Two wires A and B are made of the same material and are subjected to the same stretching force. The ratio of the length of wire B to the length of wire A is 2:1 , and the ratio of the radius of wire A to the radius of wire B is 2:1 . If the elastic potential energy stored in wire A is U_A and that in wire B is U_B , then the value of U_B U_A is _____.

Step-by-step solution

The elastic potential energy stored in a wire is given by U = 1 2 F L . Since the extension is L = FL AY = FL r^2 Y , we can write the energy as: U = F^2 L 2 r^2 Y Given that both wires are made of the same material ( Y is constant) and are subjected to the same stretching force ( F is constant), the stored energy is proportional to L r^2 . Therefore, the ratio of the stored energies is: U_B U_A = ( L_B L_A ) ( r_A r_B )^2 Given L_B L_A = 2 and r_A r_B = 2 , we have: U_B U_A = 2 (2)^2 = 8 Answer: 8