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JEE Main 2026 (5 Apr, Shift 2) · Splitting & Merging of Drops

Splitting & Merging of Drops question from JEE Main 2026 (5 Apr, Shift 2)

Q5JEE Main 2026 · 5 APR · SHIFT 2Easy

Splitting & Merging of Drops

Eight mercury drops, each of radius $r$ , coalesce to form a bigger drop. The surface energy released in this process is ______ - ( $S$ is the surface tension of mercury).

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Answer: (b)

Volume conserved: $R = 8^{1/3}r = 2r$ .
Initial energy $= 8 \times 4\pi r^2 S = 32\pi r^2 S$ ; final $= 4\pi(2r)^2 S = 16\pi r^2 S$ .
Energy released $= 32\pi r^2 S - 16\pi r^2 S = 16\pi r^2 S$
So the answer is (b) $16\pi r^2 S$ .

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