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 such that its velocity remains constant when it enters a liquid. The forces acting on the ball when it enters the liquid must balance out.

### Step-by-step Solution:

1. **Identify the forces acting on the ball:**
- The weight of the ball \( W = mg \) acts downward.
- The buoyant force \( F_B \) acts upward, given by \( F_B = \sigma V \), where \( V \) is the volume of the ball.
- The viscous force \( F_v \) also acts upward, given by \( F_v = 6 \pi \eta r v \), where \( v \) is the velocity of the ball.

2. **Set up the equation for forces:**
When the ball enters the liquid and its velocity does not change, the net force acting on it must be zero:
\[ mg - F_B - F_v = 0 \]

3. **Express the forces in terms of density and volume:**
The volume \( V \) of the ball is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the buoyant force becomes:
\[ F_B = \sigma \left(\frac{4}{3} \pi r^3\right) \]

4. **Substituting the forces into the equation:**
Substituting \( W \), \( F_B \), and \( F_v \) into the force balance equation gives:
\[ \rho \left(\frac{4}{3} \pi r^3\right) g - \sigma \left(\frac{4}{3} \pi r^3\right) - 6 \pi \eta r v = 0 \]

5. **Simplifying the equation:**
Factor out \( \frac{4}{3} \pi r^3 \):
\[ \frac{4}{3} \pi r^3 (\rho g - \sigma) - 6 \pi \eta r v = 0 \]
Rearranging gives:
\[ \rho g - \sigma = \frac{6 \eta v}{\frac{4}{3} r^2} \]

6. **Solving for velocity \( v \):**
Rearranging the equation gives:
\[ v = \frac{3}{8} r^2 (\rho g - \sigma) / \eta \]

7. **Using kinematic equations to relate height \( h \) and velocity \( v \):**
The velocity of the ball just before it hits the liquid can be calculated using the third equation of motion:
\[ v^2 = u^2 + 2gh \]
Since the initial velocity \( u = 0 \):
\[ v^2 = 2gh \]

8. **Equating the two expressions for \( v \):**
Setting the two expressions for \( v \) equal gives:
\[ \frac{3}{8} r^2 \frac{(\rho g - \sigma)}{\eta} = \sqrt{2gh} \]

9. **Squaring both sides:**
\[ \left(\frac{3}{8} r^2 \frac{(\rho g - \sigma)}{\eta}\right)^2 = 2gh \]

10. **Rearranging to find \( h \):**
\[ h = \frac{9 r^4 (\rho g - \sigma)^2}{128 \eta^2 g} \]

### Final Result:
Thus, the height \( h \) in terms of \( r \), \( \sigma \), \( \rho \), \( \eta \), and \( g \) is:
\[ h = \frac{9 r^4 (\rho - \sigma)^2}{128 \eta^2 g} \]