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MHT CET 2026 (19 Apr, Shift 1) · Surface Energy & Work Done

Surface Energy & Work Done question from MHT CET 2026 (19 Apr, Shift 1)

Q3MHT CET 2026 · 19 APR · SHIFT 1Medium

Surface Energy & Work Done

One large soap bubble of diameter 'D' breaks into 6464 bubbles having surface tension 'T'. The change in surface energy is

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Answer: (c)

  1. Surface energy of a bubble =8πr2T= 8\pi r^2 T. Big bubble radius R=D/2R = D/2: E1=8π(D/2)2T=2πTD2E_1 = 8\pi (D/2)^2 T = 2\pi T D^2.
  2. Volume conserved: 64r3=R3⇒r=R/4=D/864 r^3 = R^3 \Rightarrow r = R/4 = D/8.
  3. E2=64×8π(D/8)2T=8πTD2E_2 = 64 \times 8\pi (D/8)^2 T = 8\pi T D^2
  4. ΔE=8πTD2−2πTD2=6πTD2\Delta E = 8\pi TD^2 - 2\pi TD^2 = 6\pi TD^2
  5. So the answer is (c) 6πTD26\pi TD^2.
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