JEE Main 2026 — Mechanical Properties of Fluids Question with Solution

The terminal velocity of a spherical drop of radius $r$ falling through a viscous medium is given by Stokes' law:

$v=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$

From this relation, the terminal velocity is directly proportional to the square of the radius of the drop:

$v\propto r^2$

When the large drop of radius $R$ is broken into $64$ identical droplets of radius $r$, the total volume remains conserved:

$\frac{4}{3}\pi R^3=64\times \frac{4}{3}\pi r^3$

$R^3=64r^3$

$R=4r$

The ratio of the terminal velocities of the original drop and the smaller droplet is:

$\frac{v_1}{v_2}=\frac{R^2}{r^2}$

Substituting $R=4r$:

$\frac{v_1}{v_2}=\frac{(4r{)}^2}{r^2}=16$

Answer: $16$