Two solids A and B float in water. It is observed that A floats with `1/2` of its body immersed in water and B floats with `1/4` of its volume above the water level. The ratio of the density of A to that of B is

Options

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

Answer

To solve the problem, we need to find the ratio of the densities of solids A and B based on their behavior when floating in water.

### Step-by-Step Solution:

1. **Understanding the Situation**:
- Solid A floats with \( \frac{1}{2} \) of its body immersed in water.
- Solid B floats with \( \frac{1}{4} \) of its volume above the water level, which means \( \frac{3}{4} \) of its volume is immersed.

2. **Applying Archimedes' Principle**:
- According to Archimedes' principle, the buoyant force (up thrust) acting on a floating object is equal to the weight of the liquid displaced by the object.
- For solid A:
- Volume of A immersed in water = \( \frac{1}{2} V_A \)
- Buoyant force on A = Weight of water displaced = \( \frac{1}{2} V_A \cdot \rho_W \cdot g \)
- Weight of solid A = \( V_A \cdot \rho_A \cdot g \)

Setting the buoyant force equal to the weight of solid A:
\[ \frac{1}{2} V_A \cdot \rho_W \cdot g = V_A \cdot \rho_A \cdot g \]
- Canceling \( V_A \) and \( g \) from both sides:
\[ \frac{1}{2} \rho_W = \rho_A \quad \text{(Equation 1)} \]

3. **Analyzing Solid B**:
- For solid B:
- Volume of B immersed in water = \( \frac{3}{4} V_B \)
- Buoyant force on B = Weight of water displaced = \( \frac{3}{4} V_B \cdot \rho_W \cdot g \)
- Weight of solid B = \( V_B \cdot \rho_B \cdot g \)

Setting the buoyant force equal to the weight of solid B:
\[ \frac{3}{4} V_B \cdot \rho_W \cdot g = V_B \cdot \rho_B \cdot g \]
- Canceling \( V_B \) and \( g \) from both sides:
\[ \frac{3}{4} \rho_W = \rho_B \quad \text{(Equation 2)} \]

4. **Finding the Ratio of Densities**:
- Now, we have:
- From Equation 1: \( \rho_A = \frac{1}{2} \rho_W \)
- From Equation 2: \( \rho_B = \frac{3}{4} \rho_W \)

- To find the ratio \( \frac{\rho_A}{\rho_B} \):
\[ \frac{\rho_A}{\rho_B} = \frac{\frac{1}{2} \rho_W}{\frac{3}{4} \rho_W} \]
- Simplifying this:
\[ \frac{\rho_A}{\rho_B} = \frac{1/2}{3/4} = \frac{1}{2} \cdot \frac{4}{3} = \frac{4}{6} = \frac{2}{3} \]

### Final Answer:
The ratio of the density of solid A to that of solid B is:
\[ \frac{\rho_A}{\rho_B} = \frac{2}{3} \]