A steel rod that is 1 metre long with a square cross-section is subjected to a direct tensile load of 8 tonnes and undergoes an extension of 1 mm. Given Esteel = 2 × 10^6 kg/cm^2, determine the side length of the square cross-section (express the final answer in centimetres).

Introduction / Context:
This problem applies axial deformation relations to a prismatic steel bar under a tensile load. It tests your ability to combine stress–strain definitions with unit consistency to compute an unknown cross-sectional dimension for a given elongation. The bar is square, so once the required area is found, the side length follows directly from area = side^2.

Given Data / Assumptions:

• Length L = 1 metre = 100 cm.
• Axial load P = 8 tonnes = 8000 kgf.
• Extension δ = 1 mm = 0.1 cm.
• Young's modulus for steel, E = 2 × 10^6 kg/cm^2.
• Square cross-section with side a (cm), area A = a^2.
• Linear elasticity, small strain, uniform stress state.

Concept / Approach:
Axial stress σ = P / A. Axial strain ε = δ / L. Hooke's law for linear elastic behavior gives σ = E * ε. Combining these relations eliminates σ and ε and allows solving for A. Finally, take the square root to obtain the side a of the square cross-section.

Step-by-Step Solution:
1) Compute strain: ε = δ / L = 0.1 / 100 = 0.001.2) Compute stress from Hooke's law: σ = E * ε = (2 × 10^6) * 0.001 = 2,000 kg/cm^2.3) Relate stress and area: σ = P / A ⇒ A = P / σ = 8000 / 2000 = 4 cm^2.4) For a square section, a = sqrt(A) = sqrt(4) = 2 cm.

Verification / Alternative check:
Dimensional check: P in kgf and E in kg/cm^2 are consistent with area in cm^2. Recomputing σ = 8000 / 4 = 2000 kg/cm^2 and ε = σ / E = 2000 / (2 × 10^6) = 0.001 gives δ = εL = 0.001 × 100 cm = 0.1 cm = 1 mm, verifying the result.

Why Other Options Are Wrong:

• 1 cm, 1.5 cm: These give areas (1 or 2.25 cm^2) too small, producing stresses above the calculated 2000 kg/cm^2 and extensions larger than 1 mm.
• 2.5 cm: Area 6.25 cm^2 reduces stress too much and would predict an extension less than 1 mm.

Common Pitfalls:

• Mistaking millimetres for centimetres when computing strain, which changes the answer by a factor of 10.
• Forgetting that the area for a square is a^2 (not 4a, etc.).

Final Answer:
2 cm.