Question

Eight droplets of water each of radius 0.2 mm coalesce into a single drop. Find the decrease in the surface area.

Hint:

For coalescing droplets, use volume conservation to find the new radius; surface area decreases due to \( 4 \pi r^2 \).

Answer 1

Volume conservation: 8 droplets of radius \( r = 0.2 \, \text{mm} = 0.0002 \, \text{m} \) form one drop of radius \( R \).
\[ 8 \cdot \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \quad \Rightarrow \quad 8 r^3 = R^3 \quad \Rightarrow \quad R = 2^{1/3} r. \] \[ R = 2^{1/3} \times 0.0002 \approx 1.2599 \times 0.0002 \approx 0.000252 \, \text{m}. \] Surface area of one droplet: \( 4 \pi r^2 \).
Total initial surface area (8 droplets): \[ 8 \cdot 4 \pi (0.0002)^2 = 32 \pi \times 4 \times 10^{-8} = 1.28 \pi \times 10^{-6} \, \text{m}^2. \] Surface area of final drop: \[ 4 \pi R^2 = 4 \pi (2^{1/3} \times 0.0002)^2 = 4 \pi \cdot 2^{2/3} \times 4 \times 10^{-8} \approx 4 \pi \times 1.5874 \times 4 \times 10^{-8} \approx 7.9577 \times 10^{-7} \, \text{m}^2. \] Decrease in surface area: \[ 1.28 \pi \times 10^{-6} - 7.9577 \times 10^{-7} \approx 4.016 \times 10^{-6} - 7.9577 \times 10^{-7} \approx 3.2203 \times 10^{-6} \, \text{m}^2 \approx 0.00322 \, \text{cm}^2. \] Answer: \( 0.00322 \, \text{cm}^2 \).