Eight raindrops each of radius R fall through air with teminal velocity 6 cm `s^(-1)`. What is the terminal velocity of the bigger drop formed by coalescing these drops together ?

Options

A. 18 cm `s^(-1)`
B. 24 cm `s^(-1)`
C. 15 cm `s^(-1)`
D. 20 cm `s^(-1)`

Answer

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

### Step 1: Determine the Volume of the Smaller Drops
The volume \( V \) of a single raindrop with radius \( R \) is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Since there are 8 smaller drops, the total volume of the 8 drops is:
\[ V_{\text{total}} = 8 \times \frac{4}{3} \pi R^3 = \frac{32}{3} \pi R^3 \]

### Step 2: Set the Volume of the Bigger Drop
Let the radius of the bigger drop be \( r \). The volume of the bigger drop is:
\[ V_{\text{big}} = \frac{4}{3} \pi r^3 \]
Since the volume of the bigger drop is equal to the total volume of the smaller drops, we have:
\[ \frac{4}{3} \pi r^3 = \frac{32}{3} \pi R^3 \]

### Step 3: Simplify the Volume Equation
We can cancel \(\frac{4}{3} \pi\) from both sides:
\[ r^3 = 8 R^3 \]

### Step 4: Solve for the Radius of the Bigger Drop
Taking the cube root of both sides gives:
\[ r = 2R \]

### Step 5: Relate Terminal Velocity to Radius
The terminal velocity \( V_t \) of a drop is given by the formula:
\[ V_t = \frac{2}{9} \frac{r^2 g}{\eta} (\rho_{\text{body}} - \rho_{\text{fluid}}) \]
From this equation, we see that terminal velocity is directly proportional to the square of the radius:
\[ V_t \propto r^2 \]

### Step 6: Set Up the Proportionality for Terminal Velocities
Let \( V_{t1} \) be the terminal velocity of the smaller drops and \( V_{t2} \) be the terminal velocity of the bigger drop. We can write:
\[ \frac{V_{t1}}{V_{t2}} = \frac{R^2}{(2R)^2} \]
This simplifies to:
\[ \frac{V_{t1}}{V_{t2}} = \frac{R^2}{4R^2} = \frac{1}{4} \]
Thus, we can express \( V_{t2} \) in terms of \( V_{t1} \):
\[ V_{t2} = 4 V_{t1} \]

### Step 7: Substitute the Known Terminal Velocity
Given that \( V_{t1} = 6 \, \text{cm/s} \):
\[ V_{t2} = 4 \times 6 \, \text{cm/s} = 24 \, \text{cm/s} \]

### Final Answer
The terminal velocity of the bigger drop is:
\[ \boxed{24 \, \text{cm/s}} \]