Physics - Fluid Question with Solution | TestHub

Change in surface energy is given by $E=T(\Delta A)---(1)$ The initial area is given by, $\mathrm{A}=\left(4\pi r^2\right)n$ The final area is given by, $\mathrm{a}=4\pi R^2$ Therefore, the change in area is given by, $\begin{array}{rl}& \Delta A=a-A\\ & \Rightarrow \Delta A=4\pi \left(n{r}^2-R^2\right)---(2)\end{array}$ Now, using volume conservation: $\left(\frac{4}{3}\pi r^2\right)n=\frac{4}{3}{R}^3$ $\therefore n=\frac{R^3}{r^3}---(3)$ $\Delta A=4\pi \left[\frac{R^3}{r^3}\cdot r^2-R^2\right]=4\pi \left[\frac{R^3}{r}-\frac{R^3}{R}\right]=\left(\frac{4}{3}\pi R^3\right)3\left[\frac{1}{r}-\frac{1}{R}\right]=3V\left[\frac{1}{r}-\frac{1}{R}\right]$ Introducing above value in equation (1) $E=3VT\left[\frac{1}{r}-\frac{1}{R}\right]$