Two unknown resistances are connected in two gaps of a meter-bridge. The null point is obtained at 40 cm from left end. A `30Omega` resistance is connected in series with the smaller of the two resistances, the null point shifts by 20 cm to the right end. The value of smaller resistance in `Omega` is

Options

A. 12
B. 24
C. 36
D. 48

Answer

B

Explanation

In the first case, `(X)/(R )=(l_(X))/(l_(R ))=(40)/(60)=(2)/(3)`
In the second case, X is replaced by `(X+30) Omega` and the null point is at 40+20= 60 cm
`therefore (X+30)/(R )=(60)/(40)=(3)/(2)`
`therefore 2 X+60=3 R`
`therefore 2 X+60 = 3 R`
`therefore R= (2X+60)/(3)`
Using (2) in (1), we get
`((X)/(2X+60))/(3)=(2)/(3)`
`therefore (3X)/(2X+60)=(2)/(3)`
`therefore 4X+120=9 X`
`therefore 5 X= 120 therefore X= 24 Omega`