$A$ small metal sphere of radius $a$ is falling with a velocity $v$ through a vertical column of a viscous liquid. If the coefficient of viscosity of the liquid is $\eta$, then the sphere encounters an opposing force of

$A$ 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 $\text{(Given, } g = 9.8 \ m/s^2 \text{ and } \pi = \frac{22}{7}\text{)}$.

The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to

Two spheres of the same material, but of radii $R$ and $3R$ are allowed to fall vertically downwards through a liquid of density $\rho$. The ratio of their terminal velocities is

$A$ metal sphere of radius $R$ and density $\varrho_{1}$ moves with terminal velocity $v_{1}$ through a liquid of density $\sigma$. Another sphere of the same radius but of density $\varrho_{2}$ moves through the same liquid. Its terminal velocity will be:

$A$ 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 $= 10\,ms^{-2}$)