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JEE MAIN_2026
04-04-26_S2
Question
The surface tension of a soap solution is $3.5 \times 10^{-2} \mathrm{~N} / \mathrm{m}$. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $\alpha \times 10^{-6} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ . $(\pi=22 / 7)$
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\begin{aligned} & \text { W.D. }=2 \mathrm{~T}\left[4 \pi \mathrm{R}_2^2-4 \pi \mathrm{R}_1^2\right] \\ & =2 \times 3.5 \times 10^{-2} \times 4 \times \frac{22}{7}\left[2^2-1^2\right] \times 10^{-4} \\ & =2 \times 132 \times 10^{-6} \\ & =264 \times 10^{-6} \mathrm{~J}=\alpha \times 10^{-6} \mathrm{~J} \\ & \alpha=264 \end{aligned}
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