Answer:
ΔL = 3.12 x 10⁻⁴ m = 0.312 mm
Explanation:
First, we will find the stress applied by the mass on the rod:
[tex]\sigma = \frac{F}{A}[/tex]
where,
σ = stress = ?
F = Force applied by the mass = weight = mg = (10 kg)(9.81 m/s²) = 98.1 N
A = cross-sectional area = πr² = π(1 mm)² = π(0.001 m)² = 3.14 x 10⁻⁶ m²
Therefore,
[tex]\sigma = \frac{98.1\ N}{3.14\ x\ 10^{-6}\ m^2}\\\\[/tex]
σ = 3.12 x 10⁷ Pa
Now, we will find the value of strain:
[tex]E = \frac{\sigma}{\epsilon}\\\\\epsilon = \frac{\sigma}{E}\\\\\epsilon = \frac{3.12\ x\ 10^7\ Pa}{200\ x\ 10^9\ Pa}[/tex]
∈ = 1.56 x 10⁻⁴
Now, the change in length can be given by the formula of strain:
[tex]\epsilon = \frac{\Delta L}{L}\\\\\Delta L = (\epsilon)(L)[/tex]
where,
L = Original Length = 2 m
ΔL = Elongation = ?
Therefore,
ΔL = (1.56 x 10⁻⁴)(2 m)
ΔL = 3.12 x 10⁻⁴ m = 0.312 mm
