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

Answer: (a). Two surfaces: W = 8π T(R₂² - R₁²). R₂² - R₁² = 0.05² - 0.03² = 16×10⁻⁴ m². W = 8π × 0.03 × 16×10⁻⁴ = 3.84×10⁻⁴π J ≈ 0.4π mJ. So the answer is (a).

Work done in increasing the size of a soap bubble from radius of 3 cm… | MHT CET 2021 (23 Sep, Shift 1)

Q24MHT CET 2021 · 23 SEP · SHIFT 1Easy

Surface Energy & Work Done

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

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Answer: (a)

Two surfaces: $W = 8\pi T(R_2^2 - R_1^2)$ .
$R_2^2 - R_1^2 = 0.05^2 - 0.03^2 = 16\times10^{-4}\text{ m}^2$ .
$W = 8\pi \times 0.03 \times 16\times10^{-4} = 3.84\times10^{-4}\pi\text{ J} \approx 0.4\pi\text{ mJ}$ .
So the answer is (a).

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Surface Energy & Work Done question from MHT CET 2021 (23 Sep, Shift 1)

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