A ball floats on the surface of water in a container exposed to the atmosphere. Volume `V_(1)` of its volume is inside the water. The container is now covered and the air is pumped out. Now let `V_(2)` be the volume immersed in water. Then

Options

A. `V_(1) = V_(2)`
B. `V_(1) gt V_(2)`
C. `V_(2) gt V_(1)`
D. `V_(2) = 0`

Answer

To solve the problem, we need to analyze the situation of the ball floating on water before and after the air is pumped out of the container.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- Initially, the ball is floating on the surface of the water with a volume \( V_1 \) submerged.
- According to Archimedes' principle, the buoyant force acting on the ball is equal to the weight of the water displaced by the submerged volume \( V_1 \).

2. **Buoyant Forces**:
- There are two buoyant forces acting on the ball:
- \( F_{B1} \): the buoyant force due to the air pressure above the ball.
- \( F_{BW} \): the buoyant force due to the water.
- The total buoyant force when the ball is floating is given by:
\[ F_{B} = F_{BW} + F_{B1} \]

3. **Condition After Pumping Air Out**:
- When the air is pumped out, the buoyant force due to the air \( F_{B1} \) is removed.
- Now, only the buoyant force due to the water \( F_{BW} \) acts on the ball.

4. **Effect of Removing Air**:
- With the removal of air, the ball experiences a decrease in the total buoyant force acting on it.
- To maintain equilibrium, the ball must displace more water to balance the forces acting on it.

5. **New Submerged Volume**:
- Let \( V_2 \) be the new volume of the ball submerged in water after the air has been pumped out.
- Since the buoyant force due to the water must now equal the weight of the ball, and the only force acting is the buoyant force from the water, we have:
\[ F_{BW} = \text{Weight of the ball} \]
- Hence, the new volume \( V_2 \) must be greater than the initial volume \( V_1 \) to provide the necessary buoyant force.

6. **Conclusion**:
- Therefore, we conclude that:
\[ V_2 > V_1 \]

### Final Answer:
The volume of the ball immersed in water after the air is pumped out, \( V_2 \), is greater than the volume \( V_1 \) that was initially submerged.