An equilateral triangular plate of side a is immersed in water with one side at free water surface as in the given figure. The centre of pressure is below the water surface at a depth of :
Try it first: pick an option, then read the explanation.
Asked in SSC JE 11-12-2020 (Evening)
Detailed Explanation
When a triangular plane area is submerged vertically with its base resting along the free liquid surface and its vertex pointing downward, the depth to its centre of pressure can be found from the area moment of inertia about the centroid.
1. Geometry of an equilateral triangle of side \(a\):
- Base \(b = a\)
- Altitude (height) \(h = a \sin 60^\circ = a \frac{\sqrt{3}}{2}\)
- Area \(A = \frac{1}{2} b h = \frac{1}{2} a \left(\frac{a\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4} a^2\)
2. Centroid depth (\(\bar{h}\)) for a triangle with base at the free surface:
\[ \bar{h} = \frac{h}{3} = \frac{1}{3} \left(\frac{a\sqrt{3}}{2}\right) = \frac{a\sqrt{3}}{6} = \frac{a}{2\sqrt{3}} \]
3. Centroidal moment of inertia of a triangle:
\[ I_G = \frac{b h^3}{36} \]
4. Calculate the depth of the centre of pressure (\(h^*\)):
\[ h^* = \bar{h} + \frac{I_G}{A \bar{h}} = \frac{h}{3} + \frac{\frac{b h^3}{36}}{\left(\frac{1}{2} b h\right) \left(\frac{h}{3}\right)} = \frac{h}{3} + \frac{h^3 / 36}{h^2 / 6} = \frac{h}{3} + \frac{h}{6} = \frac{h}{2} \]
5. Substitute altitude \(h = \frac{a\sqrt{3}}{2}\):
\[ h^* = \frac{1}{2} h = \frac{1}{2} \left(\frac{a\sqrt{3}}{2}\right) = \frac{\sqrt{3} a}{4} \]
Why “√3a / 4” Is Correct
For any vertical triangular plate with its base at the water surface, the centre of pressure always lies at exactly half the total altitude (\(h^* = h/2\)). Substituting the altitude of an equilateral triangle \(h = a\sqrt{3}/2\) gives \(h^* = \frac{\sqrt{3}a}{4}\).
Why the Other Options Are Wrong
- A. a / (2√3): a / (2√3) is the depth to the centroid (h̄ = h/3 = a√3/6 = a/(2√3)), not the centre of pressure.
- B. a / 3: a / 3 confuses the side dimension with the height fraction.
- D. a / √3: a / √3 is twice the centroid depth (2h/3), which applies when the apex is at the surface, not the base.
Key Point
Triangle with base at free surface: h̄ = h/3 and h* = h/2; for equilateral triangle h* = √3a/4.
Key Concepts for Students
- If the apex is at the surface and base below: h̄ = 2h/3 and h* = 3h/4.
- Centre of pressure for base-at-surface triangle bisects the total vertical height (h* = 0.50 h).
- Moment of inertia about base is I_base = bh³/12; about centroid is I_G = bh³/36.
Practice more: this is question 11 in Hydrostatic Pressure MCQs – Part 1. Try the whole set and check every answer instantly.