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

Options

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

Answer

C

Explanation

Let S be the large and R be the smaller resistance.FromFormula for meter bridge
`s=((100-l)/(l))R=(100-20)/(20)R=4R`
Again,`S=((100-l)/(l))(R+15)`
`=(100-40)/40(R+15)`
`(3)/(2)(R+15)`
` therefore4R=(3)/(2)(R+15)`
`=(100-40)/40(R+15)`
`(3)/(2)(R+15)`
`:. 4R=(3)/(2)(R+15)`