An air bubble of radius 4.5 cm initially at a depth of 12 m below the water surface rises to the surface. If…

Analyze the question: A bubble rises from 12 m depth to the surface. Pressure drops, so the volume grows (Boyle's law, assuming constant temperature). We need the new radius.

Concept: Boyle's law: P₁V₁ = P₂V₂, and for a sphere V = (4/3)πr³, so

P₁r₁³ = P₂r₂³

Solution:

Pressures in metres of water:

• At depth: P₁ = 10.34 + 12 = 22.34 m
• Near the surface: P₂ = 10.34 m

r₂³ = (P₁/P₂) × r₁³ r₂³ = (22.34 / 10.34) × (4.5)³

• (4.5)³ = 91.125
• 22.34 / 10.34 = 2.1605
• r₂³ = 2.1605 × 91.125 = 196.88

r₂ = ∛196.88 ≈ 5.82 cm (since 5.8³ = 195.1 and 5.82³ ≈ 197.1)

Answer: The radius just before reaching the surface is about 5.82 cm, which is option D.

Tip: In Boyle's law questions involving bubbles, always add the atmospheric pressure to the water pressure at depth. Also remember that volume depends on r³, so you must take a cube root at the end. Forgetting this gives a wrong answer like 9.72 cm (the ratio times r without the cube root).