Sixty four spherical rain drops of equal size are falling vertically through air with a terminal velocity `1.5ms^(-1)`. If these drops coalesce to form a big spherical drop, then terminal velocity of big drop is:

Options

A. 8 `ms^(-1)`
B. 16 `ms^(-1)`
C. 24 `ms^(-1)`
D. 32 `ms^(-1)`

Answer

To find the terminal velocity of the big drop formed by the coalescence of 64 smaller drops, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have 64 small spherical raindrops, each with a terminal velocity of \( V_1 = 1.5 \, \text{m/s} \).
- When these drops coalesce, they form a larger spherical drop.

2. **Volume of the Small Drops**:
- The volume \( V \) of a single small drop is given by the formula for the volume of a sphere:
\[ V_1 = \frac{4}{3} \pi r_1^3 \]
- Therefore, the total volume of 64 small drops is:
\[ V_{\text{total}} = 64 \times V_1 = 64 \times \frac{4}{3} \pi r_1^3 = \frac{256}{3} \pi r_1^3 \]

3. **Volume of the Big Drop**:
- Let the radius of the big drop be \( r_2 \). The volume of the big drop is:
\[ V_2 = \frac{4}{3} \pi r_2^3 \]

4. **Setting the Volumes Equal**:
- Since the total volume of the small drops equals the volume of the big drop, we can set the two volumes equal:
\[ \frac{256}{3} \pi r_1^3 = \frac{4}{3} \pi r_2^3 \]
- Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ 64 r_1^3 = r_2^3 \]

5. **Finding the Relationship Between Radii**:
- Taking the cube root of both sides:
\[ r_2 = 4 r_1 \]

6. **Using the Terminal Velocity Formula**:
- The terminal velocity \( V \) of a sphere falling through a fluid is given by:
\[ V = \frac{2}{9} \frac{r^2 (\rho - \rho_f) g}{\eta} \]
- Here, \( \rho \) is the density of the drop, \( \rho_f \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( \eta \) is the coefficient of viscosity.

7. **Relating Terminal Velocities**:
- Since the terminal velocity is proportional to the square of the radius, we can write:
\[ \frac{V_1}{V_2} = \frac{r_1^2}{r_2^2} \]
- Substituting \( r_2 = 4 r_1 \):
\[ \frac{V_1}{V_2} = \frac{r_1^2}{(4 r_1)^2} = \frac{r_1^2}{16 r_1^2} = \frac{1}{16} \]
- Therefore, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = 16 V_1 \]

8. **Calculating the Terminal Velocity of the Big Drop**:
- Substituting \( V_1 = 1.5 \, \text{m/s} \):
\[ V_2 = 16 \times 1.5 = 24 \, \text{m/s} \]

### Final Answer:
The terminal velocity of the big drop is \( \boxed{24 \, \text{m/s}} \).