Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly (Surface tension of soap = 0.03 N m⁻¹, (a) 0.2π mJ (b) 2π mJ (c) 0.4π mJ (d) 4π mJ

Answer: (c). W = 8π T (r₂² - r₁²) W = 8π × 0.03 × (25 - 9)×10⁻⁴ = 3.84π×10⁻⁴ J W = 0.384π mJ ≈ 0.4π mJ So the answer is (c) 0.4π mJ.

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Q30JEE Main 2011 · AIEEEMedium

Surface Energy & Work Done

Work done in increasing the size of a soap bubble from a radius of $3\text{ cm}$ to $5\text{ cm}$ is nearly (Surface tension of soap $= 0.03\text{ N m}^{-1},$

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Answer: (c)

$W = 8\pi T (r_2^2 - r_1^2)$
$W = 8\pi \times 0.03 \times (25 - 9)\times10^{-4} = 3.84\pi\times10^{-4}\text{ J}$
$W = 0.384\pi\text{ mJ} \approx 0.4\pi\text{ mJ}$
So the answer is (c) $0.4\pi\text{ mJ}$ .

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Surface Energy & Work Done question from JEE Main 2011 (AIEEE)

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