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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.
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Solution
Here,
\[ V_1 = \frac{4}{3}\pi {(2.0 \times {10}^{-3 })}^3 \]
h = 3.3 m
P1 = Po + 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
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