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:

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

$A$ small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change and it continues to fall with the same constant velocity inside the water,then the value of $h$ will be $m$. (Given $g = 10 \,m \,s^{-2}$,viscosity of water $\eta = 1.0 \times 10^{-5} \,N \,s \,m^{-2}$,density of water $\rho_w = 10^3 \,kg \,m^{-3}$)

Two drops of equal radius are falling with a terminal velocity of $5 \, cm/s$. If they coalesce to form a single larger drop,what will be the terminal velocity of the new drop?

$A$ spherical liquid drop of radius $r$ acquires the terminal velocity $v_1$ when falling through a gas of viscosity $\eta$. Now the drop is broken into $64$ identical droplets and each droplet acquires terminal velocity $v_2$ falling through the same gas. The ratio of terminal velocities $v_1/v_2$ is . . . . . . .

$A$ small rigid spherical ball of mass $M$ is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball,then the viscous force acting on the ball will be (consider $g$ as acceleration due to gravity).