Eight rain drops of radius `1mm` each falling downwards with a terminal velocity of `5 cm s^(-1)` collapse to form a bigger drop. Find the terminal velocity of bigger drop.

Explanation

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

### Step 1: Understand the relationship between radius and terminal velocity
The terminal velocity \( v \) of a drop is directly proportional to the square of its radius \( r \). This can be expressed mathematically as:
\[ v \propto r^2 \]

### Step 2: Calculate the volume of the smaller drops
The volume \( V \) of a single small drop with radius \( r = 1 \text{ mm} = 0.1 \text{ cm} \) can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
Substituting \( r = 0.1 \text{ cm} \):
\[ V = \frac{4}{3} \pi (0.1)^3 = \frac{4}{3} \pi (0.001) = \frac{4\pi}{3000} \text{ cm}^3 \]

### Step 3: Calculate the total volume of the eight smaller drops
Since there are 8 smaller drops, the total volume \( V_{total} \) of the smaller drops is:
\[ V_{total} = 8 \times V = 8 \times \frac{4}{3} \pi (0.1)^3 = \frac{32}{3} \pi (0.001) = \frac{32\pi}{3000} \text{ cm}^3 \]

### Step 4: Find the radius of the bigger drop
The volume of the bigger drop formed from the smaller drops is equal to the total volume of the smaller drops. Let \( R \) be the radius of the bigger drop. The volume of the bigger drop can be expressed as:
\[ V_{big} = \frac{4}{3} \pi R^3 \]
Setting this equal to the total volume of the smaller drops:
\[ \frac{4}{3} \pi R^3 = \frac{32}{3} \pi (0.001) \]
Cancelling \( \frac{4}{3} \pi \) from both sides:
\[ R^3 = 8 \times 0.001 \]
\[ R^3 = 0.008 \]
Taking the cube root:
\[ R = \sqrt[3]{0.008} = 0.2 \text{ cm} = 2 \text{ mm} \]

### Step 5: Calculate the terminal velocity of the bigger drop
Since the terminal velocity is proportional to the square of the radius, we can find the terminal velocity \( V_{big} \) of the bigger drop using the terminal velocity of the smaller drops. Given that the terminal velocity of the smaller drops \( v_{small} = 5 \text{ cm/s} \) and the radius of the smaller drops \( r_{small} = 0.1 \text{ cm} \):
\[ \frac{v_{big}}{v_{small}} = \left(\frac{R}{r_{small}}\right)^2 \]
Substituting the values:
\[ \frac{v_{big}}{5} = \left(\frac{0.2}{0.1}\right)^2 = 2^2 = 4 \]
Thus,
\[ v_{big} = 5 \times 4 = 20 \text{ cm/s} \]

### Final Answer
The terminal velocity of the bigger drop is \( 20 \text{ cm/s} \).

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