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

$A$ lead shot of $1 \, mm$ diameter falls through a long column of glycerine. The variation of its velocity $v$ with distance covered is represented by:

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)

Eight identical small drops of water are falling down vertically through a medium,each with terminal velocity $v$. If they combine to form a single drop,then its terminal velocity will be (in $v$)

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).

The velocity of a small ball of mass $0.3 \, g$ and density $8 \, g/cc$ when dropped in a container filled with glycerine becomes constant after some time. If the density of glycerine is $1.3 \, g/cc$,then the value of the viscous force acting on the ball will be $x \times 10^{-4} \, N$. The value of $x$ is [use $g = 10 \, m/s^2$]. (in $.125$)