Twenty-seven drops of radius 2 mm each falling downwards with a terminal velocity of `4 ms^(-1)` coalesce to form a bigger drop. What will be the terminal velocity of the bigger drop?

Answer

To solve the problem of finding the terminal velocity of a bigger drop formed by the coalescence of 27 smaller drops, we can follow these steps:

### Step 1: Calculate the Volume of the Smaller Drops
The volume \( V \) of a single drop can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the drop. Given that the radius of each smaller drop is \( 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \), the volume of one small drop is:
\[ V_{\text{small}} = \frac{4}{3} \pi (2 \times 10^{-3})^3 \]

### Step 2: Calculate the Total Volume of the 27 Smaller Drops
Since there are 27 drops, the total volume \( V_{\text{total}} \) of the smaller drops is:
\[ V_{\text{total}} = 27 \times V_{\text{small}} = 27 \times \frac{4}{3} \pi (2 \times 10^{-3})^3 \]

### Step 3: Calculate the Volume of the Bigger Drop
The volume of the bigger drop formed by the coalescence of the smaller drops is equal to the total volume of the smaller drops:
\[ V_{\text{big}} = V_{\text{total}} \]

### Step 4: Relate the Volume of the Bigger Drop to its Radius
The volume of the bigger drop can also be expressed as:
\[ V_{\text{big}} = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the bigger drop. Setting the two volumes equal gives:
\[ \frac{4}{3} \pi R^3 = 27 \times \frac{4}{3} \pi (2 \times 10^{-3})^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides results in:
\[ R^3 = 27 \times (2 \times 10^{-3})^3 \]

### Step 5: Solve for the Radius of the Bigger Drop
Taking the cube root of both sides, we find:
\[ R = (27 \times (2 \times 10^{-3})^3)^{1/3} = 3 \times (2 \times 10^{-3}) = 6 \times 10^{-3} \, \text{m} = 6 \, \text{mm} \]

### Step 6: Use the Terminal Velocity Formula
The terminal velocity \( V_t \) of a drop is given by:
\[ V_t \propto r^2 \]
This means that the terminal velocity of the smaller drop \( V_{t1} \) and the bigger drop \( V_{t2} \) can be related as follows:
\[ \frac{V_{t1}}{V_{t2}} = \frac{r_1^2}{R^2} \]
where \( r_1 \) is the radius of the smaller drop and \( R \) is the radius of the bigger drop.

### Step 7: Substitute Known Values
Given \( V_{t1} = 4 \, \text{m/s} \) and \( r_1 = 2 \times 10^{-3} \, \text{m} \), we can substitute:
\[ \frac{4}{V_{t2}} = \frac{(2 \times 10^{-3})^2}{(6 \times 10^{-3})^2} \]

### Step 8: Solve for \( V_{t2} \)
This simplifies to:
\[ \frac{4}{V_{t2}} = \frac{4 \times 10^{-6}}{36 \times 10^{-6}} = \frac{1}{9} \]
Thus,
\[ V_{t2} = 4 \times 9 = 36 \, \text{m/s} \]

### Final Answer
The terminal velocity of the bigger drop is \( 36 \, \text{m/s} \).

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