A wooden block of rectangular section 1·25 m wide, 2 m deep, 4 m long floats horizontally in sea water. The specific gravity of wood is 0·64 and water weighs 1000 kg(f)/m³. Under this situation, the position of centre of buoyancy of wood is at :
- A1·28 m above the base
- B0·64 m above the baseCorrect
- C0·32 m above the base
- D0·96 m above the base
Solution
Concept
Depth of immersion and centre of buoyancy
Formula
Depth of immersion = S × depth; B = half of the draught above the base
Step-by-step solution
1. Depth of immersion = 0.64 × 2 = 1.28 m (with 2 m vertical). 2. Centre of buoyancy is at the centroid of the displaced volume: 1.28 / 2 = 0.64 m above the base.
Explanation
The body floats with the 2 m depth vertical. The submerged part is a rectangular block 1.28 m deep, whose centroid is at half of that height from the base.
Why the other options are wrong
1.28 m is the draught, 0.32 m is half of 0.64, and 0.96 m is not related to the draught.
Shortcut
B = draught / 2 (for a prismatic body).
Common mistake
Giving the draught instead of half the draught.
Correct answer: B) 0·64 m above the base