If a number of little droplets of water of surface tension `sigma` all of the same radius r combine to form a single drop of radius R and the energy released in converted into kinetic energy find the velocity acquired by the bigger drop.

Answer

To solve the problem of finding the velocity acquired by a bigger drop formed from smaller droplets of water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Surface Areas**:
- The surface area of one small droplet of radius \( r \) is given by:
\[ A_{\text{small}} = 4\pi r^2 \]
- If there are \( n \) small droplets, the total initial surface area is:
\[ A_{\text{initial}} = n \cdot A_{\text{small}} = n \cdot 4\pi r^2 \]
- The surface area of the larger droplet of radius \( R \) is:
\[ A_{\text{final}} = 4\pi R^2 \]

2. **Calculate the Change in Surface Area**:
- The change in surface area \( \Delta A \) when the droplets combine is:
\[ \Delta A = A_{\text{initial}} - A_{\text{final}} = n \cdot 4\pi r^2 - 4\pi R^2 \]
- Simplifying this gives:
\[ \Delta A = 4\pi (n r^2 - R^2) \]

3. **Calculate the Energy Released**:
- The energy released due to the change in surface area is given by:
\[ \Delta U = \sigma \cdot \Delta A = \sigma \cdot 4\pi (n r^2 - R^2) \]

4. **Relate Energy Released to Kinetic Energy**:
- The energy released is converted into kinetic energy of the larger droplet:
\[ \Delta U = \frac{1}{2} m v^2 \]
- The mass \( m \) of the larger droplet can be expressed as:
\[ m = \rho \cdot V = \rho \cdot \left(\frac{4}{3}\pi R^3\right) \]
- Thus, we have:
\[ \Delta U = \sigma \cdot 4\pi (n r^2 - R^2) = \frac{1}{2} \left(\rho \cdot \frac{4}{3}\pi R^3\right) v^2 \]

5. **Equate and Solve for Velocity \( v \)**:
- Equating the two expressions for energy:
\[ \sigma \cdot 4\pi (n r^2 - R^2) = \frac{1}{2} \left(\rho \cdot \frac{4}{3}\pi R^3\right) v^2 \]
- Cancel \( 4\pi \) from both sides:
\[ \sigma (n r^2 - R^2) = \frac{1}{2} \cdot \frac{\rho}{3} R^3 v^2 \]
- Rearranging gives:
\[ v^2 = \frac{6\sigma(n r^2 - R^2)}{\rho R^3} \]
- Taking the square root:
\[ v = \sqrt{\frac{6\sigma(n r^2 - R^2)}{\rho R^3}} \]

### Final Answer:
The velocity acquired by the bigger drop is:
\[ v = \sqrt{\frac{6\sigma(n r^2 - R^2)}{\rho R^3}} \]