A ball of radius r and density ρ falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is `eta` the value of h is given by
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Options

A. `2/9r^2((1-rho)/eta)g`
B. `2/(81)r^2((rho-1)/eta)g`
C. `2/81r^4((rho-1)/eta)^2g`
D. `2/9r^4((rho-1)/eta)^2g`

Answer

To solve the problem step by step, we will analyze the motion of the ball as it falls freely under gravity and then enters the water.

### Step 1: Determine the velocity of the ball just before it enters the water
The ball falls freely under gravity from a height \( h \). We can use the kinematic equation for uniformly accelerated motion:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) = final velocity
- \( u \) = initial velocity (which is 0)
- \( a \) = acceleration due to gravity (\( g \))
- \( s \) = distance fallen (\( h \))

Substituting the values, we have:

\[ v^2 = 0 + 2gh \]

Thus, we find:

\[ v = \sqrt{2gh} \]

### Step 2: Understand the terminal velocity in water
When the ball enters the water, it reaches a terminal velocity, which remains constant. The formula for terminal velocity \( V_t \) of a sphere falling through a viscous fluid is given by:

\[ V_t = \frac{2}{9} \cdot \frac{r^2 g (\rho - \sigma)}{\eta} \]

Where:
- \( r \) = radius of the ball
- \( g \) = acceleration due to gravity
- \( \rho \) = density of the ball
- \( \sigma \) = density of the fluid (water)
- \( \eta \) = viscosity of the fluid

### Step 3: Set the velocities equal
Since the velocity of the ball does not change upon entering the water, we can equate the velocity just before entering the water to the terminal velocity:

\[ \sqrt{2gh} = \frac{2}{9} \cdot \frac{r^2 g (\rho - \sigma)}{\eta} \]

### Step 4: Square both sides to eliminate the square root
Squaring both sides gives:

\[ 2gh = \left(\frac{2}{9} \cdot \frac{r^2 g (\rho - \sigma)}{\eta}\right)^2 \]

### Step 5: Rearranging the equation
Now, we can rearrange this equation to solve for \( h \):

\[ h = \frac{1}{2g} \left(\frac{2}{9} \cdot \frac{r^2 g (\rho - \sigma)}{\eta}\right)^2 \]

### Step 6: Simplifying the expression
Now, simplify the right-hand side:

\[ h = \frac{1}{2g} \cdot \frac{4}{81} \cdot \frac{r^4 g^2 (\rho - \sigma)^2}{\eta^2} \]

### Step 7: Final expression for \( h \)
Now, we can simplify further:

\[ h = \frac{2}{81} \cdot \frac{r^4 g}{\eta^2 (\rho - \sigma)^2} \]

This gives us the final expression for \( h \):

\[ h = \frac{2}{81} \cdot \frac{r^4 g}{\eta^2 (\rho - \sigma)^2} \]