Answer to Question #112831 in Physics for amber chloe taylor

Question #112831
A bucket of capacity 10 litres weighs the same as 2 litres of water when
empty. The bucket is inverted and held below the surface of a deep
pond, trapping all the air. Below what depth will the bucket sink of its
own accord? Assume that the bucket, air and pond are all at the same
temperature; that the inverted position is stable; and that the material of
the bucket has negligible volume. You may also assume that one
atmosphere of pressure is equivalent to a 10 metre depth of water.
1
Expert's answer
2020-05-08T16:15:07-0400

The mass of the empty bucket is 2 kg. According to Newton's second law, bucket floats when the force of gravity equals the buoyancy force:


mg=ρgV2,mg=\rho gV_2,


where mm - mass of the bucket, ρ\rho - density of water.

This corresponds to the volume of the air bubble of


V2=mgρg=mρ,V_2=\frac{mg}{\rho g}=\frac{m}{\rho},


When the volume of the air bubble becomes smaller than this value, the bucket sinks.

The initial pressure of the air bubble in the bucket is the atmospheric pressure p1p_1. As the bucket goes deeper, the volume decreases because the water pressure compresses it. Calculate what pressure can compress the air bubble from V1=10 LV_1=10 \text{ L} to V assuming the temperature does not change:


p1V1=p2V2,p2=p1V1V2=p1V1ρm.p_1V_1=p_2V_2,\\ p_2=p_1\frac{V_1}{V_2}=p_1\frac{V_1\rho}{m}.

At what depth will the air bubble experience pressure p2p_2? The answer is


p2=ρgh.ρgh=p1(V1ρm), h=p1ρg(V1ρm)= =10132510009.8(10310002)=5.2 m.p_2=\rho gh.\\ \rho gh=p_1\bigg(\frac{V_1\rho}{m}\bigg),\\ \space\\ h=\frac{p_1}{\rho g}\bigg(\frac{V_1\rho}{m}\bigg)=\\ \space\\ =\frac{101325}{1000\cdot9.8}\bigg(\frac{10^{-3}\cdot1000}{2}\bigg)=5.2\text{ m}.

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