Eight equal drops of water are falling through air with a steady velocity of $10 \,cm \,s^{-1}$. If the drops combine to form a single drop, then the terminal velocity of this big drop is:

$A$ 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:

$A$ rain drop of radius $r$ is falling through air,starting from rest. The work done by all the forces on the drop,when it attains terminal velocity,is proportional to

$A$ 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:

An iron sphere of mass $20 \times 10^{-3} \ kg$ falls through a viscous liquid with terminal velocity $0.5 \ ms^{-1}$. The terminal velocity (in $ms^{-1}$) of another iron sphere of mass $54 \times 10^{-2} \ kg$ is (in $.5$)

$A$ small drop of water falls from rest through a large height $h$ in air; the final velocity is ................