Using the concepts of Young's modulus and Hooke's Law, the elongation of a wire with doubled diameter and length subject to the same force is 0.005 meters or 5.0 mm.
To solve this problem, we will use the concept of Young's modulus, which is a measure of the stiffness of an elastic material and is defined by the formula:
Young's modulus (E) = (F/A) / (ΔL/L0)
Here, F is the applied force, A is the cross-sectional area, ΔL is the change in length (elongation), and L0 is the original length of the wire. The cross-sectional area depends on the diameter (d) of the wire, with the area being π(d/2)^2.
According to Hooke's Law, the elongation is proportional to the force applied and inversely proportional to both the original cross-sectional area and the original length. If the diameter and length of the wire are doubled (d' = 2d, L'0 = 2L0), the new cross-sectional area becomes A' = π(d'/2)^2 = 4A. The new length is twice the original, L'0 = 2L0.
From the formula of Young's modulus, we know that:
(F/A) / (ΔL/L0) = (F/A') / (ΔL'/L'0)
Given that the force is the same in both cases, we can simplify to:
(1/A) / (ΔL/L0) = (1/A') / (ΔL'/L'0)
So: (1/A) / (ΔL) = (1/4A) / (ΔL'/2)
Which simplifies to: ΔL' = ΔL/2 = 0.01m/2 = 0.005m
Therefore, the elongation of the wire with double diameter and length when the same force is applied will be 0.005 meters or 5.0 mm.