Question

An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure = 1.0 × 10⁵ Pa and density of water = 1000 kg m⁻³.

Answer 1

Here,
\[ V_1 = \frac{4}{3}\pi {(2.0 \times {10}^{-3 })}^3 \]
h = 3.3 m
P₁ = P_o + ρgh
⇒ P₁ = 1.0  × 10⁵ + 1000  × 9.8 × 3.3
⇒ P₁ = 1.32 × 10⁵ Pa
    P₂ = 1.0 × 10⁵ Pa
Since temperature remains the same, applying Boyle's law we get
P₁ V₁ = P₂ V₂ 
⇒ V₂ = \[ \frac {P_1 V_1}{P_2} \]
⇒ V₂ = \[ \frac{1.32 × {10}^5 × \frac{4}{3}\pi {(2.0 ×{10}^{-3} )}^3}{1.0 × {10}^5} \]
Let R₂ be the new radius. Then, 
\[ \frac{4}{3}\pi R_2^3 \] = \[ \frac{1.32 × {10}^5 × \frac{4}{3}\pi {(2.0 × {10}^{-3} )}^3}{1.0 ×{10}^5} \]
⇒ \[ R_2^3 = \frac{1.32 × {10}^5 × {(2.0 × {10}^{-3 })}^3}{1.0 × {10}^5} \]
⇒ \[ R_3 = \sqrt[3]{\frac{1.32 × {10}^5× {(2.0 × {10}^{-3} )}^3}{1.0 × {10}^5}} \]
⇒ R₃ = 2.2 × 10^-3 m