Two bars of different materials have the same size and carry the same tensile force. If their unit elongations are in the ratio 4 : 7, the ratio of their moduli of elasticity is: (1) 4 : 7 (2) 4 : 10 (3) 16 : 49
Try it first: pick an option, then read the explanation.
Correct answer: D. None of these
Asked in SSC JE 22/01/2018 (Morning)
Detailed Explanation
Same force and same area means the same stress in both bars. Then E = stress/strain depends only on the strain.
\[ \frac{E_1}{E_2} = \frac{\sigma/\varepsilon_1}{\sigma/\varepsilon_2} = \frac{\varepsilon_2}{\varepsilon_1} = \frac{7}{4} \]
Why “None of these” Is Correct
The ratio is 7 : 4 (the stiffer bar stretches less). None of the given ratios matches, so the answer is (d).
Why the Other Options Are Wrong
- A. 1 only: 4 : 7 is the strain ratio; E is inversely proportional to strain.
- B. 2 only: 4 : 10 has no basis.
- C. 3 only: 16 : 49 squares the ratio, which is wrong.
Key Point
Same stress → E inversely proportional to strain.
Key Concepts for Students
- Same strain (compound bar) → stress proportional to E.
- Stiffer material = smaller elongation for the same stress.
- Read whether the ratio asked is direct or inverse.
Practice more: this is question 40 in Stress and Strain MCQs – Part 3. Try the whole set and check every answer instantly.