2.Calculate work done in blowing a soap bubble of radius $r=5\text{ cm}$ if surface tension $S = 0.03\text{ N/m}$.
Sol: Soap bubble has 2 free surfaces: $W = 2 \times S \times 4\pi r^2 = 8\pi (0.03) (0.05)^2 = 8\pi (0.03)(0.0025) \approx 1.885 \times 10^{-3}\text{ Joules}$.
3.Eight mercury drops of radius $r=1\text{ mm}$ coalesce to form a single large drop. Find change in surface energy.
Sol: Volume conservation: $\frac{4}{3}\pi R^3 = 8 \left(\frac{4}{3}\pi r^3\right) \implies R = 2r$. $\Delta A = 8(4\pi r^2) - 4\pi R^2 = 16\pi r^2$. Energy released $\Delta U = S \Delta A = 16\pi r^2 S$.
