The moment of inertia of a triangular section (height h, base b) about its base is:
Try it first: pick an option, then read the explanation.
Correct answer: C. bh³/12
Asked in SSC JE 24/01/2018 (Evening)
Detailed Explanation
For a triangle, the centroidal moment of inertia about an axis parallel to the base is bh³/36. The centroid lies h/3 above the base, so the parallel axis theorem moves us to the base.
\[ I_{base} = I_G + A d^2 = \frac{bh^3}{36} + \frac{bh}{2}\left(\frac{h}{3}\right)^2 = \frac{bh^3}{36} + \frac{bh^3}{18} = \frac{bh^3}{12} \]
Why “bh³/12” Is Correct
The result is bh³/12, option (c). Note the cube on the dimension measured perpendicular to the axis (h).
Why the Other Options Are Wrong
- A. bh²/12: bh²/12 has dimension L³, not L⁴.
- B. b²h/12: b²h/12 also has dimension L³.
- D. b³h/12: b³h/12 cubes the wrong dimension; it would be about an axis perpendicular to the base.
Key Point
Triangle: I_G = bh³/36; I_base = bh³/12; I_apex = bh³/4.
Key Concepts for Students
- Parallel axis theorem: I = I_G + A·d².
- Rectangle: bd³/12 at centroid, bd³/3 at base.
- Cube the dimension perpendicular to the axis.
Practice more: this is question 11 in Moment MCQs – Part 1. Try the whole set and check every answer instantly.