JEE Main 2026 · 4 Apr · Shift 1 · Physics Q47

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The surface tension of a soap solution is $3.5\times10^{-2}$ N/m. The work required to increase the radius of a soap bubble from 1 cm to 2 cm is $\alpha\times10^{-6}$ J. The value of $\alpha$ is ________. ( $\pi=22/7$ )

Your answer

Answer and worked solution

Official answer: 264

A bubble has two surfaces.
$W=2T\cdot4\pi\left(r_2^2-r_1^2\right)=2(0.035)\left(4\cdot\tfrac{22}7\right)(3\times10^{-4})=2.64\times10^{-4}\text{ J}.$

Value. $\alpha=264$ .

Answer: 264

Topics:Surface tension

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JEE Main 2026 · 4 Apr · Shift 1, Physics question 47