Eight equal drops of water are falling through air with a steady velocity of 10 `cms^(-1)`. If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is

Options

A. 80 `cms^(-1)`
B. 30 `cms^(-1)`
C. 10 `cms^(-1)`
D. 40 `cms^(-1)`

Explanation

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

### Step 1: Understand the relationship between terminal velocity and radius
The terminal velocity (V) of a drop is directly proportional to the square of its radius (R). This can be expressed as:
\[ V \propto R^2 \]
This means that if we know the radius of the drops, we can relate their terminal velocities.

### Step 2: Calculate the volume of the drops
The volume of a single spherical drop is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
If we have eight small drops, the total volume of the eight drops combined is:
\[ V_{\text{total}} = 8 \times \frac{4}{3} \pi R_1^3 = \frac{32}{3} \pi R_1^3 \]

### Step 3: Set the volume of the large drop equal to the total volume of the small drops
When the eight small drops combine to form a single large drop, the volume of the large drop (with radius \( R_2 \)) is:
\[ V_{\text{large}} = \frac{4}{3} \pi R_2^3 \]
Setting these two volumes equal gives us:
\[ \frac{32}{3} \pi R_1^3 = \frac{4}{3} \pi R_2^3 \]

### Step 4: Simplify the equation
We can cancel out \( \frac{4}{3} \pi \) from both sides:
\[ 32 R_1^3 = R_2^3 \]
From this, we can express the relationship between the radii:
\[ R_2^3 = 32 R_1^3 \]
Taking the cube root of both sides:
\[ R_2 = 2 R_1 \]

### Step 5: Relate the terminal velocities of the small and large drops
Using the relationship of terminal velocities:
\[ \frac{V_1}{V_2} = \left(\frac{R_1}{R_2}\right)^2 \]
Where \( V_1 \) is the terminal velocity of the small drops (10 cm/s), and \( V_2 \) is the terminal velocity of the large drop. Substituting \( R_2 = 2 R_1 \):
\[ \frac{V_1}{V_2} = \left(\frac{R_1}{2 R_1}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 6: Solve for the terminal velocity of the large drop
Substituting the known values:
\[ \frac{10 \, \text{cm/s}}{V_2} = \frac{1}{4} \]
Cross-multiplying gives:
\[ 10 = \frac{1}{4} V_2 \]
Thus:
\[ V_2 = 10 \times 4 = 40 \, \text{cm/s} \]

### Final Answer
The terminal velocity of the large drop is:
\[ V_2 = 40 \, \text{cm/s} \]

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