A hollow sphere of volume V is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water?

Options

A. `V//2`
B. `V//3`
C. `V//4`
D. `V`

Explanation

To solve the problem of how much minimum volume of water should be poured inside a hollow sphere so that it sinks into the water, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- The hollow sphere has a volume \( V \) and is floating on the water surface with half of it immersed. This means that the volume of water displaced by the sphere is \( \frac{V}{2} \).

2. **Applying Archimedes' Principle**:
- According to Archimedes' principle, the buoyant force (upthrust) acting on the sphere is equal to the weight of the water displaced. Therefore, we can express this as:
\[ \text{Buoyant Force} = \text{Weight of Water Displaced} = \frac{V}{2} \cdot \rho_L \cdot g \]
- Here, \( \rho_L \) is the density of the water and \( g \) is the acceleration due to gravity.

3. **Weight of the Sphere**:
- The weight of the hollow sphere can be expressed as:
\[ \text{Weight of Sphere} = V \cdot \rho_S \cdot g \]
- Where \( \rho_S \) is the density of the sphere.

4. **Equating Forces**:
- Since the sphere is floating, the buoyant force equals the weight of the sphere:
\[ \frac{V}{2} \cdot \rho_L \cdot g = V \cdot \rho_S \cdot g \]
- Simplifying this gives:
\[ \frac{\rho_L}{2} = \rho_S \]

5. **Condition for Sinking**:
- To make the sphere sink, we need to add a volume \( V' \) of water inside the sphere. When the sphere is fully submerged, the weight of the water displaced will equal the total weight of the sphere plus the added water:
\[ V \cdot \rho_L \cdot g = V \cdot \rho_S \cdot g + V' \cdot \rho_L \cdot g \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ V \cdot \rho_L = V \cdot \rho_S + V' \cdot \rho_L \]
- This can be simplified to:
\[ V' \cdot \rho_L = V \cdot (\rho_L - \rho_S) \]

7. **Substituting for \( \rho_S \)**:
- We know from step 4 that \( \rho_S = \frac{\rho_L}{2} \). Substituting this into the equation gives:
\[ V' \cdot \rho_L = V \cdot \left(\rho_L - \frac{\rho_L}{2}\right) \]
- Simplifying this results in:
\[ V' \cdot \rho_L = V \cdot \frac{\rho_L}{2} \]

8. **Solving for \( V' \)**:
- Dividing both sides by \( \rho_L \) (assuming \( \rho_L \neq 0 \)):
\[ V' = \frac{V}{2} \]

### Conclusion:
The minimum volume of water that should be poured inside the sphere so that it sinks is \( \frac{V}{2} \).