When a mercury drop of radius 'R', breaks into 'n' droplets of equal size, the radius 'r' of each droplet is

When a mercury drop of radius 'R', breaks into 'n' droplets of equal size, the radius 'r' of each droplet is
  1. $\mathrm{r}=\frac{\mathrm{R}}{\sqrt{\mathrm{n}}}$
  2. $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}}$
  3. $\mathrm{r}=\frac{\mathrm{R}}{\mathrm{n}^{\frac{1}{3}}}$
  4. $\mathrm{r}=\mathrm{R} \mathrm{n}^{\frac{1}{3}}$

Solution

Volume remains constant $\therefore \mathrm{n} \times \frac{4}{3} \pi \mathrm{r}^{3}=\frac{4}{3} \pi \mathrm{R}^{3}$ $\therefore \mathrm{nr}^{3}=\mathrm{R}^{3}$ $\therefore \quad n^{\frac{1}{3}} r=R \quad \therefore r=\frac{R}{n^{\frac{1}{3}}}$ .

Asked in: MHT CET 2020 (12 Oct Shift 2)

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