A ball of mass 1 kg falls from a height of 5m above the free surface of water. The relative density of the solid ball is `s=2//3`. The ball travels a distance of 2m under water and becomes stationary. The work done by the resistive forces of water is

Options

A. `-50 J`
B. `-20 J`
C. `-40 J`
D. `-30 J`

Explanation

To find the work done by the resistive forces of water on a ball that falls into it, we can follow these steps:

### Step 1: Calculate the gravitational potential energy (GPE) of the ball before it falls.
The gravitational potential energy when the ball is at a height \( h \) is given by the formula:
\[ \text{GPE} = mgh \]
where:
- \( m = 1 \, \text{kg} \) (mass of the ball)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 5 \, \text{m} \) (height from which the ball falls)

Substituting the values:
\[ \text{GPE} = 1 \times 10 \times 5 = 50 \, \text{J} \]

### Step 2: Determine the buoyant force acting on the ball.
The buoyant force \( F_B \) can be calculated using the formula:
\[ F_B = \rho_{water} \cdot V \cdot g \]
where:
- \( \rho_{water} = 1000 \, \text{kg/m}^3 \) (density of water)
- \( V \) is the volume of the ball

To find the volume \( V \), we can use the relative density \( s \) of the ball:
\[ s = \frac{\rho_{ball}}{\rho_{water}} \Rightarrow \rho_{ball} = s \cdot \rho_{water} = \frac{2}{3} \cdot 1000 = \frac{2000}{3} \, \text{kg/m}^3 \]

The volume of the ball can be calculated using:
\[ V = \frac{m}{\rho_{ball}} = \frac{1}{\frac{2000}{3}} = \frac{3}{2000} \, \text{m}^3 \]

Now substituting \( V \) into the buoyant force formula:
\[ F_B = 1000 \cdot \frac{3}{2000} \cdot 10 = 15 \, \text{N} \]

### Step 3: Calculate the work done by the buoyant force.
The work done by the buoyant force \( W_B \) when the ball travels \( 2 \, \text{m} \) under water is given by:
\[ W_B = F_B \cdot d \]
where \( d = 2 \, \text{m} \):
\[ W_B = 15 \cdot 2 = 30 \, \text{J} \]

### Step 4: Apply the work-energy theorem.
According to the work-energy theorem:
\[ \text{Work done by all forces} = \Delta KE \]
Since the ball comes to rest, the change in kinetic energy \( \Delta KE = 0 \). Therefore, the total work done by all forces (gravitational force, buoyant force, and resistive force) is zero:
\[ W_{gravity} + W_B + W_R = 0 \]

### Step 5: Calculate the work done by the gravitational force.
The work done by gravity \( W_{gravity} \) when the ball falls \( 5 \, \text{m} \) is:
\[ W_{gravity} = mgh = 50 \, \text{J} \]

### Step 6: Set up the equation to find the work done by resistive forces.
Substituting the known values into the work-energy theorem:
\[ 50 + 30 + W_R = 0 \]
\[ W_R = -80 \, \text{J} \]

### Conclusion
The work done by the resistive forces of water is:
\[ W_R = -80 \, \text{J} \]