1. Given data

• Edge length of aluminium cube: $L = 1\,\text{m}$
• Volume of cube: $V = L^3 = 1^3 = 1\,\text{m}^3$
• Bulk modulus of aluminium: $B = -V\frac{dP}{dV} = 70\,\text{GPa} = 70 \times 10^9\,\text{Pa}$
• Depth of ocean: $h = 5\,\text{km} = 5000\,\text{m}$
• Density of water: $\rho = 10^3\,\text{kg m}^{-3}$
• Acceleration due to gravity: $g = 10\,\text{m s}^{-2}$

2. Increase in pressure at depth $h$

Hydrostatic pressure increase is $\Delta P = \rho g h$ So, $\Delta P = 10^3 \times 10 \times 5000 = 5 \times 10^7\,\text{Pa}$

3. Use bulk modulus relation

For small changes, $B = -\frac{\Delta P}{\Delta V/V}$ Hence, $\frac{\Delta V}{V} = -\frac{\Delta P}{B}$ Substitute values: $\frac{\Delta V}{V} = -\frac{5 \times 10^7}{70 \times 10^9} = -\frac{1}{1400}$

Thus, the fractional decrease in volume is $\left|\frac{\Delta V}{V}\right| = \frac{1}{1400}$

4. Relate volume change to edge change

If the cube edge changes from $L$ to $L+\Delta L$, then for small changes, $\frac{\Delta V}{V} = 3\frac{\Delta L}{L}$ Therefore, $\frac{\Delta L}{L} = \frac{1}{3}\frac{\Delta V}{V} = -\frac{1}{3}\cdot\frac{1}{1400} = -\frac{1}{4200}$

Since $L = 1\,\text{m}$, $\Delta L = -\frac{1}{4200}\,\text{m}$

5. Convert to mm

$\left|\Delta L\right| = \frac{1}{4200}\times 1000\,\text{mm} = \frac{1000}{4200}\,\text{mm}$ $\left|\Delta L\right| \approx 0.238\,\text{mm}$

So the decrease in edge length is approximately $0.24\,\text{mm}$

6. Final answer

The change in edge length is $0.24$ mm (decrease).