If T is the surface tension of a liquid, the energy needed to break a liquid drop of radius R into 64 drops is

Options

A. `6 pi R^(2)T`
B. `pi R^(2)T`
C. `12 pi R^(2)T`
D. `8 pi R^(2)T`

Explanation

To solve the problem of finding the energy needed to break a liquid drop of radius \( R \) into 64 smaller drops, we can follow these steps:

### Step 1: Determine the volume of the original drop
The volume \( V \) of a sphere (the original drop) is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]

### Step 2: Set the volume of the original drop equal to the total volume of the smaller drops
When the original drop breaks into 64 smaller drops, the total volume of these smaller drops must equal the volume of the original drop. Let the radius of each smaller drop be \( r \). Thus, we have:
\[ 64 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \]

### Step 3: Simplify the equation to find the radius of the smaller drops
Canceling \( \frac{4}{3} \pi \) from both sides gives:
\[ 64 r^3 = R^3 \]
Now, solving for \( r \):
\[ r^3 = \frac{R^3}{64} \]
Taking the cube root of both sides, we find:
\[ r = \frac{R}{4} \]

### Step 4: Calculate the energy of the original drop
The energy associated with the surface tension of a drop is given by:
\[ E = T \times A \]
where \( A \) is the surface area. The surface area \( A \) of the original drop is:
\[ A = 4 \pi R^2 \]
Thus, the energy of the original drop is:
\[ E_1 = T \times 4 \pi R^2 \]

### Step 5: Calculate the energy of the smaller drops
The surface area of one smaller drop with radius \( r = \frac{R}{4} \) is:
\[ A_{\text{small}} = 4 \pi r^2 = 4 \pi \left(\frac{R}{4}\right)^2 = 4 \pi \frac{R^2}{16} = \frac{\pi R^2}{4} \]
The total energy of the 64 smaller drops is:
\[ E_2 = 64 \times \frac{\pi R^2}{4} \times T = 16 \pi R^2 T \]

### Step 6: Calculate the energy required to break the original drop into smaller drops
The energy required to break the original drop into smaller drops is the difference in energy:
\[ \Delta W = E_2 - E_1 \]
Substituting the values we found:
\[ \Delta W = 16 \pi R^2 T - 4 \pi R^2 T = 12 \pi R^2 T \]

### Final Answer
Thus, the energy needed to break a liquid drop of radius \( R \) into 64 drops is:
\[ \Delta W = 12 \pi R^2 T \]

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