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Physics · Mechanical Properties of Matter

JEE Main 2026 — 5 April, Evening Shift — Question 8

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

  1. Option A:

    8πr2S8\pi r^2 S

  2. Option B:

    16πr2S16\pi r^2 S

    Correct
  3. Option C:

    64πr2S64\pi r^2 S

  4. Option D:

    4πr2S4\pi r^2 S

Answer: B

Step-by-step solution

Volume conservation: 8×43πr3=43πR3⇒R=2r8 \times \frac43 \pi r^3 = \frac43 \pi R^3 \Rightarrow R=2r. Initial surface energy Ui=8×4πr2S=32πr2SU_i = 8 \times 4\pi r^2 S = 32\pi r^2 S. Final Uf=4π(2r)2S=16πr2SU_f = 4\pi (2r)^2 S = 16\pi r^2 S. Released energy = Ui−Uf=16πr2SU_i - U_f = 16\pi r^2 S.

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Matter
Topic
Surface Tension and Surface Energy

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