The velocity of a small spherical ball of mass $m$ and density $d_1$,when dropped in a container filled with glycerine,becomes constant after some time. The viscous force acting on the ball if the density of glycerine is $d_2$ is

$A$ small steel ball is dropped from a height of $1.5 \,m$ into a glycerine jar. The ball reaches the bottom of the jar $1.5 \,s$ after it was dropped. If the retardation in the glycerine is $2.66 \,m/s^2$, the height of the glycerine in the jar is about (acceleration due to gravity $g = 9.8 \,m/s^2$) (in $\,m$)

$A$ ball of radius $r$ and density $\rho$ falls freely under gravity through a distance $h$ before entering water. The velocity of the ball does not change even on entering the water. If the viscosity of water is $\eta$ and the density of water is $\sigma$ (assumed to be $1$ unit for calculation),the value of $h$ is given by:

Eight identical drops of same radii are falling through air with steady velocity $6 \ cm/s$. If these drops coalesce,then the terminal velocity would be .......... $cm/s$.

The terminal velocity of a small-sized spherical body of radius $r$ falling vertically in a viscous liquid is given by the proportionality:

$A$ solid metal sphere released in a vertical liquid column has attained terminal velocity in the downward direction. The magnitudes of viscous force,buoyant force,and gravitational force acting on it are $F_{v}$,$F_{B}$,and $F_{W}$ respectively. Then,the correct relation between them is: