Correct Answer - Option 4 : 1.75

**Concept:**

Stress in axial loading is given by;

\(σ=\frac{F}{A}\)

where σ = Normal stress, F = Load/Weight, A = Area of cross-section

\(A=\frac{\pi d^2}{4}\)

where d = Diameter

**Calculation:**

**Given:**

F = 2200 N, d = 4 cm = 40 mm

\(A=\frac{\pi\times (40)^2}{4}=1256.63\ mm^2\)

\(σ=\frac{2200}{1256.63}\)

**σ = 1.75 N/mm**^{2}

**Hence the stress induced in the metal rod will be 1.75 N/mm**^{2}.