Question

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

Answer 1

Area of the flat surface of the coin:

$$A=\pi d^2/4.$$

Initial volume:

$$V=Ah=\frac{\pi d^2h}{4}.$$

Pressure at 770 m:

$$p=\rho gh.$$

Force on the flat surface of the coin (assuming it is parallel to the ground):

$$F=pA=\frac14\rho \pi gh^2d^2.$$

Stress caused by the force:

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

Strain in the coin:

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

the volume has shrunk by

$$\Delta V=\Delta hA,\\\space\\ \Delta V=\frac{\rho \pi gh^2d^2}{4E}=1.49·10^{-15}\text{ m}^3.$$

The density of water is 1030 kg, Young's modulus for gold is 79 GPa.