The surface tension of soap is 3.5 × 10⁻² N m⁻¹. The amount of work done required to increase the radius of soap bubble from 10 cm to 20 cm is _________ × 10⁻⁴ J. (take π = 22/7) Numerical answer type. Work it out, then open the answer.

Answer: 264. W = 8π T (r₂² - r₁²) W = 8 × 22/7 × 3.5×10⁻² × (0.04 - 0.01) W = 0.88 × 0.03 = 0.0264 J = 264×10⁻⁴ J So the answer is 264.

The surface tension of soap is 3.5 × 10⁻² N m⁻¹. The amount of work… | JEE Main 2023 (11 Apr, Shift 2)

Q21JEE Main 2023 · 11 APR · SHIFT 2Medium

Surface Energy & Work Done

The surface tension of soap is $3.5 \times 10^{-2}\text{ N m}^{-1}$ . The amount of work done required to increase the radius of soap bubble from $10\text{ cm}$ to $20\text{ cm}$ is _________ $\times 10^{-4}\text{ J}$ .

$(\text{take } \pi = 22/7)$

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

Show answer and solution

Answer: 264

$W = 8\pi T (r_2^2 - r_1^2)$
$W = 8 \times \frac{22}{7} \times 3.5\times10^{-2} \times (0.04 - 0.01)$
$W = 0.88 \times 0.03 = 0.0264\text{ J} = 264\times10^{-4}\text{ J}$
So the answer is 264.

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