Discussion
A wooden box of rectangular section 1.25 m wide, 2 m deep, 4 m long floats horizontally in sea water. If the specific gravity of wood is 0.64 and for sea water = 1.025. Then the position of center of buoyancy is given by:
Answer: B
The centre of buoyancy is 0.62 m above the base.
A floating body displaces its own weight of liquid. With the box 2 m deep, the depth immersed \(h\) satisfies:
\[ \begin{aligned} 0.64 \times 2 &= 1.025 \times h \\ h &= \frac{1.28}{1.025} = 1.249\ \text{m} \end{aligned} \]
The centre of buoyancy is the centroid of the displaced volume, which for a rectangular section is half the immersed depth: \(1.249/2 = 0.62\) m above the base. 1.24 m is the immersed depth itself, not its centroid, and 0.25 m and 0.84 m match no step of the working. The width and length cancel out because both sides of the balance use the same plan area.
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