A load of 1960 N is raised at the end of a steel wire. The minimum diameter of the wire so that the stress does not exceed 100 N/mm² is:
Try it first: pick an option, then read the explanation.
Correct answer: C. 5.0 mm
Asked in SSC JE 04/03/2017 (Morning)
Detailed Explanation
The area must be large enough that load ÷ area does not exceed the permissible stress.
\[ A = \frac{P}{\sigma} = \frac{1960}{100} = 19.6\ \text{mm}^2 \]
\[ d = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4 \times 19.6}{3.1416}} = \sqrt{24.96} \approx 5.0\ \text{mm} \]
Why “5.0 mm” Is Correct
The minimum diameter is 5.0 mm.
Why the Other Options Are Wrong
- A. 4.0 mm: 4.0 mm gives A = 12.6 mm² and stress ≈ 156 N/mm².
- B. 4.5 mm: 4.5 mm gives A = 15.9 mm² and stress ≈ 123 N/mm².
- D. 5.5 mm: 5.5 mm is safe but not the minimum.
Key Point
Design for stress: A = P/σ_allow, then d = √(4A/π).
Key Concepts for Students
- 1960 N is the weight of a 200 kg mass.
- Round up to the next available size.
- Dynamic lifting would need an impact factor.
Practice more: this is question 44 in Stress and Strain MCQs – Part 3. Try the whole set and check every answer instantly.