A liquid drop having surface energy ' $E$ ' is spread into 512 droplets of same size. The final surface energy of the droplets is

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Aicharya · Accepted Answer

Surface area of drop, $\mathrm{A}_1=4 \pi \mathrm{R}^2$ Surface area of 512 droplets, $\mathrm{A}_2=512\left(4 \pi \mathrm{r}^2ight)$ volume of drop $=\mathrm{n} 	imes$ (volume of droplet) $	herefore \quad \frac{4}{3} \pi \mathrm{R}^3=512 	imes \frac{4}{3} \pi \mathrm{r}^3$ $	herefore \quad \mathrm{R}=8 \mathrm{r}$ $	herefore \quad \mathrm{A}_2=\frac{512\left(4 \pi \mathrm{R}^2ight)}{64}$ $	herefore \quad \mathrm{A}_2=8\left(4 \pi \mathrm{R}^2ight)$ Surface energy $\propto$ Area $	herefore \quad \frac{E_2}{E_1}=\frac{A_2}{A_1}=\frac{8\left(4 \pi R^2ight)}{4 \pi R^2}$ $	herefore \quad \mathrm{E}_2=8 \mathrm{E}_1=8 \mathrm{E}$ $\ldots .\left(\because E_1=Eight)$

A liquid drop having surface energy ' $E$ ' is spread into 512 droplets of same size. The final surface…

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