Question

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.

Answer 1

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:

$$mg=\rho gV_2,$$

where $m$ - mass of the bucket, $\rho$ - density of water.

This corresponds to the volume of the air bubble of

$$V_2=\frac{mg}{\rho g}=\frac{m}{\rho},$$

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 $p_1$. As the bucket goes deeper, the volume decreases because the water pressure compresses it. Calculate what pressure can compress the air bubble from $V_1=10 \text{ L}$ to V assuming the temperature does not change:

$$p_1V_1=p_2V_2,\\ p_2=p_1\frac{V_1}{V_2}=p_1\frac{V_1\rho}{m}.$$

At what depth will the air bubble experience pressure $p_2$? The answer is

$$p_2=\rho gh.\\ \rho gh=p_1\bigg(\frac{V_1\rho}{m}\bigg),\\ \space\\ h=\frac{p_1}{\rho g}\bigg(\frac{V_1\rho}{m}\bigg)=\\ \space\\ =\frac{101325}{1000\cdot9.8}\bigg(\frac{10^{-3}\cdot1000}{2}\bigg)=5.2\text{ m}.$$