Eight spherical drops of equal size are falling vertically through air with a terminal velocity `0.1m//s`. If the drops coalesce to form a large spherical drop it is terminal velocity would be.

Options

A. `0.2 m//s`
B. `0.1 m//s`
C. `0.4 m//s`
D. `0.005 m//s`

Explanation

To solve the problem, we need to find the terminal velocity of a large spherical drop formed by the coalescence of eight smaller spherical drops, each with a terminal velocity of 0.1 m/s.

### Step-by-Step Solution:

1. **Understand the Volume Relationship:**
- The volume of a single small drop (V) is given by the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
- For eight small drops, the total volume (V_total) is:
\[ V_{total} = 8 \times V = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]

2. **Volume of the Large Drop:**
- Let the radius of the large drop be \( R \). The volume of the large drop is:
\[ V_{large} = \frac{4}{3} \pi R^3 \]
- Setting the total volume of the small drops equal to the volume of the large drop:
\[ \frac{32}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
- Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ 8r^3 = R^3 \]

3. **Finding the Radius of the Large Drop:**
- Taking the cube root of both sides:
\[ R = 2r \]

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

5. **Proportionality of Terminal Velocity to Radius:**
- The terminal velocity is proportional to the square of the radius:
\[ V_t \propto r^2 \]
- Therefore, if the radius changes, the terminal velocity changes according to the square of the ratio of the new radius to the old radius.

6. **Calculating the New Terminal Velocity:**
- Let \( V' \) be the terminal velocity of the large drop. We can express this as:
\[ \frac{V'}{V} = \left(\frac{R}{r}\right)^2 \]
- Substituting \( R = 2r \):
\[ \frac{V'}{V} = \left(\frac{2r}{r}\right)^2 = 4 \]
- Therefore, \( V' = 4V \).

7. **Substituting the Known Terminal Velocity:**
- Given that the terminal velocity \( V \) of the small drops is 0.1 m/s:
\[ V' = 4 \times 0.1 \, \text{m/s} = 0.4 \, \text{m/s} \]

### Final Answer:
The terminal velocity of the large spherical drop is **0.4 m/s**.