When a mercury drop of radius ' $R$ ' splits up into 1000 droplets of radius ' $r$ ', the change in surface…

When a mercury drop of radius ' $R$ ' splits up into 1000 droplets of radius ' $r$ ', the change in surface energy is ( $T=$ surface tension of mercury)
  1. $8 \pi \mathrm{R}^2 \mathrm{~T}$
  2. $16 \pi \mathrm{R}^2 \mathrm{~T}$
  3. $\quad 34 \pi \mathrm{R}^2 \mathrm{~T}$
  4. $\quad 36 \pi R^2 T$

Solution

As volume remains constant, $\mathrm{R}^3=1000 \mathrm{r}^3$ $\therefore \quad \mathrm{n}^{\prime}=10 \mathrm{r}$ or $\mathrm{r}=\frac{\mathrm{R}}{10}$ Change in surface area $\begin{aligned} & =\left(1000 \times 4 \pi r^2\right)-4 \pi R^2 \\ & =4 \pi\left(1000 \times \frac{R^2}{100}-R^2\right)=36 \pi R^2 \end{aligned}$
Surface energy, $T \Delta A=36 \pi R^2 T$

Asked in: MHT CET 2024 (15 May Shift 1)

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