A small ball of density `rho` is immersed in a liquid of density `sigma(gtrho)` to a depth `h` and released. The height above the surface of water up to which the ball will jump is

Options

A. `(sigma/rho-1)h`
B. `(rho/sigma-1)h`
C. `(rho/sigma+1)h`
D. `(sigma/rho+1)h`

Explanation

To solve the problem step by step, we will analyze the forces acting on the ball when it is released in the liquid and determine how high it will jump above the surface of the liquid.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Ball:**
- The ball experiences two main forces when submerged: the buoyant force (upward) and its weight (downward).
- The buoyant force \( F_b \) is given by:
\[ F_b = V \cdot \sigma \cdot g \]
where \( V \) is the volume of the ball, \( \sigma \) is the density of the liquid, and \( g \) is the acceleration due to gravity.
- The weight \( W \) of the ball is given by:
\[ W = V \cdot \rho \cdot g \]
where \( \rho \) is the density of the ball.

2. **Calculate the Net Force and Acceleration:**
- The net force \( F_{net} \) acting on the ball when it is released is:
\[ F_{net} = F_b - W = V \cdot \sigma \cdot g - V \cdot \rho \cdot g \]
- This simplifies to:
\[ F_{net} = V \cdot g (\sigma - \rho) \]
- Using Newton's second law, \( F = ma \), we can express the acceleration \( A \) of the ball:
\[ A = \frac{F_{net}}{m} = \frac{V \cdot g (\sigma - \rho)}{V \cdot \rho} = \frac{g (\sigma - \rho)}{\rho} \]

3. **Determine the Velocity of the Ball at the Surface:**
- When the ball reaches the surface of the liquid, we can use the kinematic equation:
\[ V^2 = U^2 + 2AS \]
where \( U = 0 \) (initial velocity), \( A = \frac{g (\sigma - \rho)}{\rho} \), and \( S = h \) (the depth from which it was released).
- Thus, we have:
\[ V^2 = 0 + 2 \cdot \frac{g (\sigma - \rho)}{\rho} \cdot h \]
- This gives:
\[ V^2 = \frac{2g (\sigma - \rho) h}{\rho} \]

4. **Calculate the Maximum Height Above the Surface:**
- When the ball reaches its maximum height above the surface, its velocity becomes zero. Using the same kinematic equation:
\[ 0 = V^2 - 2gh_A \]
- Rearranging gives:
\[ h_A = \frac{V^2}{2g} \]
- Substituting for \( V^2 \) from the previous step:
\[ h_A = \frac{\frac{2g (\sigma - \rho) h}{\rho}}{2g} = \frac{(\sigma - \rho) h}{\rho} \]

5. **Final Expression for Maximum Height:**
- Therefore, the height above the surface of the liquid that the ball will jump is:
\[ h_A = h \cdot \frac{\sigma - \rho}{\rho} \]

### Final Answer:
The height above the surface of the water up to which the ball will jump is:
\[ h_A = h \cdot \frac{\sigma - \rho}{\rho} \]