The graph between terminal velocity $(v_t)$ (along $y$-axis) and the square of the radius $(r^2)$ (along $x$-axis) of a spherical body of density $\rho$ falling through a fluid of density $\sigma$ is:

$A$ water drop of radius $1\,\mu m$ falls in a situation where the effect of buoyant force is negligible. The coefficient of viscosity of air is $1.8 \times 10^{-5}\,Nsm^{-2}$ and its density is negligible as compared to that of water $10^{3}\,kgm^{-3}$. The terminal velocity of the water drop is . . . . . . $\times 10^{-6}\,ms^{-1}$. (Take acceleration due to gravity $= 10\,ms^{-2}$)

$A$ spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta$. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v_t)$ is called the time constant $(\tau)$. Dimensionally,$\tau$ can be represented by:

Two equal drops are falling through air with a steady velocity of $5 \, cm/s$. If these two drops coalesce,then the new terminal velocity will be ......... $cm/s$.

Eight equal drops of water,each of radius $r = 2 \ mm$,are falling through air with a terminal velocity of $16 \ cm/s$. The eight drops combine to form a single big drop. The terminal velocity of the bigger drop will be $.... \ cm/s$.

$A$ spherical ball of radius $1 \ mm$ and density $10.5 \ g/cc$ is dropped in glycerine of coefficient of viscosity $9.8 \ \text{poise}$ and density $1.5 \ g/cc$. The viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} \ N$. The value of $x$ is $\text{(Given, } g = 9.8 \ m/s^2 \text{ and } \pi = \frac{22}{7}\text{)}$.