Two equal drops of water are falling through air with a steady velocity v. If the drops coalesced, what will be the new velocity?

Options

A. 2v
B. `sqrt(2)v`
C. `2^(2//3)v`
D. `(v)/(sqrt(2))`

Explanation

To solve the problem of two equal drops of water coalescing and finding the new velocity, we can follow these steps:

### Step 1: Understand the initial conditions
We have two equal drops of water, each with a radius \( r \), falling through the air with a steady velocity \( v \).

### Step 2: Calculate the volume of the drops
The volume \( V \) of a single drop can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Since there are two drops, the total volume \( V_{total} \) of the two drops is:
\[ V_{total} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 3: Determine the radius of the new drop after coalescence
When the two drops coalesce, they form a single larger drop. Let the radius of the new drop 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 of the two original drops equal to the volume of the new drop gives us:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
We can simplify this equation by canceling \( \frac{4}{3} \pi \) from both sides:
\[ 2r^3 = R^3 \]

### Step 4: Solve for the radius of the new drop
Taking the cube root of both sides, we find:
\[ R = (2)^{1/3} r \]

### Step 5: Relate terminal velocity to radius
The terminal velocity \( v \) of a drop is proportional to the square of its radius:
\[ v \propto R^2 \]
Thus, we can express the terminal velocity of the new drop \( v' \) in terms of the original velocity \( v \):
\[ v' = k R^2 \]
where \( k \) is a constant of proportionality.

### Step 6: Substitute the new radius into the velocity equation
Substituting \( R = (2)^{1/3} r \) into the equation for \( v' \):
\[ v' = k \left((2)^{1/3} r\right)^2 = k \cdot (2^{2/3}) r^2 \]
Since \( v = k r^2 \), we can express \( v' \) in terms of \( v \):
\[ v' = (2^{2/3}) v \]

### Conclusion
Thus, the new velocity \( v' \) of the coalesced drop is:
\[ v' = 2^{2/3} v \]