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JEE Main 2026 (2 Apr, Shift 2) · Surface Energy & Work Done

Surface Energy & Work Done question from JEE Main 2026 (2 Apr, Shift 2)

Q5JEE Main 2026 · 2 APR · SHIFT 2Medium

Surface Energy & Work Done

The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is απ×10−4\alpha\pi \times 10^{-4} J. The value of α\alpha is _________. (Take π=3.14\pi = 3.14)

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Answer: (c)

  1. Radii: r1=1 cmr_1 = 1\text{ cm}, r2=3 cmr_2 = 3\text{ cm}. W=8πT(r22−r12)W = 8\pi T (r_2^2 - r_1^2).
  2. W=8π×0.03×(9−1)×10−4W = 8\pi \times 0.03 \times (9 - 1)\times10^{-4}
  3. W=1.92π×10−4 JW = 1.92\pi \times 10^{-4}\text{ J}, so α=1.92\alpha = 1.92.
  4. So the answer is (c) 1.92.
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