Density of a liquid varies with depth as `rho=alpha`h. A small ball of density `rho_(0)` is released from the free surface of the liquid. Then

Options

A. the ball will execute SHM of amplitude `(rho_(0))/(alpha)`
B. the mean position of the ball will be at a depth `(rho_(0))/(2alpha)`
C. the ball will sink to a maximum depth of `(rho_(0))/alpha`
D. The ball sink to a maximum depth of `(rho_(0))/alpha`

Answer

To solve the problem of a small ball released in a liquid with varying density, we can follow these steps:

### Step 1: Understand the Density Variation
The density of the liquid varies with depth according to the equation:
\[ \rho = \alpha h \]
where \( \rho \) is the density of the liquid at depth \( h \), and \( \alpha \) is a constant.

### Step 2: Identify the Forces Acting on the Ball
When the ball of density \( \rho_0 \) is released from the surface, two main forces act on it:
1. The gravitational force (weight) acting downward:
\[ F_g = m g = \rho_0 V g \]
where \( V \) is the volume of the ball.

2. The buoyant force (upthrust) acting upward, which is given by:
\[ F_b = \rho V g \]
Here, \( \rho \) is the density of the liquid at the depth where the ball is submerged.

### Step 3: Establish the Condition for Equilibrium
The ball will reach an equilibrium position when the gravitational force equals the buoyant force:
\[ F_g = F_b \]
Substituting the expressions for the forces:
\[ \rho_0 V g = \rho V g \]
Cancelling \( V g \) from both sides (assuming \( V \neq 0 \)):
\[ \rho_0 = \rho \]
Substituting the density variation:
\[ \rho_0 = \alpha h \]
From this, we can find the depth \( h \):
\[ h = \frac{\rho_0}{\alpha} \]

### Step 4: Analyze the Motion of the Ball
When the ball is released, it will initially accelerate downward due to the weight being greater than the buoyant force. As it moves downward, the buoyant force increases (since the density of the liquid increases with depth).

### Step 5: Determine the Maximum Depth
The ball will oscillate about the equilibrium position. The maximum depth it reaches can be determined by analyzing the forces at the extreme position. When the ball is at maximum depth, the forces will again balance, but at a greater depth.

The maximum depth occurs when:
\[ \rho_0 = \alpha h_{max} \]
At the maximum depth, the ball will reach a position where:
\[ h_{max} = \frac{2\rho_0}{\alpha} \]

### Step 6: Calculate the Amplitude of SHM
The amplitude of the simple harmonic motion (SHM) can be found by taking the difference between the maximum depth and the mean position:
\[ \text{Amplitude} = h_{max} - h_{mean} = \frac{2\rho_0}{\alpha} - \frac{\rho_0}{\alpha} = \frac{\rho_0}{\alpha} \]

### Conclusion
Based on the analysis, we conclude:
- The ball will execute SHM with an amplitude of \( \frac{\rho_0}{\alpha} \).
- The maximum depth the ball will sink to is \( \frac{2\rho_0}{\alpha} \).

### Final Answer
- The ball will sink to a maximum depth of \( \frac{2\rho_0}{\alpha} \) (Option D is correct).

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