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

Let S be larger and R be smaller resistances connected in two gaps of meter bridge.∴     S = 100−ll R=   100−2020 R=4R            ….(i)When 15  Ω resistance is added to resistance R, thenS=   100−4040 R+15=64R+15                  ….(ii)From equations (i) and (ii),  R =9 Ω