Q 11-09-091JEE MainJEE Main 2023 (11 Apr, Shift 2)Medium
The surface tension of soap solution is $3.5\times10^{-2}\ \text{N m}^{-1}$. The amount of work done required to increase the radius of soap bubble from $10\ \text{cm}$ to $20\ \text{cm}$ is ______ $\times10^{-4}$ J. (take $\pi=\frac{22}{7}$)
Numerical value type. Enter your answer.
Answer: 264
Two surfaces: $W=2T\times4\pi(r_2^2-r_1^2)=2\times3.5\times10^{-2}\times4\times\dfrac{22}{7}\times0.03=0.0264\ \text{J}=264\times10^{-4}\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics