , 6 Omega`
D. `10 Omega , 3 Omega`

Answer

B

Explanation

Let I be the distance of balancing point from left gap, then
` X/Y = l/ (100 - I) = 80/20 = 4`
or X = 4Y
Again in parallel, the net resistances is
` X = (10X)/(10+X)`
So , ` X/Y = 50/(100 -50) =1 or (10X)/(10 +X) = Y`
` or 10X = 10Y + XY`
`or 40 Y = 10Y + 4Y^(2)`
`or y = 7.5 Omega`
Putting Eq. (i) . we get
` X = 30 Omega`