The ratio of diameters of two wires of the same material is n:1. The length of each wire is 4 m. On applying the same load, the increase in length of thin wire will be

Explanation

When a load is applied to a wire, the elongation is given by the formula ΔL = (F × L) / (A × Y), where F is the force, L is the original length, A is the cross-sectional area, and Y is the Young's modulus of the material. Since the wires are made of the same material and under the same load, Y and F are constant. The cross-sectional area A is proportional to the square of the diameter. Therefore, if the diameter of the thin wire is 1/n times the diameter of the thicker wire, its area will be (1/n2) times smaller, leading to an elongation that is n2 times greater. Thus, the correct answer is n2 times. The other options are incorrect as they do not accurately reflect the relationship between diameter and elongation.