`n` drops of water, each of radius `2 mm`, fall through air at a terminal velocity of `8 cms^(-1)` If they coalesce to form a single drop, then the terminal velocity of the combined drop is `32 cms^(-1)` The value of `n` is

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between the volumes of the drops
Let the radius of the smaller drops be \( r \) and the radius of the larger drop be \( R \). The volume of a single small drop is given by:
\[ V_{\text{small}} = \frac{4}{3} \pi r^3 \]
If there are \( n \) small drops, the total volume of the smaller drops is:
\[ V_{\text{total}} = n \cdot V_{\text{small}} = n \cdot \frac{4}{3} \pi r^3 \]
The volume of the larger drop is:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Since the total volume remains constant when the drops coalesce, we have:
\[ n \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]
Cancelling out the common terms, we get:
\[ n r^3 = R^3 \tag{1} \]

### Step 2: Relate the terminal velocities of the drops
The terminal velocity \( v \) of a drop is given by:
\[ v \propto r^2 \]
This means that the terminal velocity of the larger drop \( v' \) and the smaller drop \( v \) can be expressed as:
\[ \frac{v'}{v} = \frac{R^2}{r^2} \tag{2} \]
Given that \( v' = 32 \, \text{cm/s} \) and \( v = 8 \, \text{cm/s} \), we can substitute these values into equation (2):
\[ \frac{32}{8} = \frac{R^2}{r^2} \]
This simplifies to:
\[ 4 = \frac{R^2}{r^2} \]
Taking the square root of both sides gives:
\[ \frac{R}{r} = 2 \tag{3} \]

### Step 3: Substitute equation (3) into equation (1)
From equation (3), we can express \( R \) in terms of \( r \):
\[ R = 2r \]
Substituting this into equation (1):
\[ n r^3 = (2r)^3 \]
This simplifies to:
\[ n r^3 = 8r^3 \]
Dividing both sides by \( r^3 \) (assuming \( r \neq 0 \)):
\[ n = 8 \]

### Conclusion
The value of \( n \) is \( 8 \).

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