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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 rr, coalesce to form a bigger drop. The surface energy released in this process is ______ - ( SS is the surface tension of mercury).

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Answer: (b)

  1. Volume conserved: R=81/3r=2rR = 8^{1/3}r = 2r.
  2. Initial energy =8×4πr2S=32πr2S= 8 \times 4\pi r^2 S = 32\pi r^2 S; final =4π(2r)2S=16πr2S= 4\pi(2r)^2 S = 16\pi r^2 S.
  3. Energy released =32πr2S−16πr2S=16πr2S= 32\pi r^2 S - 16\pi r^2 S = 16\pi r^2 S
  4. So the answer is (b) 16πr2S16\pi r^2 S.
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