Eight identical drops of same radii are falling through air with steady velocity $6 \ cm/s$. If these drops coalesce,then the terminal velocity would be .......... $cm/s$.

Two spherical rain drops reach the surface of the earth with terminal velocities having ratio $16: 9$. The ratio of their surface area is

$A$ stone is projected vertically up from the bottom of a water tank. Assuming no water resistance,it will go up and come down in the same time. However,if water drag (resistance) is present,then the time it takes to go up,$t_{up}$,and the time it takes to come down,$t_{down}$,are related as:

$A$ water drop whose radius is $0.0015 \, mm$ is falling through the air. If the coefficient of viscosity of air is $1.8 \times 10^{-5} \, kg/(m \cdot s)$,then assuming the buoyancy force is negligible,the terminal velocity of the drop will be:

$125$ small water drops of same size fall through air with constant terminal velocity $4 \,cm/s$. They coalesce to form a big drop. The terminal velocity of the big drop is: (in $\,m/s$)

$A$ sphere is dropped under gravity through a fluid of viscosity $\eta$. If the average acceleration is half of the initial acceleration,the time to attain the terminal velocity is ($\rho$ = density of sphere; $r$ = radius).