Two wires A and B of equal lengths are connected in left and right gap respectively of a metre bridge, null point is obtained at 40 cm from left end. Diameters of the wires A and B are in the ratio 3:1 respectively, the ratio of specific resistance of A to that of B is

Step-by-step solution

Using the principle of a metre bridge, the ratio of resistances in the left and right gaps is given by R_A R_B = l 100 - l Substituting l = 40 cm, we get R_A R_B = 40 100 - 40 = 40 60 = 2 3 The resistance of a wire is given by R = L A = 4L d^2 , where is the specific resistance, L is the length, and d is the diameter. Since the lengths of wires A and B are equal, the ratio of their resistances is R_A R_B = ( _A _B ) ( d_B d_A )^2 Given that the ratio of diameters is d_A d_B = 3 1 , we have d_B d_A = 1 3 . Substitut