Two spherical raindrops of equal size are falling vertically through air with a velocity of `1 m//s`. What would be the terminal speed if these two drops were to coalesce to form a large spherical drop?

Explanation

To solve the problem of finding the terminal speed of a larger spherical raindrop formed by the coalescence of two smaller raindrops, we can follow these steps:

### Step 1: Understand the relationship between terminal velocity and radius
The terminal velocity \( V_t \) of a spherical object falling through a fluid is directly proportional to the square of its radius \( r \). Mathematically, this can be expressed as:
\[ V_t \propto r^2 \]

### Step 2: Calculate the volume of the two smaller raindrops
Let the radius of each smaller raindrop be \( r \). The volume \( V \) of a single spherical raindrop is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Since there are two raindrops, the total volume of the two smaller raindrops is:
\[ V_{total} = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 3: Determine the radius of the larger raindrop
When the two smaller raindrops coalesce, they form a larger raindrop. Let the radius of the larger raindrop be \( R \). The volume of the larger raindrop is:
\[ V_{large} = \frac{4}{3} \pi R^3 \]
Setting the total volume of the two smaller raindrops equal to the volume of the larger raindrop gives:
\[ \frac{8}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides, we get:
\[ 2r^3 = R^3 \]

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

### Step 5: Relate the terminal velocities of the smaller and larger raindrops
Using the relationship between terminal velocity and radius, we can express the terminal velocity of the larger raindrop \( V_{t}' \) in terms of the terminal velocity of the smaller raindrop \( V_t \):
\[ \frac{V_{t}'}{V_t} = \left(\frac{R}{r}\right)^2 \]
Substituting \( R = (2)^{1/3} r \) into the equation gives:
\[ \frac{V_{t}'}{V_t} = \left(\frac{(2)^{1/3} r}{r}\right)^2 = (2)^{2/3} \]

### Step 6: Calculate the terminal velocity of the larger raindrop
Now we can express \( V_{t}' \):
\[ V_{t}' = V_t \cdot (2)^{2/3} \]
Given that the terminal velocity \( V_t \) of the smaller raindrops is \( 1 \, \text{m/s} \):
\[ V_{t}' = 1 \cdot (2)^{2/3} \]
Calculating \( (2)^{2/3} \):
\[ (2)^{2/3} \approx 1.587 \]
Thus, the terminal velocity of the larger raindrop is approximately:
\[ V_{t}' \approx 1.587 \, \text{m/s} \]

### Final Answer
The terminal speed of the larger spherical drop formed by the coalescence of the two smaller raindrops is approximately \( 1.587 \, \text{m/s} \).

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