An air bubble of volume 4cm3 is formed 20m under water. What will be its volume when it rises to just below…

Analyze the question: A bubble rises from depth 20 m to just below the surface. The pressure drops, so the volume increases (Boyle's law, at constant temperature).

Concept: Boyle's law: P₁V₁ = P₂V₂

Careful with the data: The question says atmospheric pressure is equivalent to 10 cm of water. That is physically unrealistic, because the real value is about 10 m of water. This is most likely a typo in the question, with 10 m intended.

Solution using the question as printed (10 cm):

• P₁ = 10 cm + 20 m = 0.1 m + 20 m = 20.1 m of water
• P₂ = 0.1 m of water
• V₂ = P₁V₁/P₂ = (20.1 × 4)/0.1 = 804 cm³

This matches none of the options.

Solution using the intended value (10 m of water):

• P₁ = 10 + 20 = 30 m of water
• P₂ = 10 m of water
• V₂ = (30 × 4)/10 = 12 cm³

Answer: The question as printed gives 804 cm³, which is not an option, so the "10 cm" must be a typo for 10 m. With atmospheric pressure equal to 10 m of water, the volume is 12 cm³, which is option E.

Tip: In Boyle's law problems involving depth, total pressure = atmospheric pressure + depth (in the same units). Always check that the units match. Here the pressure ratio is 30/10 = 3, so the volume triples: 4 × 3 = 12 cm³.