Four wires of same material and of dimensions as under as stretched by a load of same magnitude separately. Which one of them will be elongated maximum?

1. The elongation (ΔL) of a wire under a load (F) is given by the formula ΔL = (F * L) / (A * Y), where L is the original length, A is the cross-sectional area, and Y is Young's modulus (constant as it's the same material). Area A = π * (d/2)^2 = (π/4) * d^2, where d is the diameter. 2. So, ΔL is proportional to L / d^2. We need to find the maximum value of L / d^2 for the given options. 3. (A) L=1 m, d=2 mm. L/d^2 = 1 / (2^2) = 1/4 = 0.25 (ignoring units/constants). 4. (B) L=2 m, d=2 mm. L/d^2 = 2 / (2^2) = 2/4 = 0.5. 5. (C) L=3 m, d=1.5 mm. L/d^2 = 3 / (1.5^2) = 3 / 2.25 ≈ 1.33. 6. (D) L=1 m, d=1 mm. L/d^2 = 1 / (1^2) = 1. 7. The maximum value of L/d^2 occurs for option (C). Therefore, the wire of 3 m length and 1.5 mm diameter will be elongated maximum.