Q 11-09-057JEE MainJEE Main 2024 (4 Apr, Shift 1)Medium
A soap bubble is blown to a diameter of $7\ \text{cm}$. $36960\ \text{erg}$ of work is done in blowing it further. If the surface tension of the soap solution is $40\ \text{dyne/cm}$, then the new radius is ______ cm. (Take $\pi = \dfrac{22}{7}$)
Numerical value type. Enter your answer.
Answer: 7
A bubble has two surfaces, so $W = 2T\times4\pi(R^2 - r^2)$ with $r = 3.5\ \text{cm}$:
$$36960 = 2\times40\times4\times\frac{22}{7}(R^2 - 12.25) \Rightarrow R^2 - 12.25 = \frac{36960\times7}{7040} = 36.75$$
$$R^2 = 49 \Rightarrow R = 7\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics