Surface tension of a soap bubble is 2.0 × 10⁻² N m⁻¹. Work done to increase the radius of soap bubble from 3.5 cm to 7 cm will be: Take [π = 22/7] (a) 18.48 × 10⁻⁴ J (b) 5.76 × 10⁻⁴ J (c) 0.72 × 10⁻⁴ J (d) 9.24 × 10⁻⁴ J

Answer: (a). W = 8π T (r₂² - r₁²) (two surfaces). r₂² - r₁² = 49 - 12.25 = 36.75 cm² = 36.75×10⁻⁴ m² W = 8 × 22/7 × 0.02 × 36.75×10⁻⁴ = 18.48×10⁻⁴ J So the answer is (a) 18.48×10⁻⁴ J.

Surface tension of a soap bubble is 2.0 × 10⁻² N m⁻¹. Work done to… | JEE Main 2023 (29 Jan, Shift 1)

Q20JEE Main 2023 · 29 JAN · SHIFT 1Medium

Surface Energy & Work Done

Surface tension of a soap bubble is $2.0 \times 10^{-2}\text{ N m}^{-1}$ . Work done to increase the radius of soap bubble from $3.5\text{ cm}$ to $7\text{ cm}$ will be:

Take $\left[\pi = \frac{22}{7}\right]$

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Answer: (a)

$W = 8\pi T (r_2^2 - r_1^2)$ (two surfaces).
$r_2^2 - r_1^2 = 49 - 12.25 = 36.75\text{ cm}^2 = 36.75\times10^{-4}\text{ m}^2$
$W = 8 \times \frac{22}{7} \times 0.02 \times 36.75\times10^{-4} = 18.48\times10^{-4}\text{ J}$
So the answer is (a) $18.48\times10^{-4}\text{ J}$ .

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Surface Energy & Work Done question from JEE Main 2023 (29 Jan, Shift 1)

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