A circular shaft ABC fixed at A has diameter D over AB (half its length) and diameter D/2 over BC (the other half), with torque T at the free end C. If the rotation of B relative to A is 0.1 rad, what is the rotation of C relative to B?
Try it first: pick an option, then read the explanation.
Correct answer: C. 1.6 rad
Asked in SSC JE 03/03/2017 (Morning)
Detailed Explanation
Both segments carry the same torque T and have the same length l/2. The twist of each is TL/(GJ), with J ∝ d⁴.
\[ \frac{\theta_{BC}}{\theta_{AB}} = \frac{J_{AB}}{J_{BC}} = \left(\frac{D}{D/2}\right)^4 = 16 \;\Rightarrow\; \theta_{BC} = 16\times0.1 = 1.6\ \text{rad} \]
Why “1.6 rad” Is Correct
C rotates 1.6 rad relative to B.
Why the Other Options Are Wrong
- A. 0.4 rad: 0.4 rad assumes θ ∝ 1/d² — wrong power.
- B. 0.8 rad: 0.8 rad assumes θ ∝ 1/d³ (that is the stress ratio, not the twist ratio).
- D. 3.2 rad: 3.2 rad doubles the correct value.
Key Point
Same T and L: θ ∝ 1/d⁴. Halving d multiplies twist by 16 (stress by 8).
Key Concepts for Students
- Total twist at C = 0.1 + 1.6 = 1.7 rad.
- Shafts in series carry the same torque.
- Shafts in parallel share the same twist.
Practice more: this is question 26 in Torsion MCQs – Part 2. Try the whole set and check every answer instantly.