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 $Pa \ s$):

$A$ 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:

In an experiment to verify Stokes' law,a small spherical ball of radius $r$ and density $\rho$ falls under gravity through a distance $h$ in air before entering a tank of water. If the terminal velocity of the ball inside water is the same as its velocity just before entering the water surface,then the value of $h$ is proportional to: (ignore viscosity of air)

The speed of a ball of radius $2 \,cm$ in a viscous liquid is $20 \,cm / s$. What will be the speed of a ball of radius $1 \,cm$ in the same liquid (in $\,cm / s$)?

$A$ spherical solid ball of volume $V$ is made of a material of density $\rho_1$. It is falling through a liquid of density $\rho_2$ $(\rho_2 < \rho_1)$. Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$,i.e.,$F_{viscous} = -kv^2$ $(k > 0)$. The terminal speed of the ball is:

$A$ liquid disturbed by stirring comes to rest after some time due to its property of