An air bubble of radius $1 \ cm$ rises from the bottom portion through a liquid of density $1.5 \ g/cc$ at a constant speed of $0.25 \ cm \ s^{-1}$. If the density of air is neglected,the coefficient of viscosity of the liquid is approximately,(in $Pa \ s$):

$A$ steel ball of radius $0.05 \,cm$ and density $7.8 \,g \,cm^{-3}$ is dropped into a tank of water. The terminal velocity of the steel ball is (Density of water $= 1 \,g \,cm^{-3}$ and viscosity of water $= 0.001 \,Pa \,s$) (in $\,m/s$)

Consider two solid spheres $P$ and $Q$ each of density $8 \ g \ cm^{-3}$ and diameters $1 \ cm$ and $0.5 \ cm$,respectively. Sphere $P$ is dropped into a liquid of density $0.8 \ g \ cm^{-3}$ and viscosity $\eta = 3 \ \text{poiseuille}$. Sphere $Q$ is dropped into a liquid of density $1.6 \ g \ cm^{-3}$ and viscosity $\eta = 2 \ \text{poiseuille}$. The ratio of the terminal velocities of $P$ and $Q$ is:

$A$ 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).

The terminal velocity of a copper ball of radius $2.0 \; mm$ falling through a tank of oil at $20 \; ^{\circ}C$ is $6.5 \; cm \; s^{-1}$. Compute the viscosity of the oil at $20 \; ^{\circ}C$. Density of oil is $1.5 \times 10^{3} \; kg \; m^{-3}$,density of copper is $8.9 \times 10^{3} \; kg \; m^{-3}$.

Two solid spheres of the same metal but of mass $M$ and $8M$ fall simultaneously in a viscous liquid. If their terminal velocities are $v$ and $nv$,then the value of $n$ is: