A ball rises to surface at a constant velocity in a liquid whose density is $4$ times greater than that of the material of the ball. The ratio of the force of friction acting on the rising ball and its weight is

Consider two solid spheres $\mathrm{P}$ and $\mathrm{Q}$ each of density $8 \mathrm{gm} \mathrm{cm}^{-3}$ and diameters $1 \mathrm{~cm}$ and $0.5 \mathrm{~cm}$, respectively. Sphere $\mathrm{P}$ is dropped into a liquid of density $0.8 \mathrm{gm} \mathrm{cm}^{-3}$ and viscosity $\eta=3$ poiseulles. Sphere $Q$ is dropped into a liquid of density $1.6 \mathrm{gm} \mathrm{cm}^{-3}$ and viscosity $\eta=2$ poiseulles. The ratio of the terminal velocities of $\mathrm{P}$ and $\mathrm{Q}$ is 

Two drops of same radius are falling through air with steady velocity of $v $ $cm/s$. If the two drops coalesce, what would be the terminal velocity?

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 water drop of radius $1\,\mu m$ falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is $1.8 \times 10^{-5}\,Nsm ^{-2}$ and its density is negligible as compared to that of water $10^{6}\,gm ^{-3}$. Terminal velocity of the water drop is________ $\times 10^{-6}\,ms ^{-1}$

(Take acceleration due to gravity $=10\,ms ^{-2}$ )