Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of 15 ohms 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 larger and R be smaller resistance connected in two gaps of meter bridge
:.`S=((100-l)/(l))R=(100-20)/(20)R=4R...(i)`
When `15 Omega` resistance is added to resistance R, then
`S=((100-40)/(40))(R+15)=(6)/(4)(R+15)`
`implies4R=(6)/(4)(R+15)`
`implies16R=6R+90implies10R-90`
`R=9Omega`