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 with a relative density of 0.8 will sink into 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 hits the water.

The ball is in free fall, so we can use the equation of motion:

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

Where:
- \( v \) = final velocity
- \( u \) = initial velocity = 0 (since it starts from rest)
- \( a \) = acceleration due to gravity \( g \)
- \( s \) = height fallen = 2 m

Substituting the values:

\[ v^2 = 0 + 2g(2) \]
\[ v^2 = 4g \]
\[ v = \sqrt{4g} = 2\sqrt{g} \]

### Step 2: Determine the forces acting on the ball when it is submerged in water.

When the ball is submerged, two main forces act on it:
1. The weight of the ball acting downwards: \( W = mg \)
2. The buoyant force acting upwards: \( F_b = \sigma V g \)

Where:
- \( \sigma \) = density of water = 1 g/cm³ (or 1000 kg/m³)
- \( V \) = volume of the ball
- \( m \) = mass of the ball = \( \rho V \) (where \( \rho \) is the density of the ball)

### Step 3: Calculate the net force acting on the ball.

The net force \( F_{net} \) acting on the ball can be expressed as:

\[ F_{net} = F_b - W = \sigma V g - \rho V g \]

Substituting \( \rho = 0.8 \sigma \):

\[ F_{net} = \sigma V g - 0.8 \sigma V g = 0.2 \sigma V g \]

### Step 4: Calculate the acceleration of the ball in water.

Using Newton's second law, \( F = ma \):

\[ ma = 0.2 \sigma V g \]

Where \( m = \rho V \):

\[ \rho V a = 0.2 \sigma V g \]

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

\[ \rho a = 0.2 \sigma g \]

Substituting \( \rho = 0.8 \sigma \):

\[ 0.8 \sigma a = 0.2 \sigma g \]

Dividing both sides by \( \sigma \):

\[ 0.8 a = 0.2 g \]

Thus, the acceleration \( a \) is:

\[ a = \frac{0.2 g}{0.8} = \frac{g}{4} \]

### Step 5: Calculate the depth to which the ball will sink.

When the ball sinks, it will eventually come to rest. We can use the equation of motion again to find the distance \( H \) it sinks:

Using the equation:

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

Where:
- Final velocity \( v = 0 \) (when it stops)
- Initial velocity \( u = 2\sqrt{g} \)
- Acceleration \( a = -\frac{g}{4} \) (acting upwards)

Substituting the values:

\[ 0 = (2\sqrt{g})^2 + 2 \left(-\frac{g}{4}\right) H \]
\[ 0 = 4g - \frac{g}{2} H \]

Rearranging gives:

\[ \frac{g}{2} H = 4g \]
\[ H = \frac{4g}{\frac{g}{2}} = 8 \text{ meters} \]

### Conclusion:

The depth to which the ball will sink in the water is **8 meters**.

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