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 of the work done in pushing a spherical ball of radius \( r \) and relative density \( 0.5 \) down so that the whole of it is just immersed in water, we can follow these steps:

### Step 1: Understand the Concept of Relative Density
Relative density (or specific gravity) is defined as the ratio of the density of a substance to the density of a reference substance (usually water). Given that the relative density of the ball is \( 0.5 \), we can express the density of the ball as:
\[ \rho_{ball} = 0.5 \cdot \rho_{water} \]
where \( \rho_{water} \) is the density of water.

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

### Step 3: Calculate the Weight of the Ball
The weight \( W \) of the ball can be calculated using the formula:
\[ W = \text{Volume} \times \text{Density} \times g = V \cdot \rho_{ball} \cdot g \]
Substituting the volume of the ball:
\[ W = \left(\frac{4}{3} \pi r^3\right) \cdot (0.5 \cdot \rho_{water}) \cdot g = \frac{2}{3} \pi r^3 \rho_{water} g \]

### Step 4: Calculate the Buoyant Force
When the ball is floating with half of it immersed, the volume of the water displaced is half of the volume of the ball:
\[ V_{displaced} = \frac{1}{2} V = \frac{1}{2} \left(\frac{4}{3} \pi r^3\right) = \frac{2}{3} \pi r^3 \]
The buoyant force \( F_b \) acting on the ball is equal to the weight of the water displaced:
\[ F_b = V_{displaced} \cdot \rho_{water} \cdot g = \left(\frac{2}{3} \pi r^3\right) \cdot \rho_{water} \cdot g \]

### Step 5: Determine the Work Done
When the ball is pushed down so that it is fully immersed, the buoyant force will now equal the weight of the ball. The work done \( W_{done} \) in pushing the ball down is equal to the change in buoyant force times the distance moved. The distance moved is equal to the radius of the ball \( r \).

The change in buoyant force when the ball is fully immersed compared to when it is half-immersed is:
\[ \Delta F_b = F_{b, fully \, immersed} - F_{b, half \, immersed} \]
Calculating the buoyant force when fully immersed:
\[ F_{b, fully \, immersed} = V \cdot \rho_{water} \cdot g = \left(\frac{4}{3} \pi r^3\right) \cdot \rho_{water} \cdot g \]
Calculating the buoyant force when half-immersed:
\[ F_{b, half \, immersed} = \left(\frac{2}{3} \pi r^3\right) \cdot \rho_{water} \cdot g \]
Thus, the change in buoyant force is:
\[ \Delta F_b = \left(\frac{4}{3} \pi r^3 \rho_{water} g\right) - \left(\frac{2}{3} \pi r^3 \rho_{water} g\right) = \frac{2}{3} \pi r^3 \rho_{water} g \]

The work done in pushing the ball down is:
\[ W_{done} = \Delta F_b \cdot r = \left(\frac{2}{3} \pi r^3 \rho_{water} g\right) \cdot r = \frac{2}{3} \pi r^4 \rho_{water} g \]

### Final Answer
The work done in pushing the ball down so that the whole of it is just immersed in water is:
\[ W_{done} = \frac{2}{3} \pi r^4 \rho_{water} g \]