A ball of density `rho` is released from deep inside of a liquid of density `2 rho`. It will move up

Options

A. with an increasing acceleration
B. with a decreasing acceleration
C. with a constant acceleration
D. with zero acceleration

Answer

To solve the problem of a ball of density \( \rho \) released from deep inside a liquid of density \( 2\rho \), we can analyze the forces acting on the ball and determine its motion. Here’s a step-by-step solution:

### Step 1: Identify the Forces Acting on the Ball
When the ball is submerged in the liquid, two main forces act on it:
1. **Weight of the Ball (W)**: This is the gravitational force acting downwards.
\[ W = m \cdot g = V \cdot \rho \cdot g \]
where \( V \) is the volume of the ball, \( \rho \) is the density of the ball, and \( g \) is the acceleration due to gravity.

2. **Buoyant Force (F_B)**: This is the upward force exerted by the liquid on the ball.
\[ F_B = V \cdot (2\rho) \cdot g \]
where \( 2\rho \) is the density of the liquid.

### Step 2: Calculate the Net Force
The net force \( F_{net} \) acting on the ball can be calculated as:
\[ F_{net} = F_B - W \]
Substituting the expressions for \( F_B \) and \( W \):
\[ F_{net} = (V \cdot (2\rho) \cdot g) - (V \cdot \rho \cdot g) \]
\[ F_{net} = V \cdot g \cdot (2\rho - \rho) = V \cdot g \cdot \rho \]

### Step 3: Determine the Direction of Motion
Since \( F_{net} \) is positive (as \( V \cdot g \cdot \rho \) is positive), the net force is directed upwards. This means that the ball will accelerate upwards.

### Step 4: Analyze the Acceleration
The acceleration \( a \) of the ball can be found using Newton's second law:
\[ F_{net} = m \cdot a \]
Substituting \( m = V \cdot \rho \):
\[ V \cdot g \cdot \rho = (V \cdot \rho) \cdot a \]
Dividing both sides by \( V \cdot \rho \) (assuming \( V \cdot \rho \neq 0 \)):
\[ g = a \]
This indicates that the ball will accelerate upwards with an acceleration equal to \( g \).

### Step 5: Conclusion
Since the ball is less dense than the liquid, it will rise to the surface of the liquid. The acceleration of the ball remains constant as it moves upward.

### Final Answer
The ball will move up with a constant acceleration equal to \( g \).

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