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JEE Main 2026 (24 Jan, Shift 2) · Surface Energy & Work Done

Surface Energy & Work Done question from JEE Main 2026 (24 Jan, Shift 2)

Q7JEE Main 2026 · 24 JAN · SHIFT 2Medium

Surface Energy & Work Done

A soap bubble of surface tension 0.04 N/m0.04\text{ N/m} is blown to a diameter of 7 cm. If (15000−x) μJ(15000 - x)\,\mu\text{J} of work is done in blowing it further to make its diameter 14 cm, then the value of xx is ‾ . (π=22/7)\underline{\qquad}\ .\ (\pi = 22/7)

Numerical answer type. Work it out, then open the answer.

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Answer: 11304

  1. Radii: 3.5 cm→7 cm3.5\text{ cm} \to 7\text{ cm}. W=8πT(r22−r12)W = 8\pi T (r_2^2 - r_1^2).
  2. W=8×227×0.04×(49−12.25)×10−4W = 8 \times \frac{22}{7} \times 0.04 \times (49 - 12.25)\times10^{-4}
  3. W=36.96×10−4 J=3696 μJW = 36.96\times10^{-4}\text{ J} = 3696\ \mu\text{J}
  4. 15000−x=3696⇒x=1130415000 - x = 3696 \Rightarrow x = 11304
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