Question

A gold doubloon 6.1 cm in diameter and 2.0 mm thick is dropped over the side of a pirate sheep. When it comes to rest on the ocean floor at a depth of 770 m, how much has its volume changed?​

Answer 1

Assume the water pressure compresses the coin only in vertical direction along the axis of the cylinder.

The pressure:

$$p=\rho gh.$$

The force that compresses the coin:

$$F=pA.$$

The stress in the material of the coin caused by the applied force:

$$\sigma=\frac FA=p.$$

The strain:

$$\epsilon=\frac{\Delta h}h=\frac p E,\\\space\\ \Delta h=h\frac pE=\frac{\rho gh^2}{E}.$$

The volume change is

$$\Delta V=\Delta hA,\\\space\\ \Delta V=\frac{\rho \pi gh^2d^2}{4E}=1.5·10^{-15}\text{ m}^3.$$

Note: we took sea water density as 1030 kg/m³, the young's modulus of gold is 79 GPa.