A light rod of length 1 m is suspended from ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of material X and is of cross-section 0.1 cm ^2 and the other of material Y of cross-section 0.3 cm ^2 . A weight is hung from the wire at a point to produce equal strain in the wires. The ratio of Young's moduli of wires A to B is 3:1. The location of the point fr

Step-by-step solution

Let the length of the rod be L = 1 m . Let the weight be suspended at a distance x from the wire of material X. For rotational equilibrium of the rod, the torques about the point of suspension must balance: T_X x = T_Y (1 - x) The strain in a wire is given by = T AY . Since the strains in both wires are equal: T_X A_X Y_X = T_Y A_Y Y_Y T_X T_Y = A_X Y_X A_Y Y_Y Given A_X = 0.1 cm ^2 , A_Y = 0.3 cm ^2 , and Y_X Y_Y = 3 1 . Substituting the values: T_X T_Y = ( 0.1 0.3 ) ( 3 1 ) = 1 This gives T_X = T_Y . Substituting