$A$ spherical metal ball of radius $r$ falls through a viscous liquid with terminal velocity $V$. Another metal ball of the same material but of radius $\frac{r}{3}$ falls through the same liquid. What will be its terminal velocity?

$A$ small sphere of mass $m$ is dropped from a great height. After it has fallen $100 \; m$,it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first $100 \; m$ of fall is

The velocity of a small ball of mass $M$ and density $d_1$ when dropped in a container filled with glycerin becomes constant after some time. If the density of glycerin is $d_2$,the viscous force acting on the ball is ($g$ = acceleration due to gravity).

$A$ solid sphere of radius $R$ acquires a terminal velocity $\nu_1$ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\eta$. The sphere is broken into $27$ identical solid spheres. If each of these spheres acquires a terminal velocity $\nu_2$ when falling through the same fluid,the ratio $(\nu_1/\nu_2)$ equals:

$A$ 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 $(v)$ is called the time constant $(\tau)$. Dimensionally,$\tau$ can be represented by:

$A$ metal sphere of radius $R$ and density $\rho_1$ is dropped in a liquid of density $\sigma$ and moves with terminal velocity $V$. Another metal sphere of the same radius and density $\rho_2$ is dropped in the same liquid. Its terminal velocity will be: