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- Subject
- Strength of Materialsmechanical-engineering-mcqs › strength-of-materials-mechanical-engineering-mcqs
- Published
- 25 Jan 2019
- Last updated
- 28 May 2026
Explanation
Using the formula for distance under constant acceleration starting from rest, s = 0.5 × a × t², where a = 5 m/s² and t = 5 s, the distance covered is 0.5 × 5 × 25 = 62.5 meters.
More Strength of Materials MCQs
Practice related questions from the same subject.
- 1.Up to which point does Hooke's law remain valid?
- 2.What is the unit of Young's modulus?
- 3.When a body is subjected to equal and opposite forces that attempt to stretch it, the resulting stress is known as______________________?
- 4.What is the modulus of rigidity defined as the ratio of?
- 5.How does the ultimate tensile strength of mild steel compare to its ultimate compressive strength?
- 6.When a thin mild steel wire is subjected to gradually increasing loads in equal steps until it fractures, how does the extension change with the applied loads?
- 7.What happens to the Young's modulus of a wire if its radius is increased to twice its original size under the same load?
- 8.How is the tensile strength of a material calculated during a tensile test?
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