A ball is dropped from height h. If falls on liquid surface. Its velocity does not change when its enters in liquid. Find height h in terms of r = radius of ball, `sigma` = density of liquid, `rho` = density of ball `eta` = coefficient of viscosity and g = acceleration due to gravity :

Options

A. `2 r ^4 g (rho - sigma)^2 / 81 eta ^2`
B. `2 r ^4 g (rho - sigma)^2 /50 eta ^2`
C. `2 r ^4 g (rho - sigma)^2 / 25 eta ^2`
D. `2 r ^4 g (rho - sigma)^2 / 90 eta ^2`

Answer

To solve the problem, we need to find the height \( h \) from which a ball is dropped, given that its velocity does not change when it enters a liquid. We will use the concepts of free fall and terminal velocity.

### Step-by-Step Solution:

1. **Determine the velocity of the ball just before it enters the liquid:**
The ball is dropped from height \( h \). Using the equation of motion for free fall:
\[ v = \sqrt{2gh} \]
where \( v \) is the velocity of the ball just before it hits the liquid, \( g \) is the acceleration due to gravity, and \( h \) is the height.

2. **Identify the terminal velocity:**
When the ball enters the liquid, it reaches a terminal velocity \( V_t \). The formula for terminal velocity \( V_t \) of a sphere moving through a viscous fluid is given by:
\[ V_t = \frac{2}{9} \frac{(\rho - \sigma) g r^2}{\eta} \]
where:
- \( \rho \) is the density of the ball,
- \( \sigma \) is the density of the liquid,
- \( r \) is the radius of the ball,
- \( \eta \) is the coefficient of viscosity.

3. **Set the velocities equal:**
According to the problem, the velocity of the ball just before it enters the liquid is equal to the terminal velocity when it is in the liquid:
\[ \sqrt{2gh} = V_t \]

4. **Substitute the expression for terminal velocity:**
Substituting the expression for \( V_t \) into the equation gives:
\[ \sqrt{2gh} = \frac{2}{9} \frac{(\rho - \sigma) g r^2}{\eta} \]

5. **Square both sides to eliminate the square root:**
Squaring both sides results in:
\[ 2gh = \left(\frac{2}{9} \frac{(\rho - \sigma) g r^2}{\eta}\right)^2 \]

6. **Simplify the equation:**
Expanding the right-hand side:
\[ 2gh = \frac{4}{81} \frac{(\rho - \sigma)^2 g^2 r^4}{\eta^2} \]

7. **Cancel \( g \) from both sides:**
Dividing both sides by \( g \) (assuming \( g \neq 0 \)):
\[ 2h = \frac{4}{81} \frac{(\rho - \sigma)^2 g r^4}{\eta^2} \]

8. **Solve for \( h \):**
Rearranging gives:
\[ h = \frac{2}{81} \frac{(\rho - \sigma)^2 g r^4}{\eta^2} \]

### Final Expression:
Thus, the height \( h \) in terms of \( r \), \( \sigma \), \( \rho \), \( \eta \), and \( g \) is:
\[ h = \frac{2}{81} \frac{(\rho - \sigma)^2 g r^4}{\eta^2} \]