TGPSC HMWSSB Manager 2015 Paper 2 · Question 111
Question 111 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.
Two circular shafts of same length, weight and material are compared for strength. The first one is a solid shaft and the other is a hollow shaft of outer to inner diameter ratio as 2. The ratio of the torque carrying capacity of the hollow shaft to solid shaft considering the shear stress criterion alone is ______.
- 2.5/√3
- 5√3/6
- √3
- √3/2
Show answer
Answer: (2) 5√3/6
Equal weight and length means equal areas: D² − (D/2)² = d², so D = 2d/√3. T_hollow/T_solid = [(D⁴ − (D/2)⁴)/D]/d³ = (15/16)(2/√3)³ = 5/(2√3) = 5√3/6 ≈ 1.44. Note that option (1), 2.5/√3, is the same number written differently; the key gives (2).