Let ' $n$ ' is the number of liquid drops, each with surface energy ' $E$ '. These drops join to form single…

Let ' $n$ ' is the number of liquid drops, each with surface energy ' $E$ '. These drops join to form single drop. In this process
  1. some energy will be absorbed
  2. energy absorbed is $\left[E\left(n-n^{2 / 3}\right)\right]$
  3. energy released will be $\left[E\left(n-n^{2 / 3}\right)\right]$
  4. energy released will be $\left[\mathrm{E}\left(2^{2 / 3}-1\right)\right]$

Solution

Let $r=$ radius of each small drop and $\mathrm{R}=$ radius of a big single drop. Then, $n \times \frac{4}{3} \pi \mathrm{r}^3=\frac{4}{3} \pi \mathrm{R}^3$ $\therefore \quad \mathrm{R}=\mathrm{n}^{1 / 3} \mathrm{r}$...(i) Initial surface energy, $\mathrm{E}_1=\mathrm{n} \times 4 \pi \mathrm{r}^2 \times \mathrm{T}=\mathrm{nE}$
Final surface energy, $\begin{aligned} \mathrm{E}_2 & =4 \pi \mathrm{R}^2 \times \mathrm{T}=4 \pi \mathrm{r}^2 \mathrm{n}^{2 / 3} \times \mathrm{T} \quad \ldots[\operatorname{From}(\mathrm{i})] \\ & =\mathrm{n}^{2 / 3} \mathrm{E} \end{aligned}$ Energy released $=E_1-E_2=\left[E\left(n-n^{2 / 3}\right)\right]$ ~

Asked in: MHT CET 2024 (09 May Shift 2)

Practice more Mechanical Properties of Fluids questions on Aicharya