What is the terminal velocity of an air bubble of $1.0 \ mm$ in diameter rising in a liquid of viscosity $0.85 \ N \ s \ m^{-2}$ and density $900 \ kg \ m^{-3}$? (Air density $= 1.293 \ kg \ m^{-3}$, $g = 9 \ m \ s^{-2}$)

In an experiment to calculate the coefficient of viscosity of a liquid by measuring terminal velocity, it was observed that a ball of mass $(\frac{\pi}{2}) \ gm$ and radius $0.5 \ cm$ takes $5 \ s$ to fall steadily through a height of $50 \ cm$ inside a long column of liquid of density $1.2 \ gm/cc$. The coefficient of viscosity will be given as: (in $\text{poise}$)

If the terminal speed of a sphere of gold (density $\rho_g = 19.5 \times 10^3 \ kg/m^3$) is $0.2 \ m/s$ in a viscous liquid (density $\rho_L = 1.5 \times 10^3 \ kg/m^3$),find the terminal speed of a sphere of silver (density $\rho_s = 10.5 \times 10^3 \ kg/m^3$) of the same size in the same liquid. (in $m/s$)

The velocity of a small ball of mass $m$ and density $d_{1}$,when dropped in a container filled with glycerine,becomes constant after some time. If the density of glycerine is $d_{2}$,then the viscous force acting on the ball will be:

$A$ liquid drop of mass $m$ and radius $r$ is falling from a great height. Its terminal velocity is proportional to ............

The terminal velocity of a liquid drop of radius $r$ falling through air is $v$. If two such drops are combined to form a bigger drop,the terminal velocity with which the bigger drop falls through air is (ignore any buoyant force due to air).