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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

a
3
b
6
c
9
d
12

detailed solution

Correct option is C

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

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