Q 11-09-033JEE MainJEE Main 2026 (4 Apr, Shift 1)Easy
The surface tension of a soap solution is $3.5 \times 10^{-2}$ N/m. The work required to increase the radius of a soap bubble from $1$ cm to $2$ cm is $\alpha \times 10^{-6}$ J. The value of $\alpha$ is ______. ($\pi = 22/7$)
Numerical value type. Enter your answer.
Answer: 264
Concept. A soap bubble has two surfaces: $W=2S\times4\pi(r_2^2-r_1^2)$.
Step 1. Substitute. $W=8\times\dfrac{22}{7}\times3.5\times10^{-2}\times(4-1)\times10^{-4}=264\times10^{-6}$ J.
Answer. $264$.
Solution by Sreeraj P, M.Sc Physics
