A spherical liquid drop of radius \(R\) acquires the terminal velocity \(v_1\) when falls through a gas of viscosity \(\eta\). Now the drop is broken into \(64\) identical droplets and each droplet acquires terminal velocity \(v_2\) falling through the same gas. The ratio of terminal velocities \(v_1/v_2\) is ________.

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: \(16\)