A large number of liquid drops each of radius 'a' coalesce to form a single spherical drop of radius b. The energy released in the process is converted into kinetic energy of the big drops formed. The speed of big drop will be

Options

A. `sqrt((6T)/(rho)[1/a-1/b])`
B. `sqrt((4T)/(rho)[1/a+1/b])`
C. `sqrt((8T)/(rho)[1/a+1/b])`
D. `sqrt((5T)/(rho)[1/a+1/b])`

Explanation

To solve the problem, we need to find the speed of the large drop formed after a number of smaller drops coalesce. We will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Volume Relationship**:
- Let the number of small drops be \( N \), each with radius \( a \).
- The volume of one small drop is given by:
\[ V_{\text{small}} = \frac{4}{3} \pi a^3 \]
- Therefore, the total volume of \( N \) small drops is:
\[ V_{\text{total}} = N \cdot V_{\text{small}} = N \cdot \frac{4}{3} \pi a^3 \]
- The volume of the single large drop with radius \( b \) is:
\[ V_{\text{large}} = \frac{4}{3} \pi b^3 \]
- Setting these two volumes equal gives us:
\[ N \cdot \frac{4}{3} \pi a^3 = \frac{4}{3} \pi b^3 \]
- Simplifying, we find:
\[ N = \frac{b^3}{a^3} \]

2. **Calculating Surface Area Change**:
- The surface area of one small drop is:
\[ A_{\text{small}} = 4 \pi a^2 \]
- The total surface area of \( N \) small drops is:
\[ A_{\text{total}} = N \cdot A_{\text{small}} = \frac{b^3}{a^3} \cdot 4 \pi a^2 = 4 \pi \frac{b^3}{a} \]
- The surface area of the large drop is:
\[ A_{\text{large}} = 4 \pi b^2 \]

3. **Calculating Change in Surface Area**:
- The change in surface area \( \Delta A \) is:
\[ \Delta A = A_{\text{total}} - A_{\text{large}} = 4 \pi \frac{b^3}{a} - 4 \pi b^2 \]
- Factoring out \( 4 \pi \):
\[ \Delta A = 4 \pi \left( \frac{b^3}{a} - b^2 \right) = 4 \pi b^2 \left( \frac{b}{a} - 1 \right) \]

4. **Calculating Energy Released**:
- The energy released due to the change in surface area is given by:
\[ W = T \cdot \Delta A = T \cdot 4 \pi b^2 \left( \frac{b}{a} - 1 \right) \]

5. **Relating Energy to Kinetic Energy**:
- This energy is converted into kinetic energy of the large drop:
\[ W = \frac{1}{2} m v^2 \]
- The mass \( m \) of the large drop is:
\[ m = \rho \cdot V_{\text{large}} = \rho \cdot \frac{4}{3} \pi b^3 \]
- Substituting this into the kinetic energy equation gives:
\[ T \cdot 4 \pi b^2 \left( \frac{b}{a} - 1 \right) = \frac{1}{2} \left( \rho \cdot \frac{4}{3} \pi b^3 \right) v^2 \]

6. **Simplifying the Equation**:
- Cancel \( 4 \pi \) from both sides:
\[ T b^2 \left( \frac{b}{a} - 1 \right) = \frac{2}{3} \rho b^3 v^2 \]
- Rearranging gives:
\[ v^2 = \frac{3T}{2\rho b} \left( \frac{b}{a} - 1 \right) \]

7. **Finding the Speed**:
- Taking the square root gives the speed of the large drop:
\[ v = \sqrt{\frac{6T}{\rho} \left( \frac{1}{a} - \frac{1}{b} \right)} \]

### Final Answer:
The speed of the large drop formed after coalescence is:
\[ v = \sqrt{\frac{6T}{\rho} \left( \frac{1}{a} - \frac{1}{b} \right)} \]