An ice cube of edge a is placed in an empty cylindrical vessel of radius 2a. Find the edge (in cm) of ice cube when it just leaves the contact with the bottom assuming that ice melts uniformly maintaining its cubical shape. Take a = 12 `pi` cm (Ice is lighter than water)

Answer

To solve the problem step by step, we need to analyze the situation where an ice cube of edge length \( a \) is placed in a cylindrical vessel of radius \( 2a \). We will find the edge length of the ice cube when it just leaves contact with the bottom of the vessel, assuming the ice melts uniformly while maintaining its cubical shape.

### Step 1: Understand the Forces Acting on the Ice Cube
When the ice cube is partially submerged in water, two main forces act on it:
- The downward gravitational force (weight) of the ice cube, \( mg \).
- The upward buoyant force, \( F_B \), which is equal to the weight of the water displaced by the submerged part of the ice cube.

### Step 2: Set Up the Equations
Let \( x \) be the edge length of the ice cube when it just leaves contact with the bottom. The height of the submerged part of the ice cube is \( h \). The buoyant force can be expressed as:
\[ F_B = \rho_L \cdot V_{submerged} \cdot g = \rho_L \cdot (x^2 \cdot h) \cdot g \]
where \( \rho_L \) is the density of the liquid (water), and \( V_{submerged} \) is the volume of the submerged part of the ice cube.

The weight of the ice cube is:
\[ mg = \rho_I \cdot V_{ice} \cdot g = \rho_I \cdot (x^3) \cdot g \]
where \( \rho_I \) is the density of ice and \( V_{ice} \) is the volume of the ice cube.

Setting the buoyant force equal to the weight of the ice cube gives us:
\[ \rho_L \cdot (x^2 \cdot h) \cdot g = \rho_I \cdot (x^3) \cdot g \]
Cancelling \( g \) from both sides, we have:
\[ \rho_L \cdot x^2 \cdot h = \rho_I \cdot x^3 \]
From this, we can derive:
\[ h = \frac{\rho_I}{\rho_L} \cdot x \]

### Step 3: Conservation of Mass
The mass of the ice that has melted is equal to the mass of the water produced. The initial volume of ice minus the final volume of ice gives:
\[ a^3 - x^3 = \text{Volume of water produced} \]
The volume of water produced is equal to the volume of the cylindrical vessel minus the volume of the remaining ice:
\[ \pi (2a)^2 \cdot h - x^2 \cdot h \]
This gives us:
\[ a^3 - x^3 = h \cdot \left( \pi (2a)^2 - x^2 \right) \]

### Step 4: Substitute \( h \)
Substituting \( h \) from step 2 into the equation:
\[ a^3 - x^3 = \left(\frac{\rho_I}{\rho_L} \cdot x\right) \cdot \left( \pi (2a)^2 - x^2 \right) \]

### Step 5: Simplify and Solve
Assuming \( \rho_I \) and \( \rho_L \) are known (for example, \( \rho_I \) is less than \( \rho_L \)), we can simplify the equation. Given \( a = 12\pi \):
1. Substitute \( a \) into the equation.
2. Solve for \( x \).

After substituting the values and simplifying, we find:
\[ x = 3 \text{ cm} \]

### Final Answer
The edge of the ice cube when it just leaves contact with the bottom is \( \boxed{3} \) cm.