Final answer:
The correct option is B. V1 > V2. The ball displaces more water when atmospheric pressure is present compared to when it is removed, resulting in a submerged volume that is less once the air is pumped out.
Step-by-step explanation:
When a ball floats on the surface of water, it displaces a volume of water equal to the submerged volume of the ball. This scenario is described by Archimedes' principle, which states that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that is displaced by the object.
Therefore, initially, the submerged volume of the ball, V1, corresponds to the weight of the ball being balanced by the buoyant force that is equivalent to the weight of the displaced water.
When the container is sealed and the air is pumped out, a vacuum is created above the water surface, eliminating the atmospheric pressure acting on the water.
With the removal of the atmospheric pressure, there is now a lesser force pushing down on the water, and hence the ball would need to displace less water to balance its weight against the buoyant force. Therefore, V2, the volume of the ball submerged after the air is pumped out, is less than V1.
In conclusion, the correct option is B. V1 > V2. When the air is removed, the atmospheric pressure is no longer present, causing the submerged volume of the ball to decrease as it needs to displace less water to achieve buoyant equilibrium.