Two small spherical metal balls, having equal masses, are made from materials of densities $\rho_{1}$ and $\rho_{2}\left(\rho_{1}=8 \rho_{2}\right)$ and have radii of $1\; \mathrm{mm}$ and $2\; \mathrm{mm}$, respectively. They are made to fall vertically (from rest) in a viscous medum whose coefficient of viscosity equals $\eta$ and whose denstry is $0.1 \mathrm{\rho}_{2} .$ The ratio of their terminal velocitites would be 

A spherical ball of radius $1 \times 10^{-4} \mathrm{~m}$ and density $10^5$ $\mathrm{kg} / \mathrm{m}^3$ falls freely under gravity through a distance $h$ before entering a tank of water, If after entering in water the velocity of the ball does not change, then the value of $h$ is approximately:

(The coefficient of viscosity of water is $9.8 \times 10^{-6}$ $\left.\mathrm{N} \mathrm{s} / \mathrm{m}^2\right)$

A ball of radius $r $ and density $\rho$ falls freely under gravity through a distance $h$ before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $\eta$, the value of $h$ is given by

Write $\mathrm{SI}$ and $\mathrm{CGS}$ unit of coefficient of viscosity.

If a ball of steel (density $\rho=7.8 \;gcm ^{-3}$) attains a terminal velocity of $10 \;cms ^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text {water }}=8.5 \times 10^{-4} \;Pa - s$ ) then its terminal velocity in glycerine $\left(\rho=12 gcm ^{-3}, \eta=13.2\right)$ would be nearly

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