A spherical liquid drop of radius R acquires the terminal velocity...

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

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

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:

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

$R^3 = 64 r^3$

$R = 4r$

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

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

Substituting $R = 4r$:

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

Answer: $16$