Two solids `A` and `B` floats in water. It is observed that `A` floats with half of its volume immersed and `B` Floats with `2//3` of its volume immersed. The ration of densities of `A` and `B` is

Options

A. `4:3`
B. `2:3`
C. `3:4`
D. `1:3`

Explanation

To find the ratio of densities of solids A and B, we can follow these steps:

### Step 1: Understand the problem
We have two solids, A and B, floating in water. Solid A is half immersed, while solid B is two-thirds immersed. We need to find the ratio of their densities.

### Step 2: Apply Archimedes' Principle
According to Archimedes' Principle, the buoyant force acting on a floating object is equal to the weight of the fluid displaced by the object. For both solids A and B, we can set up the equilibrium equations.

### Step 3: Analyze Solid A
Let the density of solid A be \( \rho_A \) and its volume be \( V_A \). Since solid A is half immersed, the volume of water displaced is \( \frac{V_A}{2} \).

The weight of solid A is:
\[ \text{Weight of A} = \rho_A \cdot V_A \cdot g \]

The buoyant force acting on solid A is equal to the weight of the displaced water:
\[ \text{Buoyant Force on A} = \frac{V_A}{2} \cdot \rho_{\text{water}} \cdot g \]

Setting the weight equal to the buoyant force:
\[ \rho_A \cdot V_A \cdot g = \frac{V_A}{2} \cdot \rho_{\text{water}} \cdot g \]

Cancelling \( V_A \) and \( g \) from both sides:
\[ \rho_A = \frac{1}{2} \cdot \rho_{\text{water}} \]

### Step 4: Analyze Solid B
Let the density of solid B be \( \rho_B \) and its volume be \( V_B \). Since solid B is two-thirds immersed, the volume of water displaced is \( \frac{2V_B}{3} \).

The weight of solid B is:
\[ \text{Weight of B} = \rho_B \cdot V_B \cdot g \]

The buoyant force acting on solid B is:
\[ \text{Buoyant Force on B} = \frac{2V_B}{3} \cdot \rho_{\text{water}} \cdot g \]

Setting the weight equal to the buoyant force:
\[ \rho_B \cdot V_B \cdot g = \frac{2V_B}{3} \cdot \rho_{\text{water}} \cdot g \]

Cancelling \( V_B \) and \( g \) from both sides:
\[ \rho_B = \frac{2}{3} \cdot \rho_{\text{water}} \]

### Step 5: Find the ratio of densities
Now we have:
\[ \rho_A = \frac{1}{2} \cdot \rho_{\text{water}} \quad \text{and} \quad \rho_B = \frac{2}{3} \cdot \rho_{\text{water}} \]

To find the ratio \( \frac{\rho_A}{\rho_B} \):
\[ \frac{\rho_A}{\rho_B} = \frac{\frac{1}{2} \cdot \rho_{\text{water}}}{\frac{2}{3} \cdot \rho_{\text{water}}} \]

The \( \rho_{\text{water}} \) cancels out:
\[ \frac{\rho_A}{\rho_B} = \frac{1/2}{2/3} = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} \]

### Step 6: Conclusion
Thus, the ratio of densities of solids A and B is:
\[ \frac{\rho_A}{\rho_B} = \frac{3}{4} \]