A cubical solid aluminium (bulk modulus = $$-V \frac{dP}{dV} = 70$$ GPa) block has an edge length of 1 m on the surface of the earth. It is kept on the floor of a 5 km deep ocean. Taking the average density of water and the acceleration due to gravity to be 10$$^3$$ kg m$$^{−3}$$ and 10 ms$$^{−2}$$, respectively, the change in the edge length of the block in mm is _____.

Hydrostatic pressure at depth $$h$$ in a liquid of density $$\rho$$ is given by
$$\Delta P=\rho g h$$.

Given
$$\rho = 10^{3}\,\text{kg m}^{-3},\; g = 10\,\text{m s}^{-2},\; h = 5\,\text{km}=5\times 10^{3}\,\text{m}$$,

$$\Delta P = 10^{3}\times 10 \times 5\times 10^{3}=5\times 10^{7}\,\text{Pa}$$.

The bulk modulus relation is
$$B=-V\frac{\Delta P}{\Delta V}\;\Longrightarrow\;\frac{\Delta V}{V}=-\frac{\Delta P}{B}$$.

For a cube of edge $$L$$, $$V=L^{3}$$. A small change $$\Delta L$$ gives
$$\Delta V = 3L^{2}\,\Delta L \;\Rightarrow\; \frac{\Delta V}{V}= \frac{3L^{2}\Delta L}{L^{3}} = 3\frac{\Delta L}{L}$$.

Hence
$$3\frac{\Delta L}{L} = -\frac{\Delta P}{B}\;\Longrightarrow\;\Delta L = -\frac{\Delta P}{3B}\,L$$.

Substituting $$\Delta P = 5\times 10^{7}\,\text{Pa}$$, $$B = 70\,\text{GPa}=70\times 10^{9}\,\text{Pa}$$ and $$L = 1\,\text{m}$$,

$$\Delta L = -\frac{5\times 10^{7}}{3\times 70\times 10^{9}}\;(1)$$
$$= -\frac{5}{210}\times 10^{-2}\,\text{m}$$
$$= -2.38\times 10^{-4}\,\text{m}$$.

Magnitude in millimetres:
$$|\Delta L| = 2.38\times 10^{-4}\,\text{m}\times 10^{3}\,\frac{\text{mm}}{\text{m}} \approx 0.24\,\text{mm}$$.

Therefore, the edge shortens by about $$0.24\,\text{mm}$$.