A cube of water with side 10.0 cm floats vertically in water with 4.5 cm its length submerged. Calculate the…

Analyze the question: The question says "cube of water", but it is clearly meant to be a cube of wood, since we are asked for the density of the wood. The cube has side 10.0 cm and floats with 4.5 cm of its height submerged in water.

Concept: A floating body obeys the law of flotation: weight of the body = weight of the water displaced (upthrust).

• Weight of cube = ρ_wood × V_cube × g
• Upthrust = ρ_water × V_submerged × g

Solution:

Cross-sectional area A = 10 × 10 = 100 cm²

• V_cube = 100 × 10 = 1000 cm³
• V_submerged = 100 × 4.5 = 450 cm³

Equating weight and upthrust:

ρ_wood × 1000 = 1 × 450

ρ_wood = 450 / 1000 = 0.45 g/cm³

Equivalently, ρ_wood = (submerged depth / total height) × ρ_water = (4.5/10) × 1 = 0.45 g/cm³.

Note on the "D = 0.090 g/cm³" in the question: This value is not needed for the calculation and appears to be a leftover from the original question (possibly relating to another part). The result from the flotation principle is 0.

Answer: B. 0.450 g/cm³

Tip: For any floating object, the fraction submerged equals the ratio of the object's density to the liquid's density. Here, 4.5/10 = 0.45 of the cube is under water, so the wood's density is 0.45 times that of water.