WAEC 2012 Physics · Question 12

A block of wood of density 0.6 gcm ^ -3 , weighing 3.06N in air, floats freely in a liquid of density 0.9 gcm…

Question

A block of wood of density 0.6 gcm $^{-3}$ , weighing 3.06N in air, floats freely in a liquid of density 0.9 gcm $^{-3}$ . Calculate volume of the portion immersed (g = 10ms $^{-2}$ )

A.510cm3
B.340cm3
C.254cm3
D.170cm3

Answer: B. 340cm3

Explanation written with AI and checked against the answer key

Analyze the question: A floating block displaces liquid equal in weight to its own weight (law of flotation). We need the volume of the block that is under the liquid. Concept: For a floating body, Weight of block = Upthrust = weight of liquid displaced. Solution: Weight of block = 3.06 N, so its mass = 3.06 / 10 = 0.306 kg = 306 g. Upthrust = weight of liquid displaced: V_immersed × ρ_liquid × g = 3.06 N Convert the liquid density: 0.9 g/cm³ = 900 kg/m³ V = 3.06 / (900 × 10) = 3.06 / 9000 = 3.4 × 10⁻⁴ m³ = 340 cm³ Check using grams: V = mass / density of liquid = 306 g / 0.9 g/cm³ = 340 cm³ ✓ (The total volume of the block is 306 / 0.6 = 510 cm³. This is option A, which is the trap. The question asks for the immersed portion only. Immersed fraction = 0.6/0.9 = 2/3, and 2/3 × 510 = 340 cm³ ✓) Answer: The volume immersed is 340 cm³, which is option B. Tip: For floating objects, immersed volume = mass of object / density of liquid. Always check whether the question asks for the total volume or the immersed volume, because examiners often include the total volume as a distractor.

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