$A$ gas bubble of $2 \ cm$ diameter rises through a liquid of density $1.75 \ g \ cm^{-3}$ with a fixed speed of $0.35 \ cm \ s^{-1}$. Neglect the density of the gas. The coefficient of viscosity of the liquid is (in $\text{poise}$)

$A$ lead sphere of mass $m$ falls in a viscous liquid with terminal velocity $V_0$. Another lead sphere of mass $8m$ but of the same material will fall through the same liquid with terminal velocity:

$Assertion :$ Falling raindrops acquire a terminal velocity.
$Reason :$ $A$ constant force in the direction of motion and a velocity dependent force opposite to the direction of motion,always result in the acquisition of terminal velocity.

What is the terminal velocity of a rain drop of radius $0.02 \ mm$ (in $cm \ s^{-1}$)? [Note that the coefficient of viscosity of air is $1.8 \times 10^{-5} \ N \ s \ m^{-2}$, density of water is $1000 \ kg \ m^{-3}$. Use $g = 10 \ m \ s^{-2}$ and density of air can be neglected in comparison with the density of water.]

$A$ 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:

Two spheres of the same material, but of radii $R$ and $3R$ are allowed to fall vertically downwards through a liquid of density $\rho$. The ratio of their terminal velocities is