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

Explanation

To solve the problem step by step, we will analyze the forces acting on the ball and apply the equations of motion to find the height to which the ball will jump above the surface of the liquid.

### Step 1: Identify the forces acting on the ball
When the ball is immersed in the liquid, two main forces act on it:
1. The weight of the ball, \( W = V \cdot \rho \cdot g \), where \( V \) is the volume of the ball, \( \rho \) is the density of the ball, and \( g \) is the acceleration due to gravity.
2. The buoyant force (upthrust) acting on the ball, \( F_b = V \cdot \sigma \cdot g \), where \( \sigma \) is the density of the liquid.

### Step 2: Calculate the net downward force
The net downward force \( F_{net} \) acting on the ball can be expressed as:
\[ F_{net} = W - F_b = V \cdot \rho \cdot g - V \cdot \sigma \cdot g \]
This simplifies to:
\[ F_{net} = V \cdot g (\rho - \sigma) \]

### Step 3: Determine the net acceleration of the ball
Using Newton's second law, the net acceleration \( a \) of the ball can be calculated as:
\[ a = \frac{F_{net}}{m} = \frac{V \cdot g (\rho - \sigma)}{V \cdot \rho} = \frac{g (\rho - \sigma)}{\rho} \]

### Step 4: Calculate the velocity of the ball just before it hits the surface of the liquid
When the ball is released from a depth \( h \), we can use the kinematic equation:
\[ v^2 = u^2 + 2a h \]
Here, the initial velocity \( u = 0 \) (since it is released), so:
\[ v^2 = 0 + 2 \cdot a \cdot h = 2 \cdot \left(\frac{g (\rho - \sigma)}{\rho}\right) \cdot h \]
Thus:
\[ v^2 = \frac{2g (\rho - \sigma)}{\rho} \cdot h \]

### Step 5: Apply the equation of motion for the jump above the surface
When the ball reaches the surface of the liquid, it will jump to a height \( h_a \) in the air. At the maximum height, the final velocity is \( 0 \). Using the equation:
\[ 0 = v^2 - 2g h_a \]
We can substitute for \( v^2 \):
\[ 0 = \frac{2g (\rho - \sigma)}{\rho} \cdot h - 2g h_a \]
Rearranging gives:
\[ 2g h_a = \frac{2g (\rho - \sigma)}{\rho} \cdot h \]
Dividing both sides by \( 2g \):
\[ h_a = \frac{(\rho - \sigma)}{\rho} \cdot h \]

### Step 6: Final expression for the height above the surface
To express the height \( h_a \) in terms of \( \sigma \) and \( \rho \):
\[ h_a = \left( \frac{\sigma - \rho}{\rho} \right) \cdot h \]

Thus, the height above the surface of the liquid to which the ball will jump is:
\[ h_a = \left( \frac{\sigma - \rho}{\rho} \right) \cdot h \]