The surface tension of a soap is 3.5 × 10⁻² N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is α × 10⁻⁶ J. The value of α is ______. (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 × (4 - 1)×10⁻⁴ W = 0.88 × 3 × 10⁻⁴ = 2.64×10⁻⁴ J = 264×10⁻⁶ J So α = 264.

The surface tension of a soap is 3.5 × 10⁻² N/m. The work required to… | JEE Main 2026 (4 Apr, Shift 1)

Q6JEE Main 2026 · 4 APR · SHIFT 1Medium

Surface Energy & Work Done

The surface tension of a soap is $3.5 \times 10^{-2}\text{ N/m}$ . The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $\alpha \times 10^{-6}\text{ J}$ . The value of $\alpha$ is ______.

$(\pi = 22/7)$

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

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Answer: 264

$W = 8\pi T (r_2^2 - r_1^2)$
$W = 8 \times \frac{22}{7} \times 3.5\times10^{-2} \times (4 - 1)\times10^{-4}$
$W = 0.88 \times 3 \times 10^{-4} = 2.64\times10^{-4}\text{ J} = 264\times10^{-6}\text{ J}$
So $\alpha = 264$ .

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