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. 1. `32 cm s^(-1)`
B. 2. `30 cms^(-1)`
C. 3. `28 cm s^(-1)`
D. 4. `24 cm s^(-1)`

Explanation

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the problem
We have 8 small drops of water, each with a radius of \( r = 2 \, \text{mm} = 0.2 \, \text{cm} \), falling at a terminal velocity of \( v_1 = 8 \, \text{cm/s} \). When these drops coalesce, they form a single larger drop. We need to find the terminal velocity \( v_2 \) of this larger drop.

### Step 2: Calculate the volume of the small drops
The volume \( V \) of a single drop can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
For one small drop:
\[ V_{\text{small}} = \frac{4}{3} \pi (0.2)^3 = \frac{4}{3} \pi (0.008) = \frac{32}{3} \pi \, \text{cm}^3 \]
For 8 small drops:
\[ V_{\text{total}} = 8 \times V_{\text{small}} = 8 \times \frac{32}{3} \pi = \frac{256}{3} \pi \, \text{cm}^3 \]

### Step 3: Set the volume of the larger drop equal to the total volume of the small drops
Let the radius of the larger drop be \( R \). The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Setting the volumes equal gives:
\[ \frac{4}{3} \pi R^3 = \frac{256}{3} \pi \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 64 \]
Taking the cube root:
\[ R = 4 \, \text{cm} \]

### Step 4: Relate terminal velocities of the small and large drops
The terminal velocity \( v \) of a drop is proportional to the square of its radius:
\[ v \propto R^2 \]
Thus, we can write:
\[ \frac{v_1}{v_2} = \frac{r^2}{R^2} \]
Substituting the known values:
\[ \frac{8}{v_2} = \frac{(0.2)^2}{(4)^2} \]
Calculating the squares:
\[ \frac{8}{v_2} = \frac{0.04}{16} \]
This simplifies to:
\[ \frac{8}{v_2} = \frac{1}{400} \]

### Step 5: Solve for \( v_2 \)
Cross-multiplying gives:
\[ 8 \times 400 = v_2 \]
Thus:
\[ v_2 = 3200 \, \text{cm/s} \]

### Step 6: Final Result
The terminal velocity of the larger drop is:
\[ v_2 = 32 \, \text{cm/s} \]

### Summary
The terminal velocity of the combined drop after coalescing is \( 32 \, \text{cm/s} \).