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Two identical mercury droplets, each of radius $r$, combine to form a single larger droplet. If the surface tension is $T$, what is the surface energy of the resulting drop?
- $4\pi r^2T$
- $2\pi r^2T$
- $8^{2/3}\pi r^2T$
- $2^{5/3}\pi r^2T$
Correct answer: $8^{2/3}\pi r^2T$
Solution
When two identical drops merge, volume doubles: $\frac{4}{3}\pi R^3 = 2\cdot\frac{4}{3}\pi r^3$, so $R=2^{1/3}r$. The surface energy of the new drop is $4\pi R^2T = 4\pi(2^{2/3}r^2)T = 8^{2/3}\pi r^2T$.
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