A ball of relative density 0.8 falls into water from a height of 2 m. The depth to which the ball will sink is (neglect viscous forces)

Options

A. 8 m
B. 2 m
C. 6 m
D. 16 m

Explanation

To solve the problem of how deep a ball of relative density 0.8 will sink in water after falling from a height of 2 meters, we can follow these steps:

### Step 1: Calculate the velocity of the ball just before it strikes the water
The ball falls freely under the influence of gravity. We can use the equation of motion to find the velocity just before it hits the water.

\[ v = \sqrt{2gh} \]

Where:
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height from which the ball falls (2 m)

Substituting the values:

\[ v = \sqrt{2 \times 9.81 \times 2} \]
\[ v = \sqrt{39.24} \]
\[ v \approx 6.26 \, \text{m/s} \]

### Step 2: Determine the acceleration of the ball while it is submerged in water
The ball experiences two forces when submerged: the buoyant force and its weight. The net force acting on the ball can be expressed as:

\[ F_{\text{net}} = F_{\text{buoyant}} - F_{\text{weight}} \]

The buoyant force \( F_{\text{buoyant}} \) is given by:

\[ F_{\text{buoyant}} = V \cdot \rho_{\text{water}} \cdot g \]

The weight \( F_{\text{weight}} \) of the ball is:

\[ F_{\text{weight}} = V \cdot \rho_{\text{ball}} \cdot g \]

Where:
- \( V \) is the volume of the ball
- \( \rho_{\text{water}} \) = 1000 kg/m³ (density of water)
- \( \rho_{\text{ball}} = 0.8 \cdot 1000 = 800 \, \text{kg/m}^3 \) (density of the ball)

Thus, the net force becomes:

\[ F_{\text{net}} = V \cdot (1000 - 800) \cdot g \]
\[ F_{\text{net}} = V \cdot 200 \cdot g \]

Using Newton's second law, we can express the net force as:

\[ F_{\text{net}} = m \cdot a \]

Where \( m = V \cdot \rho_{\text{ball}} \), so:

\[ V \cdot 200 \cdot g = (V \cdot 800) \cdot a \]

Cancelling \( V \) from both sides (assuming \( V \neq 0 \)):

\[ 200g = 800a \]

Solving for \( a \):

\[ a = \frac{200g}{800} = \frac{g}{4} \]

### Step 3: Apply the second equation of motion to find the depth
Now we can use the second equation of motion to find the depth \( h \) to which the ball sinks:

\[ v^2 = u^2 + 2ah \]

Where:
- \( u = 0 \) (initial velocity when the ball starts sinking)
- \( v = 6.26 \, \text{m/s} \) (velocity just before it hits the water)
- \( a = \frac{g}{4} \)

Substituting the values:

\[ (6.26)^2 = 0 + 2 \left(\frac{g}{4}\right) h \]

Calculating \( (6.26)^2 \):

\[ 39.1876 = 2 \left(\frac{9.81}{4}\right) h \]
\[ 39.1876 = \frac{19.62}{4} h \]

Multiplying both sides by 4:

\[ 156.75 = 19.62h \]

Solving for \( h \):

\[ h = \frac{156.75}{19.62} \approx 8 \, \text{m} \]

### Conclusion
The depth to which the ball will sink is approximately **8 meters**.

---