The terminal velocity of a spherical drop of radius r falling through a viscous medium is given by Stokes' law:
v=92ηr2(ρ−σ)g
From this relation, the terminal velocity is directly proportional to the square of the radius of the drop:
v∝r2
When the large drop of radius R is broken into 64 identical droplets of radius r, the total volume remains conserved:
34πR3=64×34πr3
R3=64r3
R=4r
The ratio of the terminal velocities of the original drop and the smaller droplet is:
v2v1=r2R2
Substituting R=4r:
v2v1=r2(4r)2=16
Answer: 16