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Questions 3-23 · Q20

Q.Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m. [Ans. 1.628×10⁻⁷ J = 1.628 erg]

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Step 1. Volume is conserved: 27×43πr3=43πR3⇒R3=27r3⇒R=3r27×\tfrac43\pi r^3 = \tfrac43\pi R^3 \Rightarrow R^3 = 27r^3 \Rightarrow R = 3r. With r=0.1r = 0.1 mm =1×10−4= 1×10^{-4} m, R=3×10−4R = 3×10^{-4} m.

Step 2. Initial total surface area (27 droplets): 27×4πr2=108πr227×4\pi r^2 = 108\pi r^2. Final surface area (1 drop): 4πR2=4π(3r)2=36πr24\pi R^2 = 4\pi(3r)^2 = 36\pi r^2. …

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