Two resistances are connected in the two gaps of a meter bridge. The balance point is `20 cm` from the zero end. When a resistance `15 Omega` is connected in series with the smaller of two resistance, the null point+ shifts to `40 cm`. The smaller of the two resistance has the value.

Options

A. `5 Omega`
B. `7 Omega`
C. `9 Omega`
D. `12 Omega`

Answer

C

Explanation

Let X and R be the smaller and the larger resistances.
`l_(1)=20 cm and l_(2)=80 cm`
then `(X )/(R )=(l_(1))/(l_(2))=(20)/(80)=(1)/(4) therefore R= 4 X`
In the second case, the smaller resistance becomes X+15
then `(X+15)/(4X)=(40)/(60)=(2)/(3)`
`therefore 3 (X+15) =8 X therefore 3X+45=8X`
`therefore 5x=45 therefore X=9 Omega`