Two wires, A and B, are connected in series and pulled by a constant force. Wire A has length L , radius r , and Young's modulus Y . Wire B has length 2L , radius 3r , and Young's modulus Y 3 . The ratio of the elastic potential energy stored in wire A to that stored in wire B is:

Step-by-step solution

Since the wires are connected in series, the stretching force F is the same for both wires. The elastic potential energy stored in a stretched wire is given by: U = 1 2 F L Therefore, the ratio of the energies stored is equal to the ratio of their elongations: U_A U_B = L_A L_B Using Hooke's law, L = F L A Y = F L r^2 Y . For wire A: L_A = F L r^2 Y For wire B: L_B = F (2L) (3r)^2 ( Y 3 ) = 2 F L (9r^2) ( Y 3 ) = 2 F L 3 r^2 Y Taking the ratio: U_A U_B = L_A L_B = F L r^2 Y 2 F L 3 r^2 Y = 3 2 The ratio is 3:2 . An