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. `8 Omega`
B. `9 Omega`
C. `10 Omega`
D. `12 Omega`

Answer

B

Explanation

` (R_(1))/(R_(2))= (20)/(80)=(1)/(4)`
`:. R_(2) = 4R_1`
`:. (R_(1) + 15)/(R_(2)=(40)/(60)=2/3`
`:. (R_(1) + 15)/(4 R_(1)=( 2)/(3)`
`:. R_(1) = 9 Omega`.