Question

Work done to get ' \(n\) ' spherical drops of equal size from a single spherical drop of water, is proportional to

• A. \(\left( \frac{1}{n^{2/3}} - 1 \right)\)
• B. \(\left( \frac{1}{n^{1/3}} - 1 \right)\)
• C. \(n^{1/3} - 1\)
• D. \(n^{4/3} - 1\)

Hint:

Key: More drops $\rightarrow$ more surface area $\rightarrow$ more work

Answer 1

The Correct Option is C

Concept: Work done = increase in surface energy: \[ W \propto \Delta (\text{surface area}) \]

Step 1: Volume conservation. \[ \frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3 \Rightarrow R^3 = n r^3 \Rightarrow R = n^{1/3} r \]

Step 2: Initial surface area. \[ A_i = 4\pi R^2 = 4\pi (n^{1/3} r)^2 = 4\pi n^{2/3} r^2 \]

Step 3: Final surface area. \[ A_f = n \cdot 4\pi r^2 = 4\pi n r^2 \]

Step 4: Increase in area. \[ \Delta A = 4\pi r^2 (n - n^{2/3}) \] \[ \propto n^{1/3} - 1 \]

Step 5: Conclusion.
Work done $\propto n^{1/3} - 1$ Final Answer: Option (C)