Eight drops of water, each of radius `2mm` are falling through air at a terminal velcity of `8cms^-1`. If they coalesce to form a single drop, then the terminal velocity of combined drop will be

Options

A. `32cms^-1`
B. `30cms^-1`
C. `28cms^-1`
D. `24cms^-1`

Explanation

To find the terminal velocity of a single drop formed by the coalescence of eight smaller drops, we can follow these steps:

### Step 1: Understand the relationship between terminal velocity and radius
The terminal velocity \( v \) of a drop is given by the formula:
\[ v \propto r^2 \]
This means that the terminal velocity is directly proportional to the square of the radius of the drop.

### Step 2: Calculate the radius of the new drop
When the eight drops of radius \( r \) coalesce to form a single drop, the volume is conserved. The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
For eight drops, the total volume is:
\[ V_{\text{total}} = 8 \times \frac{4}{3} \pi r^3 \]
For the single drop of radius \( R \), the volume is:
\[ V_{\text{single}} = \frac{4}{3} \pi R^3 \]
Setting these equal gives:
\[ 8 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 8r^3 = R^3 \]
Taking the cube root of both sides, we find:
\[ R = 2r \]

### Step 3: Relate the terminal velocities of the small and large drops
From the proportionality established earlier, we can write:
\[ \frac{v}{v_c} = \left(\frac{r}{R}\right)^2 \]
Where \( v \) is the terminal velocity of the smaller drops and \( v_c \) is the terminal velocity of the combined drop.

### Step 4: Substitute the values
We know:
- \( v = 8 \, \text{cm/s} \)
- \( R = 2r \)

Substituting \( R \) into the equation:
\[ \frac{8}{v_c} = \left(\frac{r}{2r}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
Rearranging gives:
\[ v_c = 8 \times 4 = 32 \, \text{cm/s} \]

### Conclusion
The terminal velocity of the single drop formed by the coalescence of the eight smaller drops is:
\[ \boxed{32 \, \text{cm/s}} \]