Young's modulus of the material of wires X and Y are in the ratio 4: 1 and the areas of cross sections of the wires X and Y are in the ratio 2:1. If the same amount of load is applied to both the wires, the ratio of elongation produced in the wires X and Y will be: (Assume length of the wires X and Y initially are the same)

Step-by-step solution

The Young's modulus Y is defined as Y = FL A L , where F is the applied force, L is the initial length, A is the cross-sectional area, and L is the elongation. Rearranging for elongation, we get L = FL AY . Given that the force F and initial length L are the same for both wires X and Y, the elongation L is inversely proportional to the product of area A and Young's modulus Y , i.e., L 1 AY . The ratio of elongations is L_X L_Y = A_Y Y_Y A_X Y_X . Given Y_X Y_Y = 4 1 and A_X A_Y = 2 1 , we substitute these values: L