Two mercury drops, each of radius R, coalesce into one large drop. Find the ratio of the total surface energies before and after coalescence.
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Question

Two mercury drops, each of radius R, coalesce into one large drop. Find the ratio of the total surface energies before and after coalescence.

G1 Physics · Accepted Answer

Step 1: Let the large drop radius be R′. Volume conservation: 2(4πR³/3) = 4πR′³/3. Step 2: R′³ = 2R³, giving R′ = ∛2 · R. Step 3: Total initial surface energy Uᵢ = 2(4πR²T) = 8πR²T. Step 4: Final energy U𝒇 = 4πR′²T = 4π(2²ᐟ³R²)T. Step 5: Uᵢ/U𝒇 = 2/2²ᐟ³ = 2¹ᐟ³ = ∛2. Final answer: ∛2 : 1 (Option B).