A drop of liquid is broken down into `27` identical liquid drops. If the terminal velocity of original liquid drops is `V_(T)`, then find the terminal velocity of the new liquid drop thus formed.

Options

A. `(V_(T))/(3)`
B. `(V_(T))/(9)`
C. `(V_(T))/(27)`
D. `(9V_(T))/(9)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between the original drop and the smaller drops
Let the radius of the original drop be \( R \). When this drop is broken into 27 identical smaller drops, each smaller drop will have a radius \( r \).

### Step 2: Relate the volumes of the drops
The volume of the original drop is given by:
\[ V = \frac{4}{3} \pi R^3 \]
When the original drop is broken into 27 smaller drops, the total volume remains the same. Therefore, the volume of one smaller drop is:
\[ V_{\text{small}} = \frac{1}{27} V = \frac{1}{27} \left( \frac{4}{3} \pi R^3 \right) = \frac{4}{3} \pi r^3 \]

### Step 3: Set the volumes equal
Since the total volume is conserved, we can equate the volumes:
\[ \frac{4}{3} \pi R^3 = 27 \left( \frac{4}{3} \pi r^3 \right) \]
Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ R^3 = 27 r^3 \]

### Step 4: Solve for the radius of the smaller drops
Taking the cube root of both sides, we find:
\[ R = 3r \quad \Rightarrow \quad r = \frac{R}{3} \]

### Step 5: Relate terminal velocities to the radius
The terminal velocity \( V_T \) of a drop is given by the formula:
\[ V_T \propto r^2 \]
This means that the terminal velocity is directly proportional to the square of the radius of the drop.

### Step 6: Calculate the terminal velocity of the smaller drops
Let \( V_T \) be the terminal velocity of the original drop and \( V_n \) be the terminal velocity of the smaller drops. Using the relationship between terminal velocity and radius:
\[ \frac{V_n}{V_T} = \left( \frac{r}{R} \right)^2 \]
Substituting \( r = \frac{R}{3} \):
\[ \frac{V_n}{V_T} = \left( \frac{\frac{R}{3}}{R} \right)^2 = \left( \frac{1}{3} \right)^2 = \frac{1}{9} \]
Thus, we have:
\[ V_n = \frac{V_T}{9} \]

### Final Answer
The terminal velocity of the new liquid drops thus formed is:
\[ V_n = \frac{V_T}{9} \]