Physics - Fluid Question with Solution | TestHub

The volume of the mercury drop remains conserved when it breaks into $n$ smaller droplets of equal size.

Initial volume of the large drop of radius $R$ is $\frac{4}{3}\pi R^3$.

Final volume of $n$ droplets of radius $r$ is $n\times \frac{4}{3}\pi r^3$.

Equating the initial and final volumes:

$\frac{4}{3}\pi R^3=n\times \frac{4}{3}\pi r^3$

$R^3=n{r}^3$

$r^3=\frac{R^3}{n}$

$r=\frac{R}{n^{1/3}}=(n{)}^{-1/3}R$

Given that $r=xR$, comparing the two expressions gives:

$x=(n{)}^{-1/3}$

Answer: $(n{)}^{\frac{-1}{3}}$