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 coalescing eight smaller raindrops, we can follow these steps:

### Step 1: Understand the problem
We have eight raindrops, each with a radius \( R \), falling through air with a terminal velocity of \( 6 \, \text{cm/s} \). We need to find the terminal velocity of a larger drop formed by combining these eight smaller drops.

### Step 2: Use the principle of conservation of volume
The volume of the larger drop must equal the total volume of the eight smaller drops. The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
For the eight smaller drops, the total volume is:
\[ V_{\text{small}} = 8 \times \frac{4}{3} \pi R^3 = \frac{32}{3} \pi R^3 \]
Let the radius of the larger drop be \( R_0 \). The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R_0^3 \]

Setting the two volumes equal gives:
\[ \frac{4}{3} \pi R_0^3 = \frac{32}{3} \pi R^3 \]

### Step 3: Simplify the equation
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ R_0^3 = 8R^3 \]

### Step 4: Solve for \( R_0 \)
Taking the cube root of both sides:
\[ R_0 = 2R \]

### Step 5: Determine the relationship between terminal velocity and radius
The terminal velocity \( v \) of a sphere falling through a fluid is given by the formula:
\[ v = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
where \( r \) is the radius of the sphere, \( \rho \) is the density of the fluid, \( \sigma \) is the density of the sphere, \( g \) is the acceleration due to gravity, and \( \eta \) is the viscosity of the fluid.

### Step 6: Relate the terminal velocities of the smaller and larger drops
Since terminal velocity is proportional to the square of the radius, we can write:
\[ \frac{v_2}{v_1} = \left(\frac{R_0}{R}\right)^2 \]
where \( v_1 \) is the terminal velocity of the smaller drops and \( v_2 \) is the terminal velocity of the larger drop.

### Step 7: Substitute the values
We know \( R_0 = 2R \) and \( v_1 = 6 \, \text{cm/s} \):
\[ \frac{v_2}{6} = \left(\frac{2R}{R}\right)^2 = 4 \]
Thus,
\[ v_2 = 6 \times 4 = 24 \, \text{cm/s} \]

### Conclusion
The terminal velocity of the bigger drop formed by coalescing the eight smaller drops is \( 24 \, \text{cm/s} \).

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