Answer:
Let's analyze the problem step-by-step to determine the ratio \(\frac{\rho_B}{\rho_A}\).
### Step 1: Understanding the Situation for Cube A
- Cube A floats in water with a fraction \(\eta\) of its volume immersed.
- Let the volume of each cube be \(V\).
- The density of water is \(\rho_w\).
For cube A floating, the buoyant force equals the weight of cube A:
\[ \eta V \rho_w g = V \rho_A g \]
Here, \(g\) is the acceleration due to gravity. Simplifying, we get:
\[ \eta \rho_w = \rho_A \]
\[ \rho_A = \eta \rho_w \]
### Step 2: Understanding the Situation with Cube B Placed on Cube A
- Cube B is placed on top of cube A, and cube A becomes entirely immersed, while cube B stays entirely above water.
- The combined system of cubes A and B is floating.
The total weight of the system is balanced by the buoyant force:
\[ (V \rho_A + V \rho_B) g = V \rho_w g \]
Here, \(V \rho_w\) represents the buoyant force when the combined volume \(2V\) is immersed in water. Simplifying, we get:
\[ \rho_A + \rho_B = \rho_w \]
### Step 3: Substituting \(\rho_A\) and Solving for \(\rho_B\)
From Step 1, we know:
\[ \rho_A = \eta \rho_w \]
Substitute this into the equation from Step 2:
\[ \eta \rho_w + \rho_B = \rho_w \]
\[ \rho_B = \rho_w (1 - \eta) \]
### Step 4: Finding the Ratio \(\frac{\rho_B}{\rho_A}\)
Now we have expressions for \(\rho_B\) and \(\rho_A\):
\[ \rho_A = \eta \rho_w \]
\[ \rho_B = \rho_w (1 - \eta) \]
The ratio \(\frac{\rho_B}{\rho_A}\) is:
\[ \frac{\rho_B}{\rho_A} = \frac{\rho_w (1 - \eta)}{\eta \rho_w} \]
\[ \frac{\rho_B}{\rho_A} = \frac{1 - \eta}{\eta} \]
Thus, the ratio \(\frac{\rho_B}{\rho_A}\) is:
\[ \frac{\rho_B}{\rho_A} = \frac{1 - \eta}{\eta} \]
