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

Step 1: Identify the given parameters.

The physical properties of the aluminum block and the ocean environment are given.

$$L_0 = 1\text{ m}$$ $$K = 70\text{ GPa} = 7 \times 10^{10}\text{ Pa}$$ $$\rho = 10^{3}\text{ kg m}^{-3}$$ $$g = 10\text{ m s}^{-2}$$ $$h = 5\text{ km} = 5 \times 10^{3}\text{ m}$$

Step 2: Calculate the change in pressure.

The hydrostatic pressure at depth $h$ is determined by the density of water and gravity.

$$\Delta P = \rho g h$$ $$\Delta P = (10^{3})(10)(5 \times 10^{3})$$ $$\Delta P = 5 \times 10^{7}\text{ Pa}$$

Step 3: Calculate the volume strain.

The bulk modulus relates the pressure change to the fractional change in volume.

$$K = -\frac{\Delta P}{\frac{\Delta V}{V}}$$ $$\frac{\Delta V}{V} = -\frac{\Delta P}{K}$$ $$\frac{\Delta V}{V} = -\frac{5 \times 10^{7}}{7 \times 10^{10}}$$ $$\frac{\Delta V}{V} \approx -7.14 \times 10^{-4}$$

Step 4: Relate volume strain to linear strain.

For a cube of volume $V = L^3$, the fractional change in volume is three times the fractional change in edge length.

$$\frac{\Delta V}{V} = 3\frac{\Delta L}{L}$$ $$\frac{\Delta L}{L} = \frac{1}{3}\frac{\Delta V}{V}$$ $$\frac{\Delta L}{L} = \frac{1}{3}(-7.14 \times 10^{-4})$$ $$\frac{\Delta L}{L} \approx -2.38 \times 10^{-4}$$

Step 5: Calculate the change in edge length.

The change in edge length is found by multiplying the linear strain by the initial length.

$$\Delta L = L_0 \left(\frac{\Delta L}{L}\right)$$ $$\Delta L = (1)(-2.38 \times 10^{-4})\text{ m}$$ $$\Delta L = -0.238\text{ mm}$$ $$|\Delta L| \approx 0.238\text{ mm}$$

Answer: The change in the edge length of the block is $0.238\text{ mm}$ (or approximately $0.24\text{ mm}$).