The work done to split a liquid drop id radius `R` into `N` identical drops is (taken `sigma` as the surface tension of the liquid)

Options

A. `4piR^(2)(N^(-1//3)-1)sigma`
B. `4piR^(2)N sigma`
C. `4piR^(2)(N^(-1//2)-1)`
D. None of these

Explanation

To solve the problem of finding the work done to split a liquid drop of radius \( R \) into \( N \) identical drops, we can follow these steps:

### Step 1: Understand the Volume Conservation
The volume of the original drop must equal the total volume of the \( N \) smaller drops. The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
For \( N \) identical smaller drops, each with radius \( r \), the total volume is:
\[ V' = N \times \frac{4}{3} \pi r^3 \]
Setting the two volumes equal gives:
\[ \frac{4}{3} \pi R^3 = N \times \frac{4}{3} \pi r^3 \]
By simplifying, we can cancel out \( \frac{4}{3} \pi \):
\[ R^3 = N r^3 \]

### Step 2: Solve for the Radius of the Smaller Drops
From the equation \( R^3 = N r^3 \), we can solve for \( r \):
\[ r^3 = \frac{R^3}{N} \implies r = R \left(\frac{1}{N}\right)^{1/3} \]

### Step 3: Calculate the Surface Area Change
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
The surface area of the original drop is:
\[ A_{\text{original}} = 4 \pi R^2 \]
The surface area of one smaller drop is:
\[ A_{\text{small}} = 4 \pi r^2 = 4 \pi \left(R \left(\frac{1}{N}\right)^{1/3}\right)^2 = 4 \pi R^2 \left(\frac{1}{N^{2/3}}\right) \]
Thus, the total surface area of \( N \) smaller drops is:
\[ A_{\text{total}} = N \times A_{\text{small}} = N \times 4 \pi R^2 \left(\frac{1}{N^{2/3}}\right) = 4 \pi R^2 N^{1/3} \]

### Step 4: Calculate the Change in Surface Area
The change in surface area \( \Delta A \) when splitting the drop is:
\[ \Delta A = A_{\text{total}} - A_{\text{original}} = 4 \pi R^2 N^{1/3} - 4 \pi R^2 \]
Factoring out \( 4 \pi R^2 \):
\[ \Delta A = 4 \pi R^2 (N^{1/3} - 1) \]

### Step 5: Calculate the Work Done
The work done \( W \) to split the drop is given by:
\[ W = \sigma \Delta A \]
Substituting \( \Delta A \):
\[ W = \sigma \cdot 4 \pi R^2 (N^{1/3} - 1) \]

### Final Result
Thus, the work done to split a liquid drop of radius \( R \) into \( N \) identical drops is:
\[ W = 4 \sigma \pi R^2 (N^{1/3} - 1) \]