This problem combines two basic strength-of-materials ideas: direct (axial) stress and factor of safety, and axial elongation under load. The rod carries a purely tensile load, so the stress is uniform across the cross-section.

Given: diameter d = 20 mm, load F = 40 kN = 40000 N, yield strength σ_y = 450 MPa, modulus E = 200 GPa = 200000 MPa, length L = 500 mm.

Step 1 — Cross-sectional area:

• A = π·d²/4 = π × (20)²/4 = π × 400/4 = 100π ≈ 314.16 mm²

Step 2 — Working (applied) stress:

• σ = F/A = 40000 / 314.16 ≈ 127.3 MPa

Step 3 — Factor of safety on yield strength:

• FoS = σ_y / σ_applied = 450 / 127.3 ≈ 2.25
• The factor of safety tells us the applied stress is only about 44% of the yield stress, so the rod stays comfortably in the elastic region.

Step 4 — Elongation (from Hooke's law, δ = FL/AE):

• δ = (F × L) / (A × E) = (40000 × 500) / (314.16 × 200000)
• δ = 20,000,000 / 62,831,853 ≈ 0.318 mm

Both computed values — FoS ≈ 2.25 and δ ≈ 0.318 mm — match the choice giving Factor of Safety (Yield) = 2.25, Elongation = 0.318 mm.

Why the other options fail: the pair listing FoS = 3.58 uses an incorrect (too large) safety factor that does not follow from 450/127.3, so both variants carrying 3.58 are ruled out. The option pairing the correct 2.25 with an elongation of 0.5 mm is inconsistent, because substituting the given F, L, A and E into δ = FL/AE yields 0.318 mm, not 0.5 mm — an elongation of 0.5 mm would require a much larger load or a smaller area/modulus.