A cubical box is to be built from iron sheets 1 mm thick. Determine...

Let the external edge of the cubical box be $L$ and the thickness of the iron sheet be $t = 1 \text{ mm} = 0.1 \text{ cm}$.

Since the thickness $t$ is very small compared to the edge length $L$, the volume of the iron used to build the closed cubical box (which has 6 faces) can be approximated as the total surface area multiplied by the thickness:

$V_{\text{iron}} \approx 6 L^2 t$

The mass of the iron box is:

$m = \rho_{\text{iron}} V_{\text{iron}} = \rho_{\text{iron}} (6 L^2 t)$

For the cube to just float (not sink), the maximum buoyant force must be equal to the weight of the box. The maximum buoyant force is provided when the cube is completely submerged:

$F_B = \rho_{\text{water}} V_{\text{ext}} g = \rho_{\text{water}} L^3 g$

Equating the weight of the box to the maximum buoyant force:

$m g = F_B$

$\rho_{\text{iron}} (6 L^2 t) g = \rho_{\text{water}} L^3 g$

$6 \rho_{\text{iron}} t = \rho_{\text{water}} L$

Solving for the minimum external edge $L$:

$L = \dfrac{6 \rho_{\text{iron}} t}{\rho_{\text{water}}}$

Substitute the given values $\rho_{\text{iron}} = 8000 \text{ kg m}^{-3}$, $\rho_{\text{water}} = 1000 \text{ kg m}^{-3}$, and $t = 0.1 \text{ cm}$:

$L = \dfrac{6 \times 8000 \times 0.1}{1000}$

$L = 6 \times 8 \times 0.1 = 4.8 \text{ cm}$

Answer: $4.8 \text{ cm}$