A body floats with one-third of its volume outside water and `3//4` of its volume outside another liquid. The density of another liquid is :

Options

A. `9/4 g//c c`
B. `4/9 g//c c`
C. `8/3 g//c c`
D. `3/8 g//c c`

Explanation

To solve the problem, we will use the principle of buoyancy, which states that the weight of the liquid displaced by a floating body is equal to the weight of the body.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body floats with one-third of its volume above the water surface.
- The same body floats in another liquid with three-fourths of its volume above that liquid.

2. **Volume of the Body**:
- Let the total volume of the body be \( V \).
- When floating in water, the volume submerged in water is \( V - \frac{V}{3} = \frac{2V}{3} \).
- When floating in the other liquid, the volume submerged is \( V - \frac{3V}{4} = \frac{V}{4} \).

3. **Weight of the Body**:
- The weight of the body can be expressed as:
\[ W = \text{Density of the body} \times V \times g \]
- Let the density of the body be \( \rho_b \).

4. **Buoyant Force in Water**:
- The buoyant force when the body is floating in water is equal to the weight of the water displaced:
\[ \text{Buoyant Force in Water} = \text{Density of Water} \times \text{Volume submerged in Water} \times g \]
- This gives:
\[ \rho_w \times \frac{2V}{3} \times g \]
- Setting the buoyant force equal to the weight of the body:
\[ \rho_w \times \frac{2V}{3} \times g = \rho_b \times V \times g \]
- Canceling \( g \) and \( V \) from both sides:
\[ \rho_w \times \frac{2}{3} = \rho_b \]
- Thus, the density of the body is:
\[ \rho_b = \frac{2}{3} \rho_w \]

5. **Buoyant Force in Another Liquid**:
- The buoyant force when the body is floating in the other liquid is:
\[ \text{Buoyant Force in Another Liquid} = \text{Density of Another Liquid} \times \text{Volume submerged in Another Liquid} \times g \]
- This gives:
\[ \rho_l \times \frac{V}{4} \times g \]
- Setting the buoyant force equal to the weight of the body:
\[ \rho_l \times \frac{V}{4} \times g = \rho_b \times V \times g \]
- Canceling \( g \) and \( V \) from both sides:
\[ \rho_l \times \frac{1}{4} = \rho_b \]
- Substituting \( \rho_b \) from the previous step:
\[ \rho_l \times \frac{1}{4} = \frac{2}{3} \rho_w \]

6. **Finding the Density of the Other Liquid**:
- Rearranging the equation to find \( \rho_l \):
\[ \rho_l = \frac{2}{3} \rho_w \times 4 \]
\[ \rho_l = \frac{8}{3} \rho_w \]

### Final Answer:
The density of the other liquid is \( \frac{8}{3} \) times the density of water.