A 1 m long steel rod of cross-section 20 mm × 20 mm carries a tensile force of 40 kN. Find the elongation if E = 200 GPa.
Try it first: pick an option, then read the explanation.
Correct answer: B. 0.5 mm
Asked in SSC JE 23/09/2019 (Morning)
Detailed Explanation
Elongation of a bar under axial load follows directly from Hooke's law.
\(A = 400\ \text{mm}^2,\ L = 1000\ \text{mm},\ P = 40000\ \text{N},\ E = 2 \times 10^5\ \text{N/mm}^2\)
\[ \delta = \frac{PL}{AE} = \frac{40000 \times 1000}{400 \times 2\times10^5} = 0.5\ \text{mm} \]
Why “0.5 mm” Is Correct
The rod stretches by 0.5 mm. (Stress = 100 N/mm², strain = 0.0005.)
Why the Other Options Are Wrong
- A. 0.005 mm: Off by a factor of 100 — check the unit of L.
- C. 5 mm: Ten times too large.
- D. 50 mm: 100 times too large.
Key Point
δ = PL/AE (keep N and mm throughout).
Key Concepts for Students
- 1 GPa = 1000 N/mm².
- Stress = 40000/400 = 100 N/mm².
- Strain = σ/E = 0.0005.
Practice more: this is question 16 in Stress and Strain MCQs – Part 2. Try the whole set and check every answer instantly.