A rectangular pontoon is 5 m long, 3 m wide and 1·20 m high. The depth of immersion of the pontoon is 1·0 m in sea water. If the centre of gravity is 0·6 m above the bottom of the pontoon, what is the metacentric height of the same ? (Take the density of sea water as 1000 kg/m³)
- A0·65 mCorrect
- B0·85 m
- C0·75 m
- D0·50 m
Solution
Concept
Metacentric height
Formula
GM = I/V − BG
Step-by-step solution
1. Volume of displaced water V = 5 × 3 × 1 = 15 m³. 2. I (about the longitudinal axis, the smaller value) = L B³/12 = 5 × 27 / 12 = 11.25 m⁴. 3. BM = I / V = 0.75 m. 4. B is 0.5 m above the bottom, G is 0.6 m above the bottom, so BG = 0.1 m (G above B). 5. GM = BM − BG = 0.75 − 0.10 = 0.65 m.
Explanation
The pontoon is stable since GM is positive. The smaller moment of inertia (about the long axis) governs the stability, so L B³/12 is used.
Why the other options are wrong
0.85 m would result from adding BG (BM + 0.1); 0.75 is BM alone; 0.50 does not follow from the geometry.
Shortcut
GM = BM − BG with G above B.
Common mistake
Adding BG instead of subtracting.
Correct answer: A) 0·65 m