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. `2^(1/2) v`
B. `2^(2/3) v`
C. `2^(3/2) v`
D. None of these

Explanation

To solve the problem of finding the new velocity of two equal drops of water that coalesce while falling through air, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have two equal drops of water, each with a steady velocity \( v \). The drops are falling through air and are assumed to be spherical in shape.

### Step 2: Calculate the Volume of Each Drop
The volume \( V \) of a single spherical drop is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the drop.

### Step 3: Find the Total Volume of Two Drops
Since there are two equal drops, the total volume \( V_{total} \) of the two drops before coalescing is:
\[ V_{total} = 2 \times V = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 4: Determine the Volume of the New Drop After Coalescence
When the two drops coalesce, they form a new drop with a new radius \( R \). The volume of the new drop can be expressed as:
\[ V_{new} = \frac{4}{3} \pi R^3 \]
Since volume is conserved, we set the total volume equal to the volume of the new drop:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

### Step 5: Simplify the Equation
We can cancel \( \frac{4}{3} \pi \) from both sides:
\[ 2r^3 = R^3 \]

### Step 6: Solve for the New Radius
Taking the cube root of both sides gives us:
\[ R = (2)^{1/3} r \]

### Step 7: Relate Terminal Velocity to Radius
The terminal velocity \( v \) of a spherical object falling through a fluid is proportional to the square of its radius:
\[ v \propto R^2 \]
Thus, if the original drops have a velocity \( v \) and radius \( r \), the new velocity \( v' \) for the new drop with radius \( R \) can be expressed as:
\[ v' = k R^2 \]
where \( k \) is a constant of proportionality.

### Step 8: Substitute for the New Radius
Substituting \( R = (2)^{1/3} r \) into the equation for \( v' \):
\[ v' = k \left((2)^{1/3} r\right)^2 = k (2^{2/3} r^2) = 2^{2/3} (k r^2) = 2^{2/3} v \]

### Step 9: Final Expression for New Velocity
Thus, the new velocity \( v' \) after the drops coalesce is:
\[ v' = 2^{2/3} v \]

### Summary of the Solution
The new velocity of the coalesced drop is \( v' = 2^{2/3} v \).