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

$\begin{array}{l}\text{Here,}\\ {\text{V}}_1=\frac{4}{3}\pi {(2.0\times {10}^{-3})}^3\\ h=3.3\text{m}\\ P_1=P_o+\rho gh\\ \Rightarrow P_1=1.0\times {10}^5+1000\times 9.8\times 3.3\\ \Rightarrow P_1=1.32\times {10}^{5}Pa\\ P_2=1.0\times {10}^{5}Pa\\ \text{Since temperature remains the same, applying Boyle's law we get}\\ P_1{V}_1=P_2{V}_2\\ \Rightarrow V_2=\frac{P_1{V}_1}{P_2}\\ \Rightarrow V_2=\frac{1.32\times {10}^5\times \frac{4}{3}\pi {(2.0\times {10}^{-3})}^3}{1.0\times {10}^5}\\ {\text{Let R}}_2\text{be the new radius. Then,}\\ \frac{4}{3}\pi R_2^3=\frac{1.32\times {10}^5\times \frac{4}{3}\pi {(2.0\times {10}^{-3})}^3}{1.0\times {10}^5}\\ \Rightarrow R_2^3=\frac{1.32\times {10}^5\times {(2.0\times {10}^{-3})}^3}{1.0\times {10}^5}\\ \Rightarrow R_3=\sqrt[3]{\frac{1.32\times {10}^5\times {(2.0\times {10}^{-3})}^3}{1.0\times {10}^5}}\\ \Rightarrow R_3=2.2\times {10}^{-3}m\end{array}$