The work done to get 'n' smaller equal size spherical drop from a bigger size spherical drop of water is proportional to

Options

A. `l//n^(2//3) -1`
B. `l//n^(2//3)-1`
C. `n^(1//3) -1`
D. `n^(4//3) -1`

Explanation

To solve the problem of determining the work done to create 'n' smaller equal-sized spherical drops from a larger spherical drop of water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final States:**
- Let the radius of the larger spherical drop be \( R \).
- Let the radius of each smaller drop be \( r \).
- There are \( n \) smaller drops formed from the larger drop.

2. **Calculate the Surface Area:**
- The surface area of the larger drop is given by:
\[ A_{\text{initial}} = 4\pi R^2 \]
- The surface area of one smaller drop is:
\[ A_{\text{small}} = 4\pi r^2 \]
- Therefore, the total surface area of the \( n \) smaller drops is:
\[ A_{\text{final}} = n \times A_{\text{small}} = n \times 4\pi r^2 \]

3. **Calculate the Change in Surface Area:**
- The change in surface area, which is related to the work done, is:
\[ \Delta A = A_{\text{final}} - A_{\text{initial}} = n \times 4\pi r^2 - 4\pi R^2 \]

4. **Conservation of Volume:**
- The volume of the larger drop is:
\[ V_{\text{initial}} = \frac{4}{3}\pi R^3 \]
- The volume of the \( n \) smaller drops is:
\[ V_{\text{final}} = n \times \frac{4}{3}\pi r^3 \]
- Setting the initial volume equal to the final volume gives:
\[ \frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3 \]
- Simplifying this, we find:
\[ R^3 = n r^3 \quad \Rightarrow \quad r = R n^{-1/3} \]

5. **Substituting for r in the Change in Surface Area:**
- Substitute \( r = R n^{-1/3} \) into the change in surface area:
\[ \Delta A = n \times 4\pi (R n^{-1/3})^2 - 4\pi R^2 \]
- This simplifies to:
\[ \Delta A = n \times 4\pi R^2 n^{-2/3} - 4\pi R^2 = 4\pi R^2 (n^{1/3} - 1) \]

6. **Calculate the Work Done:**
- The work done \( W \) is proportional to the change in surface energy, which is given by:
\[ W \propto T \Delta A = T \times 4\pi R^2 (n^{1/3} - 1) \]
- Therefore, we conclude that the work done is proportional to:
\[ n^{1/3} - 1 \]

### Final Conclusion:
The work done to create 'n' smaller equal-sized spherical drops from a larger drop is proportional to \( n^{1/3} - 1 \).