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 step by step, we will analyze the forces acting on the ball in both scenarios: when the container is open to the atmosphere and when it is covered and air is pumped out.

### Step 1: Analyze the initial condition (open container)
In the initial condition, the ball is floating on the surface of the water, and the volume \( V_1 \) of the ball is submerged in water. The forces acting on the ball are:
- The weight of the ball, \( mg \) (acting downwards).
- The buoyant force due to water, \( F_{uW} \) (acting upwards).
- The buoyant force due to air, \( F_{uA} \) (acting upwards).

The equilibrium condition can be expressed as:
\[ mg = F_{uW} + F_{uA} \]

### Step 2: Express the buoyant forces
The buoyant force due to water can be expressed as:
\[ F_{uW} = V_1 \cdot \rho_W \cdot g \]
where \( \rho_W \) is the density of water.

The buoyant force due to air can be expressed as:
\[ F_{uA} = (V - V_1) \cdot \rho_A \cdot g \]
where \( \rho_A \) is the density of air and \( V \) is the total volume of the ball.

### Step 3: Substitute the buoyant forces into the equilibrium equation
Substituting the expressions for the buoyant forces into the equilibrium equation gives:
\[ mg = V_1 \cdot \rho_W \cdot g + (V - V_1) \cdot \rho_A \cdot g \]

### Step 4: Simplify the equation
Dividing through by \( g \) (assuming \( g \neq 0 \)):
\[ m = V_1 \cdot \rho_W + (V - V_1) \cdot \rho_A \]

### Step 5: Analyze the second condition (vacuum condition)
When the air is pumped out, the buoyant force due to air \( F_{uA} \) becomes zero. The new equilibrium condition is:
\[ mg = F_{uW} \]
Substituting for \( F_{uW} \) gives:
\[ mg = V_2 \cdot \rho_W \cdot g \]
where \( V_2 \) is the new volume of the ball submerged in water.

### Step 6: Set the equations equal
Since the weight of the ball remains constant, we can set the two expressions for \( mg \) equal:
\[ V_1 \cdot \rho_W + (V - V_1) \cdot \rho_A = V_2 \cdot \rho_W \]

### Step 7: Solve for \( V_2 \)
Rearranging the equation to solve for \( V_2 \):
\[ V_2 \cdot \rho_W = V_1 \cdot \rho_W + (V - V_1) \cdot \rho_A \]
\[ V_2 = \frac{V_1 \cdot \rho_W + (V - V_1) \cdot \rho_A}{\rho_W} \]

### Step 8: Analyze the relationship between \( V_2 \) and \( V_1 \)
From the equation derived, we can see that:
\[ V_2 = V_1 + \frac{(V - V_1) \cdot \rho_A}{\rho_W} \]
Since \( \rho_A \) (density of air) is much smaller than \( \rho_W \) (density of water), the term \( \frac{(V - V_1) \cdot \rho_A}{\rho_W} \) is positive. Therefore, we conclude that:
\[ V_2 > V_1 \]

### Conclusion
Thus, the final conclusion is that \( V_2 \) is greater than \( V_1 \).