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

Options

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

Answer

C

Explanation

`(P)/(Q)=(20)/(80)=(1)/(4)" "….(i)`
`(P+15)/(Q)=(40)/(60)=(2)/(3)" "...(ii)`
Divide (i) by (ii), we get
`therefore (P)/(P+15)=(1//4)/(2//3)=(3)/(8)`
or `8P=3P+45or5P=45orP=9Omega`