Answer:
option b) 119.35 mm
Explanation:
Given:
Internal diameter of the hollow steel tube, d = 100 mm = 0.1 m
Tensile load, P = 400 KN = 400000 N
Limiting stress, σ = 120 MN/m² = 120 × 10⁶ N/m²
let the outer diameter of the tube be 'D'
thus,
Cross-sectional area of the tube, A = [tex]\frac{\pi}{4}(D^2-d^2)[/tex]
also
stress = Load/Area
therefore,
[tex]120\times10^6=\frac{400000}{\frac{\pi}{4}(D^2-0.1^2)}[/tex]
or
[tex]\frac{\pi}{4}(D^2-0.1^2)=\frac{400000}{120\times10^6}[/tex]
or
D² - 0.1² = 4.244 × 10⁻³
or
D² = 0.014244 m
or
D = 0.119348 m
or
D = 119.348 ≈ 119.35 mm
hence. option b) is the correct answer