A solid shaft (diameter 20 cm, length 400 cm, modulus of rigidity N = 0.8 × 10⁵ N/mm²) under a twisting moment develops a maximum shear stress of 50 N/mm². The angle of twist in radians is:
Try it first: pick an option, then read the explanation.
Correct answer: C. 0.025
Asked in SSC JE 24/01/2018 (Evening)
Detailed Explanation
Cancelled question: the commission cancelled this question because the correct value, 0.025 rad, was printed as 0.0025 in option (c). Option (c) is shown here with the correct value.
Use τ/R = Nθ/L with R = 100 mm and L = 4000 mm.
\[ \theta = \frac{\tau L}{N R} = \frac{50 \times 4000}{0.8\times10^5 \times 100} = \frac{2\times10^5}{8\times10^6} = 0.025\ \text{rad} \approx 1.43^\circ \]
Why “0.025” Is Correct
The angle of twist is 0.025 rad.
Why the Other Options Are Wrong
- A. 0.001: 0.001 rad is 25 times too small.
- B. 0.002: 0.002 rad is too small.
- D. 0.004: 0.004 rad is too small (would need τ = 8 MPa).
Key Point
θ = τL/(GR) — convert all lengths to mm and use the radius, not the diameter.
Key Concepts for Students
- Using D instead of R halves the answer (0.0125).
- Book option printed 0.0025 — cancelled in the revised key.
- No torque value is needed when τmax is given.
Practice more: this is question 15 in Torsion MCQs – Part 1. Try the whole set and check every answer instantly.