NEET-XII-Physics
24: Kinetic Theory of Gases
- #25An 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.Ans : `` \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}``
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