A ball whose density is `0.4 xx 10^3 kg//m^3` falls into water from a height of 9 cm. To what depth does the ball sink ?

Options

A. 9 cm
B. 6 cm
C. 4.5 cm
D. 2.25 cm

Answer

To solve the problem of how deep the ball sinks in water after falling from a height of 9 cm, we can follow these steps:

### Step 1: Understand the scenario
The ball has a density of \(0.4 \times 10^3 \, \text{kg/m}^3\) and falls into water from a height of 9 cm. We need to find out how deep it sinks into the water.

### Step 2: Identify forces acting on the ball
When the ball is submerged in water, two main forces act on it:
1. The gravitational force (weight) acting downwards, \(F_g = mg\).
2. The buoyant force acting upwards, \(F_b = \rho_{\text{liquid}} \cdot V_{\text{submerged}} \cdot g\).

### Step 3: Calculate the mass of the ball
Assuming the volume of the ball is \(V\), the mass \(m\) of the ball can be expressed as:
\[ m = \rho_{\text{object}} \cdot V = (0.4 \times 10^3) \cdot V \]

### Step 4: Write the work-energy principle
According to the work-energy theorem, the work done by the forces equals the change in kinetic energy. Since the ball starts and ends with zero velocity, the total work done is zero:
\[ \text{Work done by gravity} + \text{Work done by buoyant force} = 0 \]

### Step 5: Calculate work done by gravity
The work done by gravity when the ball falls a distance of \(h + x\) (where \(h = 9 \, \text{cm}\) and \(x\) is the depth it sinks) is:
\[ W_g = mgh + mgx = mg(h + x) \]

### Step 6: Calculate work done by buoyant force
The buoyant force acts upwards while the ball moves downwards. Thus, the work done by the buoyant force is:
\[ W_b = -F_b \cdot x = -\rho_{\text{liquid}} \cdot V \cdot g \cdot x \]

### Step 7: Set up the equation
Substituting the expressions for work done:
\[ mg(h + x) - \rho_{\text{liquid}} \cdot V \cdot g \cdot x = 0 \]

### Step 8: Substitute values
Using \(\rho_{\text{liquid}} = 10^3 \, \text{kg/m}^3\) and substituting \(m\):
\[ (0.4 \times 10^3) \cdot V \cdot g \cdot (9 \, \text{cm} + x) - (10^3) \cdot V \cdot g \cdot x = 0 \]

### Step 9: Cancel out common terms
Since \(V\) and \(g\) are common in both terms, we can cancel them out:
\[ 0.4 \times 10^3 \cdot (9 + x) - 10^3 \cdot x = 0 \]

### Step 10: Simplify the equation
Rearranging gives:
\[ 0.4 \cdot (9 + x) = x \]
\[ 3.6 + 0.4x = x \]
\[ 3.6 = x - 0.4x \]
\[ 3.6 = 0.6x \]
\[ x = \frac{3.6}{0.6} = 6 \, \text{cm} \]

### Conclusion
The ball sinks to a depth of **6 cm** in the water.

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