A large number of liquid drops each of radius 'a' coalesce to form a single spherical drop of radish 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 of finding the speed of the big drop formed by the coalescence of smaller drops, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Volume Conservation**:
- Let the number of smaller drops be \( n \) and the radius of each smaller drop be \( a \).
- The volume of one small drop is given by:
\[ V_{\text{small}} = \frac{4}{3} \pi a^3 \]
- Therefore, the total volume of \( n \) smaller drops is:
\[ V_{\text{total}} = n \cdot \frac{4}{3} \pi a^3 \]
- The volume of the larger drop with radius \( b \) is:
\[ V_{\text{big}} = \frac{4}{3} \pi b^3 \]
- Setting the total volume of smaller drops equal to the volume of the larger drop gives:
\[ n \cdot \frac{4}{3} \pi a^3 = \frac{4}{3} \pi b^3 \]
- Simplifying this, we find:
\[ n = \frac{b^3}{a^3} \]

**Hint**: Remember that the volume of a sphere is proportional to the cube of its radius.

2. **Calculate the Change in Surface Area**:
- The surface area of one small drop is:
\[ A_{\text{small}} = 4 \pi a^2 \]
- The total surface area of \( n \) smaller drops is:
\[ A_{\text{total}} = n \cdot 4 \pi a^2 = \frac{b^3}{a^3} \cdot 4 \pi a^2 = 4 \pi \frac{b^3}{a} \]
- The surface area of the larger drop is:
\[ A_{\text{big}} = 4 \pi b^2 \]
- The change in surface area \( \Delta A \) is:
\[ \Delta A = A_{\text{total}} - A_{\text{big}} = 4 \pi \left( \frac{b^3}{a} - b^2 \right) = 4 \pi b^2 \left( \frac{b}{a} - 1 \right) \]

**Hint**: The change in surface area is crucial as it relates to the energy released during the coalescence.

3. **Calculate the Work Done**:
- The work done \( W \) due to the change in surface area is given by:
\[ W = T \cdot \Delta A = T \cdot 4 \pi b^2 \left( \frac{b}{a} - 1 \right) \]

**Hint**: Work done is related to the surface tension and the change in area.

4. **Relate Work Done to Kinetic Energy**:
- The kinetic energy \( KE \) of the larger drop is given by:
\[ KE = \frac{1}{2} mv^2 \]
- The mass \( m \) of the larger drop can be expressed as:
\[ m = \rho V_{\text{big}} = \rho \cdot \frac{4}{3} \pi b^3 \]
- Therefore, the kinetic energy becomes:
\[ KE = \frac{1}{2} \left( \rho \cdot \frac{4}{3} \pi b^3 \right) v^2 \]

**Hint**: Kinetic energy is derived from the mass and velocity of the drop.

5. **Equate Work Done to Kinetic Energy**:
- Setting the work done equal to the kinetic energy gives:
\[ T \cdot 4 \pi b^2 \left( \frac{b}{a} - 1 \right) = \frac{1}{2} \left( \rho \cdot \frac{4}{3} \pi b^3 \right) v^2 \]
- Cancel \( 4 \pi \) from both sides:
\[ T b^2 \left( \frac{b}{a} - 1 \right) = \frac{2}{3} \rho b^3 v^2 \]

**Hint**: Simplifying the equation helps isolate the velocity.

6. **Solve for Velocity \( v \)**:
- Rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{3T}{2\rho} \left( \frac{b}{a} - 1 \right) \]
- Taking the square root gives:
\[ v = \sqrt{\frac{3T}{2\rho} \left( \frac{b}{a} - 1 \right)} \]

**Hint**: The final step involves taking the square root to find the speed.

### Final Answer:
The speed of the big drop formed is:
\[ v = \sqrt{\frac{3T}{2\rho} \left( \frac{b}{a} - 1 \right)} \]