Physics - Fluid Question with Solution | TestHub

Initially area of soap bubble, $A_1=4\pi r^2$ Under isothermal condition radius becomes $2r$. $\therefore$ Area $A_2=4\pi (2r{)}^2=16\pi r^2$ Increase in surface area $$\Delta A=2\left(A_2-A_1\right)=2\left(16\pi r^2-4\pi r^2\right)=24\pi r^2$$ Energy spent, $$W=T\times \Delta A=T\cdot 24\pi r^2=24\pi T{r}^2\mathrm{J}$$