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Answer: A. decreases

Answer: (a) decreases

When \(n\) droplets of radius \(r\) join, volume is conserved: \(\frac{4}{3}\pi R^3 = n\cdot\frac{4}{3}\pi r^3\), so \(R = n^{1/3} r\). The new surface area is \(4\pi R^2 = n^{2/3}\,4\pi r^2\), which is less than the total initial area \(n\,4\pi r^2\) (because \(n^{2/3} < n\) for \(n>1\)). The surface area decreases and the surface energy released appears as heat.

Doubtly AI · 27 Oct 2026 AI-generated · not yet verified Report
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