$A$ 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:

An air bubble of radius $r$ rises steadily through a liquid of density $\rho$ with velocity $v$. The coefficient of viscosity of the liquid is:

If the terminal velocity of a metal sphere of mass $8 \ g$ falling through a liquid is $3 \ cm s^{-1}$, then the terminal velocity of another sphere of mass $64 \ g$ made of the same metal falling through the same liquid is (in $cm s^{-1}$)

Two solid spheres of radii $2 \ mm$ and $4 \ mm$ are tied to the two ends of a light string and released in a liquid of specific gravity $1.3$ and coefficient of viscosity $1 \ Pa \cdot s$. The string is just taut when the two spheres are completely submerged in the liquid. If the density of the materials of the two spheres is $2800 \ kg \cdot m^{-3}$,then the terminal velocity of the system of the spheres is (take $g = 10 \ m \cdot s^{-2}$):

What is the rate at which a trapped bubble of $2 \,mm$ diameter rises through a solution of density $13.6 \times 10^3 \,kg/m^3$ and coefficient of viscosity $1.5 \,cP$ (in $\,m/s$)? Assume the density of air is negligible and $g = 10 \,m/s^2$.

$A$ 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).