An air bubble of radius 2.0 mm forms at the bottom of a 3.3 m deep river. Determine the radius of the bubble as it reaches the surface. Assume atmospheric pressure = 1.0 10^5 Pa and density of water = 1000 kg m ⁻³ .

Step-by-step solution

Let the pressure at the surface be P₂ = 1.0 10^5 Pa and the pressure at the bottom be P₁ . Using the relation for hydrostatic pressure, P₁ = P₂ + g h . Substituting the given values and taking g = 10 m s ⁻² : P₁ = 1.0 10^5 + 1000 10 3.3 = 1.33 10^5 Pa Assuming the temperature remains constant, Boyle's law is applicable: P₁ V₁ = P₂ V₂ P₁ ( 4 3 r₁^3 ) = P₂ ( 4 3 r₂^3 ) r₂^3 = P₁ P₂ r₁^3 r₂^3 = 1.33 10^5 1.0 10^5 (2.0)^3 r₂^3 = 1.33 8 = 10.64 mm ^3 Since (2.2)^3 = 10.648 , taking the cube root gives: r₂ 2.2 mm