27 identical drops of water are falling down vertically in air each with a terminal velocity of 0.15 m/s. If they combine to form a single bigger drop , what will be its terminal velocity?

Options

A. 0 m/s
B. 1.35 m/s
C. 0.3 m/s
D. 0.45 m/s

Explanation

To solve the problem of finding the terminal velocity of a single larger drop formed by combining 27 identical smaller drops, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Volume Relationship**:
- The volume of one small drop (radius \( r \)) is given by the formula for the volume of a sphere:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
- Therefore, the total volume of 27 small drops is:
\[ V_{\text{total}} = 27 \times V_{\text{small}} = 27 \times \frac{4}{3} \pi r^3 = 36 \pi r^3 \]

2. **Volume of the Larger Drop**:
- 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 total volume of the smaller drops equal to the volume of the larger drop gives:
\[ 36 \pi r^3 = \frac{4}{3} \pi R^3 \]

3. **Simplifying the Equation**:
- Cancel \( \pi \) from both sides:
\[ 36 r^3 = \frac{4}{3} R^3 \]
- Multiply both sides by 3:
\[ 108 r^3 = 4 R^3 \]
- Dividing both sides by 4:
\[ 27 r^3 = R^3 \]
- Taking the cube root gives:
\[ R = 3r \]

4. **Finding the Terminal Velocity**:
- The terminal velocity \( V \) of a drop is given by the formula:
\[ V = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\mu} \]
- Here, \( \rho \) is the density of water, \( \sigma \) is the density of air, \( g \) is the acceleration due to gravity, and \( \mu \) is the viscosity of air.

5. **Relating the Terminal Velocities**:
- Since terminal velocity is proportional to the square of the radius:
\[ V_1 \propto r_1^2 \quad \text{and} \quad V_2 \propto R^2 \]
- We can write the ratio of the terminal velocities as:
\[ \frac{V_1}{V_2} = \frac{r_1^2}{R^2} \]
- Given \( V_1 = 0.15 \, \text{m/s} \) and \( R = 3r \), we can substitute:
\[ \frac{0.15}{V_2} = \frac{r^2}{(3r)^2} = \frac{r^2}{9r^2} = \frac{1}{9} \]

6. **Calculating \( V_2 \)**:
- Rearranging gives:
\[ V_2 = 0.15 \times 9 = 1.35 \, \text{m/s} \]

### Final Answer:
The terminal velocity of the larger drop is \( \boxed{1.35 \, \text{m/s}} \).