A sperical ball of radius r and relative density 0.5 is floating in equilibrium in water with half of it immersed in water. The work done in pushing the ball down so that whole of it is just immersed in water is [`rho` is the density of water]-

Options

A. 0.25 `rhorg`
B. 0.5 `rho` rg
C. `(4)/(3)pir^(3)rhog`
D. `(5)/(12)pir^(4)rhog`

Answer

To solve the problem, we need to determine the work done in pushing a spherical ball of radius \( r \) and relative density \( 0.5 \) downwards so that it becomes fully immersed in water.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The ball is floating in water with half of its volume submerged.
- The relative density of the ball is \( 0.5 \), which means the density of the ball is half that of water.

2. **Calculate the Volume of the Ball**:
- The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

3. **Calculate the Mass of the Ball**:
- The density of water \( \rho \) is taken as \( \rho \) (in kg/m³).
- Since the relative density of the ball is \( 0.5 \), the density of the ball \( \rho_b \) is:
\[ \rho_b = 0.5 \rho \]
- Therefore, the mass \( m \) of the ball is:
\[ m = \rho_b \cdot V = 0.5 \rho \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \rho \]

4. **Determine the Initial and Final Positions**:
- Initially, half of the ball is submerged, meaning the submerged volume \( V_s \) is:
\[ V_s = \frac{1}{2} V = \frac{1}{2} \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \]
- The center of gravity of the submerged part (hemisphere) is located at a distance of \( \frac{3}{8} r \) from the water surface downwards.

5. **Calculate the Change in Height**:
- When the ball is fully submerged, the center of gravity of the ball is at a distance of \( \frac{3}{8} r \) below the water surface.
- The change in height \( h \) when the ball is pushed down from half-immersed to fully immersed is:
\[ h = \frac{3}{8} r - \frac{r}{2} = \frac{3}{8} r - \frac{4}{8} r = -\frac{1}{8} r \]
- This negative sign indicates that the center of gravity of the ball is moving downwards.

6. **Calculate the Work Done**:
- The work done \( W \) in moving the ball down is given by:
\[ W = m \cdot g \cdot h \]
- Substituting the values:
\[ W = \left(\frac{2}{3} \pi r^3 \rho\right) \cdot g \cdot \left(-\frac{1}{8} r\right) \]
- Simplifying this gives:
\[ W = -\frac{2}{3} \cdot \frac{1}{8} \cdot \pi r^4 \rho g = -\frac{1}{12} \pi r^4 \rho g \]
- The negative sign indicates that work is done against buoyancy.

7. **Final Expression for Work Done**:
- The work done in pushing the ball down so that the whole of it is just immersed in water is:
\[ W = \frac{1}{12} \pi r^4 \rho g \]

### Final Answer:
\[ W = \frac{1}{12} \pi r^4 \rho g \]