Question

The work done in splitting a water drop of radius R into 64 droplets is ($T=$ surface tension)

• A. $6\pi TR^{2}$
• B. $24\pi TR^{2}$
• C. $12\pi TR^{2}$
• D. $16\pi TR^{2}$

Hint:

Work Done for $n$ droplets: $W = 4\pi TR^2(n^{1/3} - 1)$.

Answer 1

The Correct Option is C

Step 1: Concept
Work done $W = T \Delta A$, where $\Delta A$ is the increase in surface area.

Step 2: Analysis
Volume remains constant: $\frac{4}{3}\pi R^{3} = 64 \times \frac{4}{3}\pi r^{3}$.
$R^{3} = 64 r^{3} \implies r = R/4$.

Step 3: Calculation
$\Delta A = A_{\text{final}} - A_{\text{initial}} = 64(4\pi r^{2}) - 4\pi R^{2}$
$\Delta A = 64(4\pi (R/4)^{2}) - 4\pi R^{2} = 64(4\pi R^{2}/16) - 4\pi R^{2}$
$\Delta A = 16\pi R^{2} - 4\pi R^{2} = 12\pi R^{2}$.
$W = T(12\pi R^2)$.

Step 4: Conclusion
Hence, work done is $12\pi TR^{2}$. Final Answer: (C)