Correct Answer - Option 1 : 80 MPa and 0.1%

**Concept:**

Tensile stress = \({\sigma _T} = \frac{P}{A} \)

where, P = Load, A = Cross-sectional area

\(\% Strain = \frac{{Δ L}}{L} \times 100 \)

ΔL = Change in length, L = Original length

**Calculation:**

**Given: **

B = 25 mm, D = 50 mm, L = 2000 mm, P = 100 kN, ΔL = 2 mm

Tensile stress = \({\sigma _T} = \frac{P}{A} = \frac{{100 \times {{10}^3}}}{{25 \times 50}} = 80 \ MPa\)

\(\% Strain = \frac{{Δ L}}{L} \times 100 = \frac{2}{{2000}} \times 100 = 0.1\% \)

Hence, The tensile stress and % strain in the bar is 80 MPa and 0.1 %