Q 11-09-036JEE MainJEE Main 2026 (2 Apr, Shift 2)Easy
The surface tension of a soap bubble is $0.03$ N/m. The work done in increasing the diameter of bubble from $2$ cm to $6$ cm is $\alpha\pi\times10^{-4}$ J. The value of $\alpha$ is ______. (Take $\pi=3.14$)
Answer: (C) $1.92$
A soap bubble has two surfaces, so $W=T\times2\times\Delta(4\pi r^2)$.
Radius changes from $1$ cm to $3$ cm:
$$W=0.03\times2\times4\pi\left[(0.03)^2-(0.01)^2\right]=0.03\times8\pi\times8\times10^{-4}=1.92\pi\times10^{-4}\ \text{J}$$
So $\alpha=1.92$.
Solution by Sreeraj P, M.Sc Physics