Two drops of same radius are falling through air with steady speed v. If the two drops coalesce, what would be the terminal speed –

Options

A. v
B. `2v`
C. `3v`
D. None of these

Answer

To solve the problem of determining the terminal speed of two coalesced drops, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have two drops of the same radius \( r \) falling through air with a steady speed \( v \). When these two drops coalesce, they form a larger drop.

2. **Volume of the Drops**:
The volume \( V \) of a single spherical drop is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Therefore, the total volume of the two drops is:
\[ V_{total} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

3. **Finding the Radius of the New Drop**:
Let the radius of the new drop formed after coalescence be \( R \). The volume of the new drop can also be expressed as:
\[ V_{new} = \frac{4}{3} \pi R^3 \]
Setting the total volume equal to the new drop's volume gives:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 2r^3 = R^3 \]
Taking the cube root of both sides, we find:
\[ R = 2^{1/3} r \]

4. **Terminal Speed Relation**:
The terminal speed \( v_t \) of a drop is proportional to the square of its radius:
\[ v_t \propto R^2 \]
For the original drops, we have:
\[ v \propto r^2 \]
Therefore, we can write the ratio of the terminal speeds before and after coalescence:
\[ \frac{v_{new}}{v} = \frac{R^2}{r^2} \]

5. **Substituting for \( R \)**:
Substitute \( R = 2^{1/3} r \) into the equation:
\[ \frac{v_{new}}{v} = \frac{(2^{1/3} r)^2}{r^2} = \frac{2^{2/3} r^2}{r^2} = 2^{2/3} \]

6. **Finding the New Terminal Speed**:
Therefore, we can express the new terminal speed \( v_{new} \) as:
\[ v_{new} = v \cdot 2^{2/3} \]

### Final Answer:
The terminal speed of the coalesced drop is \( v_{new} = v \cdot 2^{2/3} \).