A prismatic bar of cross-section 30 mm × 30 mm and length 1 m carries an axial tensile force of 90 kN. If the strain is 0.0005, Young's modulus is:
Try it first: pick an option, then read the explanation.
Correct answer: B. 200 × 10³ N/mm²
Asked in SSC JE 06/11/2024 (9 am–11 am)
Detailed Explanation
Find the direct stress from load and area, then divide by the strain (Hooke's law). The length is not needed because the strain is given.
\[ A = 30 \times 30 = 900\ \text{mm}^2,\qquad \sigma = \frac{90 \times 10^3}{900} = 100\ \text{N/mm}^2 \]
\[ E = \frac{\sigma}{\varepsilon} = \frac{100}{0.0005} = 200 \times 10^3\ \text{N/mm}^2 \]
Why “200 × 10³ N/mm²” Is Correct
E = 200 × 10³ N/mm² (200 GPa) — option (b), the usual value for steel.
Why the Other Options Are Wrong
- A. 167 × 10⁴ N/mm²: 167 × 10⁴ N/mm² is about ten times too large.
- C. 230 × 10³ N/mm²: 230 × 10³ N/mm² does not follow from the data.
- D. 182 × 10² N/mm²: 182 × 10² N/mm² is about ten times too small.
Key Point
E = σ/ε = P/(Aε).
Key Concepts for Students
- Elongation here = εL = 0.0005 × 1000 = 0.5 mm.
- E of steel ≈ 2 × 10⁵ N/mm²; of concrete ≈ 2.5–3.5 × 10⁴ N/mm².
- 1 N/mm² = 1 MPa.
Practice more: this is question 15 in Bending Moment and Shear Force MCQs – Part 1. Try the whole set and check every answer instantly.