Two raindrops reach the earth with different terminal velocities having the ratio $9: 4$. The ratio of their volumes is

$A$ 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).

Sixty-four rain drops of radius $1 \ mm$ each, falling down with a terminal velocity of $10 \ cm/s$, coalesce to form a bigger drop. The terminal velocity of the bigger drop is . . . . . . $cm/s$.

$A$ small spherical solid ball is dropped from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the

$A$ small rigid spherical ball of mass $M$ is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball,then the viscous force acting on the ball will be (consider $g$ as acceleration due to gravity).

$A$ water drop of radius $1\,\mu m$ falls in a situation where the effect of buoyant force is negligible. The coefficient of viscosity of air is $1.8 \times 10^{-5}\,Nsm^{-2}$ and its density is negligible as compared to that of water $10^{3}\,kgm^{-3}$. The terminal velocity of the water drop is . . . . . . $\times 10^{-6}\,ms^{-1}$. (Take acceleration due to gravity $= 10\,ms^{-2}$)