With resistances P and Q in the left and the right gap respectively of a metre bridge, the null point divides the wire in the ratio `3:4` . When P and Q are increased by `20Omega` each, the null point divides the wire in the ratio `5:6` . The values of P and Q are

Options

A. `30 Omega, 40 Omega`
B. `20 Omega, 40Omega`
C. `30 Omega, 80Omega`
D. `20 Omega, 20Omega`

Answer

A

Explanation

For balanced bridge, `(X)/(R)=(l_(x))/(l_(r))`
`:.(P)/(Q)=(3)/(4)`
`:.3Q-4P=0` .... (i)
Now, `(P+20)/(Q+20)=5/6`
`:.5Q-6P=20` ...(ii)
From equation (i) and (ii), we have,
`P=30 Omega ` and `Q=40Omega`