In a metre bridge, the gaps are closed by two resistance `P` and `Q` and the balance point is obtained at `40 cm`. When `Q` is shunted by a resistance of `10 Omega`, the balance point shifts to `50 cm`. The values of `P` and `Q` are.
.

Options

A. `10/3 Omega, 5Omega`
B. `20 Omega,30Omega`
C. `10 Omega,15Omega`
D. `5Omega,(15)/2Omega`

Answer

A

Explanation

For meter bridge to be balanced
`P/Q = (40)/60 =2/3`
`:. P = 2/3Q`
When Q is shunted, i.e. a resistance of `10 Omega` is connected in parallel across Q, the net resistance becomes
`(10Q)/(10+Q)`
Now, the balance point shifts to 50 cm, i.e.
`P/(((10Q)/(10+Q)))=50/60`
`:. 2/3 = 10/(10 +Q)`
So we get `Q = 5Omega and P= 10/3Omega`