Question

A liquid drop having diameter $2\text{mm}$ and surface tension $0.08 \text{ N/m}$. This drop splits into 512 identical small drops. Find change in surface energy ?

• A. 4.034
• B. 5
• C. $7.034 \times 10^{-6}$
• D. 9.03

Answer 1

The Correct Option is C

Step 1: Relate the radii of the large drop ($R$) and small drops ($r$).
Volume remains constant: $\frac{4}{3}\pi R^3 = 512 \times \frac{4}{3}\pi r^3 \implies R = 8r \implies r = R/8$.

Step 2: Formula for surface energy change.
$\Delta U = T \times \Delta A = T \times (512 \cdot 4\pi r^2 - 4\pi R^2)$.

Step 3: Substitute $r = R/8$ into the formula.
$\Delta U = 4\pi T \times [512 (R/8)^2 - R^2] = 4\pi T \times (8R^2 - R^2) = 28\pi T R^2$.

Step 4: Substitute values $T = 0.08$, $R = 10^{-3} \text{ m}$.
$\Delta U = 28 \times 3.14 \times 0.08 \times (10^{-3})^2 \approx 7.034 \times 10^{-6} \text{ J}$.

Final Answer: Option (3).