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`, it is clear that `R_(1)` is smaller resistance.
`therefore R_(2) = 4R_(1)`
When `15 Omega` is connected in series with `R_(1)`, then equivalent resistance is `R_(1) + 15`, So
`therefore (R_(1) + 15)/R_(2) = 40/60 = 2/3`
`therefore (R_(1) + 15)/(4R_(1)) = 2/3`
`therefore R_(1) = 9 Omega`