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, 5 Omega`
B. `20 Omega, 30 Omega`
C. `10 Omega, 15 Omega`
D. ` 5 Omega, 15/2 Omega`

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 `(10 Q)/(10 + Q)`
Now, the balance point shifts to `50 cm`, i.e.
`p/(((10Q)/(10 + Q))) = 1`
`:. 2/3 = (10)/(10 + Q)`
`:. 20 + 2Q = 30`
`:. Q = 5 Omega`
and `P = (10)/(3) Omega`.