Melting Mechanics

Let the side length of the melting ice cube when it just starts to float be $x$ . Then its mass is $0.9x^2$ and this is the weight of the water displaced by the ice. Since water has a specific gravity of 1, the volume displaced is also $0.9x^3$ . As the ice cube floats with a face horizontal and that it has a uniform cross-sectional area of $x^2$ , the side length submerged in water is therefore $0.9x$ . This is also the depth of the water since the ice cube is just floating. Then we have:

$\begin{aligned} \text{Volume of water}+ \text{Volume of the submerged part of ice} & = \pi r^2 (0.9x) \\ 0.9(1-x^3) + 0.9x^3 & = 0.9 \pi x \\ 1 & = \pi x \\ \implies x & = \frac 1\pi \approx \boxed{0.318} \end{aligned}$

The ice will float in the container when the buoyant force from the liquid water is equal to the weight of the ice.

$F_B = mg$

$\rho_{liq}V_Dg = \rho_{ice}V_{ice}g$ ...Eq 1

Where,

$\rho$ is the density of the medium (ice or liquid water).

$g$ is the acceleration due to gravity.

$V_{ice} = a^3$ is the volume of ice at this instant.

$V_{D}$ is the volume of displaced liquid water.

$V_{D} = a^2 h$ ...Eq 2

Where,

$a^2$ is the crossectional area of the cube.

$h$ is the height of the liquid water in the container.

Substitue Eq2 $\rightarrow$ Eq1

Next. the height of the liquid can be written in terms of the net crossectional area of the container and the volume of the liquid:

$\displaystyle h= \frac{V_{liq}}{\pi r^2 - a^2}$ ...Eq 3

Substitue Eq 3 $\rightarrow$ Eq1

Mass is conserved in the phase change, and it follows that the change in mass of the ice is equivalent to the mass of the liquid.

$\Delta m_{ice} = \rho_{liq}V_{liq}$

$\rho_{ice}( {a_o}^3 - a^3 ) = \rho_{liq}V_{liq}$

or

$\displaystyle V_{liq} = \frac{\rho_{ice} ({a_o}^3 - a^3 )}{\rho_{liq}}$ ...Eq 4

Substitue Eq 4 $\rightarrow$ Eq1

After the last substitution is made, make the appropriate cancellations and we are left with,

$\displaystyle a = \frac{{a_o}^3 - a^3}{\pi r^2 - a^2}$ , which is valid for $0 < a_o \leq \sqrt{2}r$ (the cube must have some volume and fit inside the container)

$\displaystyle a = \frac{{a_o}^3}{\pi r^2}$

Notice the relationship is independent of the material densities between its solid and liquid state. However, the implicit assumption in Eq1 for a body to float is its acceleration must be $\geq 0$ , so $\rho_{liquid} \geq \rho_{solid}$ for this to hold.

Finally, substitute $a_o = \SI{1}{\meter}$ and $r = \SI{1}{\meter}$

$\displaystyle a = \frac{1}{\pi} \approx \SI{0.318}{\meter}$