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

Options

A. 3
B. 6
C. 9
D. 12

Answer

C

Explanation

`P/Q=20/(100-20) or Q=4P`….i
`:. `PltQ`
Now,` (P+15)/Q=40/(100-40)`
`or (P+15)/Q=2/3`
Solving these two equations we get
`P=9Omega`