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 the thin wire will be (n > 1):

Explanation

Let's denote the diameter of the thicker wire as D and the diameter of the thinner wire as d. Since the ratio of diameters is n:1, we can write: D/d = n Now, the cross-sectional area of a wire is proportional to the square of its diameter (assuming the wires have the same shape): Area ∝ (diameter)^2 So, the cross-sectional area of the thicker wire and the thinner wire is related as follows: A D / A d = (D2) / (d2) = (n2) When the same load is applied to both wires, they experience the same force. The tensile stress is defined as the applied force (F) divided by the cross-sectional area (A) of the wire: Stress = F / A Since the applied force is the same for both wires, we can write Stress D / Stress d = (F / A D) / (F / A d) = A d / A D = 1 / (n2) Now, the strain of a wire is directly proportional to the stress applied to it. So, the strain in the thinner wire and the strain in the thicker wire is related as follows: strain d / strain D = 1 / (n2) Now, the question asks for the increase in length of the thin wire compared to the thick wire. The increase in length is directly proportional to the strain: increase in length d / increase in length D = strain d / strain D = 1 / (n2) Since we know that the increase in length of the thicker wire is the same for both wires and is equal to 4m, we can write: increase in length d = (1 / (n2)) x increase in length D = (1 / (n2)) x 4m = 4m / n2 Therefore, the increase in length of the thin wire is n2 times smaller than the increase in length of the thick wire.