A ball of mass 5.0 gm and relative density 0.5 strikes the surface of the water with a velocity of 20 m/sec. It comes to rest at a depth of 2m. Find the work done by the resisting force in water: (take `g = 10 m//s^(2)`)

Options

A. (a)`6 J`
B. (b)` + 7. 5 J`
C. (c)-`9 J`
D. (d)`-10 J`

Explanation

To solve the problem step by step, let's break it down:

### Given Data:
- Mass of the ball (m) = 5.0 g = 0.005 kg (since 1 g = 0.001 kg)
- Relative density of the ball = 0.5
- Velocity of the ball (v) = 20 m/s
- Depth at which the ball comes to rest (d) = 2 m
- Acceleration due to gravity (g) = 10 m/s²

### Step 1: Calculate the Volume of the Ball
Relative density (RD) is defined as the ratio of the density of the object to the density of water.
Given that the relative density of the ball is 0.5, we can find the density of the ball (ρ_ball):
\[ \rho_{\text{ball}} = \text{RD} \times \rho_{\text{water}} = 0.5 \times 1000 \, \text{kg/m}^3 = 500 \, \text{kg/m}^3 \]

Now, we can calculate the volume (V) of the ball using its mass:
\[ V = \frac{m}{\rho_{\text{ball}}} = \frac{0.005 \, \text{kg}}{500 \, \text{kg/m}^3} = 1 \times 10^{-5} \, \text{m}^3 \]

### Step 2: Calculate the Work Done by Gravity
The work done by gravity (W_g) when the ball is submerged to a depth of 2 m is given by:
\[ W_g = m \cdot g \cdot d = 0.005 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 2 \, \text{m} = 0.1 \, \text{J} \]

### Step 3: Calculate the Buoyant Force
The buoyant force (F_b) acting on the ball can be calculated using Archimedes' principle:
\[ F_b = \rho_{\text{water}} \cdot V \cdot g = 1000 \, \text{kg/m}^3 \cdot 1 \times 10^{-5} \, \text{m}^3 \cdot 10 \, \text{m/s}^2 = 0.1 \, \text{N} \]

### Step 4: Calculate the Work Done by Buoyant Force
The work done by the buoyant force (W_b) is:
\[ W_b = F_b \cdot d = 0.1 \, \text{N} \cdot 2 \, \text{m} = 0.2 \, \text{J} \]

### Step 5: Calculate the Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the ball is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 0.005 \, \text{kg} \cdot (20 \, \text{m/s})^2 = 1 \, \text{J} \]

### Step 6: Apply the Work-Energy Principle
According to the work-energy principle:
\[ \text{Total Work Done} = \Delta KE = KE_{\text{final}} - KE_{\text{initial}} \]
Since the ball comes to rest, \( KE_{\text{final}} = 0 \):
\[ \Delta KE = 0 - 1 \, \text{J} = -1 \, \text{J} \]

### Step 7: Set Up the Equation for Total Work Done
The total work done can be expressed as:
\[ \text{Total Work Done} = W_g + W_b + W_r \]
Where \( W_r \) is the work done by the resisting force. Substituting the values we have:
\[ -1 \, \text{J} = 0.1 \, \text{J} + 0.2 \, \text{J} + W_r \]
\[ W_r = -1 - 0.1 - 0.2 = -1.3 \, \text{J} \]

### Conclusion
The work done by the resisting force in water is:
\[ W_r = -1.3 \, \text{J} \]