A large number of droplets, each of radius a, coalesce to form a bigger drop of radius `b`. Assume that the energy released in the process is converted into the kinetic energy of the drop. The velocity of the drop is `sigma =` surface tension, `rho =` density)

Options

A. `[(sigma)/(rho)(1/a-1/b)]^(1/2)`
B. `[(2sigma)/(rho)(1/a-1/b)]^(1/2)`
C. `[(3sigma)/(rho)(1/a-1/b)]^(1/2)`
D. `[(6sigma)/(rho)(1/a-1/b)]^(1/2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the number of droplets (n)
Given that we have small droplets of radius \( a \) coalescing to form a larger droplet of radius \( b \), we can express the volumes of the droplets. The volume of one small droplet is given by:

\[ V_a = \frac{4}{3} \pi a^3 \]

The volume of the larger droplet is:

\[ V_b = \frac{4}{3} \pi b^3 \]

Since \( n \) small droplets combine to form one larger droplet, we have:

\[ n \cdot V_a = V_b \]

Substituting the volumes, we get:

\[ n \cdot \frac{4}{3} \pi a^3 = \frac{4}{3} \pi b^3 \]

Cancelling \( \frac{4}{3} \pi \) from both sides, we find:

\[ n = \frac{b^3}{a^3} \]

### Step 2: Calculate the change in surface energy
The surface energy \( E \) is given by the formula:

\[ E = \sigma \cdot \Delta A \]

where \( \Delta A \) is the change in surface area. The surface area of one small droplet is:

\[ A_a = 4 \pi a^2 \]

The surface area of the larger droplet is:

\[ A_b = 4 \pi b^2 \]

The change in surface area when \( n \) droplets coalesce is:

\[ \Delta A = n \cdot A_a - A_b = n \cdot 4 \pi a^2 - 4 \pi b^2 \]

Substituting \( n = \frac{b^3}{a^3} \):

\[ \Delta A = \frac{b^3}{a^3} \cdot 4 \pi a^2 - 4 \pi b^2 \]

Simplifying this gives:

\[ \Delta A = 4 \pi \left( \frac{b^3}{a^3} a^2 - b^2 \right) = 4 \pi \left( \frac{b^3}{a} - b^2 \right) = 4 \pi b^2 \left( \frac{b}{a} - 1 \right) \]

### Step 3: Calculate the surface energy
Now substituting \( \Delta A \) into the surface energy formula:

\[ E = \sigma \cdot \Delta A = \sigma \cdot 4 \pi b^2 \left( \frac{b}{a} - 1 \right) \]

### Step 4: Relate surface energy to kinetic energy
The kinetic energy \( KE \) of the larger droplet can be expressed as:

\[ KE = \frac{1}{2} m v^2 \]

where \( m \) is the mass of the larger droplet:

\[ m = \rho \cdot V_b = \rho \cdot \frac{4}{3} \pi b^3 \]

Thus, the kinetic energy becomes:

\[ KE = \frac{1}{2} \left( \rho \cdot \frac{4}{3} \pi b^3 \right) v^2 \]

### Step 5: Equate surface energy and kinetic energy
Setting the surface energy equal to the kinetic energy:

\[ \sigma \cdot 4 \pi b^2 \left( \frac{b}{a} - 1 \right) = \frac{1}{2} \left( \rho \cdot \frac{4}{3} \pi b^3 \right) v^2 \]

### Step 6: Solve for \( v^2 \)
Cancelling \( 4 \pi b^2 \) from both sides:

\[ \sigma \left( \frac{b}{a} - 1 \right) = \frac{2}{3} \rho b v^2 \]

Rearranging gives:

\[ v^2 = \frac{3 \sigma}{2 \rho} \left( \frac{b}{a} - 1 \right) \]

### Step 7: Final expression for \( v \)
Taking the square root, we find:

\[ v = \sqrt{\frac{3 \sigma}{2 \rho} \left( \frac{b}{a} - 1 \right)} \]

### Final Answer
Thus, the velocity of the drop is given by:

\[ v = \sqrt{\frac{6 \sigma}{\rho} \left( \frac{1}{a} - \frac{1}{b} \right)} \]