With resistances P and Q in the left and right gaps of a metre bridge respectively, the null point divides the wire in the ratio 1 : 2 . If P and Q are increased by `20 Omega` each, the null point divides the wire in the ratio 3 : 4, then the resistances P and Q are given as :

Options

A. `P=10 Omega, Q=10 Omega`
B. `P=10 Omega, Q=20 Omega`
C. `P=40 Omega, Q=10 Omega`
D. `P=20 Omega, Q=40 Omega`

Answer

B

Explanation

`(P)/(Q)=(1)/(2)=(l_(1))/(l_(2)) therefore Q= 2P`
`(P+20)/(Q+20)=(3)/(4) therefore (P+20)/(2P+20)=(3)/(4)`
`therefore 6 P+60= 4P +80 therefore P= 10 Omega, Q= 20 Omega`