A rod of uniform cross-section A and length L elongates by δ under an axial force P. Young's modulus E is: (i) Pδ/AL (ii) Aδ/PL (iii) PL/Aδ (iv) PA/Lδ
Try it first: pick an option, then read the explanation.
Correct answer: C. (iii) only
Asked in SSC JE 27/01/2018 (Evening)
Detailed Explanation
Write stress and strain separately, then divide.
\[ \sigma = \frac{P}{A}, \qquad \varepsilon = \frac{\delta}{L}, \qquad E = \frac{\sigma}{\varepsilon} = \frac{PL}{A\delta} \]
Why “(iii) only” Is Correct
Only expression (iii), PL/Aδ, matches — option (c).
Why the Other Options Are Wrong
- A. (i) only: Pδ/AL puts δ in the numerator; units are wrong.
- B. (ii) only: Aδ/PL is the inverse of E times area-squared terms; wrong.
- D. (iv) only: PA/Lδ multiplies by area instead of dividing.
Key Point
E = PL/(Aδ); δ = PL/(AE).
Key Concepts for Students
- Axial stiffness k = AE/L.
- Check units: N·mm/(mm²·mm) = N/mm².
- Larger E or A → smaller elongation.
Practice more: this is question 27 in Stress and Strain MCQs – Part 2. Try the whole set and check every answer instantly.