An ice cube with an edge of 4 cm is placed into an empty...

Let the initial edge length of the ice cube be $a_0 = 4 \text{ cm}$.

Let the edge length of the ice cube at the given instant be $a$.

The inner cross-sectional area of the cylindrical glass is $A = \pi r^2 = \pi \times (3)^2 = 9\pi \text{ cm}^2$.

The volume of ice that has melted is $a_0^3 - a^3$.

The mass of the melted ice is $\rho_i (a_0^3 - a^3)$, where $\rho_i$ is the density of ice.

The volume of water formed from the melted ice is $V_w = \dfrac{\rho_i}{\rho_w} (a_0^3 - a^3)$, where $\rho_w$ is the density of water.

At the instant the ice cube just loses contact with the bottom of the glass, the buoyant force exactly balances its weight. Let $h$ be the depth of the water. Since the ice cube is resting on the bottom, its submerged volume is $a^2 h$.

Equating the buoyant force and the weight of the ice cube:

$\rho_w g (a^2 h) = \rho_i g a^3 \Rightarrow h = \dfrac{\rho_i}{\rho_w} a$

The volume of water in the glass can also be written as the total cylindrical volume up to height $h$ minus the submerged volume of the ice cube:

$V_w = A h - a^2 h = h(A - a^2) = \dfrac{\rho_i}{\rho_w} a (A - a^2)$

Equating the two expressions for the volume of water $V_w$:

$\dfrac{\rho_i}{\rho_w} (a_0^3 - a^3) = \dfrac{\rho_i}{\rho_w} a (A - a^2)$

$a_0^3 - a^3 = a A - a^3$

$a A = a_0^3$

$a = \dfrac{a_0^3}{A} = \dfrac{4^3}{9\pi} = \dfrac{64}{9\pi}$

$a \approx \dfrac{64}{28.274} \approx 2.26 \text{ cm}$