$A$ 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 small spherical metal balls,having equal masses,are made from materials of densities $\rho_{1}$ and $\rho_{2}$ (where $\rho_{1} = 8 \rho_{2}$) and have radii of $1 \; mm$ and $2 \; mm$,respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals $\eta$ and whose density is $0.1 \rho_{2}$. The ratio of their terminal velocities would be:

$A$ large drop of oil (density $0.8 \, g/cm^3$ and viscosity $\eta_0$) floats up through a column of another liquid (density $1.2 \, g/cm^3$ and viscosity $\eta_L$). Assuming that the two liquids do not mix,the velocity with which the oil drop rises will depend on.

While determining the coefficient of viscosity of the given liquid,a spherical steel ball sinks by a distance $h=0.9 \,m$. The radius of the ball $r=\sqrt{3} \times 10^{-3} \,m$. The time taken by the ball to sink in three trials are tabulated as follows:

Trial No.	Time taken by the ball to fall by $h$ (in second)
$1$.	$2.75$
$2$.	$2.65$
$3$.	$2.70$

The difference between the densities of the steel ball and the liquid is $7000 \,kg \,m^{-3}$. If $g=10 \,ms^{-2}$,then the coefficient of viscosity of the given liquid at room temperature is

Eight identical drops of water falling through air with a uniform terminal velocity of $10 \ cm/s$ combine to form a single big drop. The terminal velocity of the big drop will be: (in $cm/s$)

$A$ metal sphere of mass $m$ and density $\sigma_1$ falls with terminal velocity through a container containing liquid. The density of the liquid is $\sigma_2$. The viscous force acting on the sphere is: