A cube of side 4 cm is just completely immersed in a liquid A. When it is put in a liquid B, it floats with 2 cm outside the liquid. Calculate the ratio of densities of the two liquids.

Explanation

To solve the problem, we need to analyze the situation in two different liquids: Liquid A and Liquid B. Let's break it down step by step.

### Step 1: Understand the situation in Liquid A
When the cube is completely immersed in Liquid A, the buoyant force acting on it is equal to its weight.

- **Weight of the cube (W)**: The weight of the cube can be expressed as:
\[ W = \text{Density of the cube} \times \text{Volume of the cube} \times g \]
Since the cube is made of a uniform material, we can denote its density as \(\rho_c\) and its volume \(V\) as:
\[ V = \text{side}^3 = 4 \, \text{cm} \times 4 \, \text{cm} \times 4 \, \text{cm} = 64 \, \text{cm}^3 \]
Thus, the weight of the cube is:
\[ W = \rho_c \times 64 \, \text{cm}^3 \times g \]

- **Buoyant force in Liquid A (F_bA)**: The buoyant force is equal to the weight of the liquid displaced, which is:
\[ F_bA = \rho_A \times V \times g = \rho_A \times 64 \, \text{cm}^3 \times g \]

Setting the weight equal to the buoyant force gives us:
\[ \rho_c \times 64 \, g = \rho_A \times 64 \, g \]
From this, we can simplify to:
\[ \rho_c = \rho_A \]

### Step 2: Understand the situation in Liquid B
When the cube is placed in Liquid B, it floats with 2 cm of its height above the liquid. This means that 2 cm of the cube is above the surface, and the remaining part is submerged.

- **Volume submerged in Liquid B (V')**: The height submerged is:
\[ \text{Height submerged} = \text{Total height} - \text{Height above liquid} = 4 \, \text{cm} - 2 \, \text{cm} = 2 \, \text{cm} \]
The volume submerged is:
\[ V' = \text{side}^2 \times \text{height submerged} = 4 \, \text{cm} \times 4 \, \text{cm} \times 2 \, \text{cm} = 32 \, \text{cm}^3 \]

- **Buoyant force in Liquid B (F_bB)**: The buoyant force is now:
\[ F_bB = \rho_B \times V' \times g = \rho_B \times 32 \, \text{cm}^3 \times g \]

Setting the weight equal to the buoyant force gives us:
\[ \rho_c \times 64 \, g = \rho_B \times 32 \, g \]
From this, we can simplify to:
\[ \rho_c \times 64 = \rho_B \times 32 \]

### Step 3: Relate the densities of the two liquids
Now we can express the relationship between the densities of the two liquids:
\[ \rho_B = 2 \rho_c \]
Since we found earlier that \(\rho_c = \rho_A\), we can substitute:
\[ \rho_B = 2 \rho_A \]

### Step 4: Find the ratio of densities
To find the ratio of the densities of the two liquids, we have:
\[ \frac{\rho_A}{\rho_B} = \frac{\rho_A}{2 \rho_A} = \frac{1}{2} \]

### Final Answer
The ratio of the densities of the two liquids A and B is:
\[ \frac{\rho_A}{\rho_B} = \frac{1}{2} \]