Mercury is filled in a tube of radius $2 \mathrm{~cm}$ up to a height of $30 \mathrm{~cm}$. The force exerted by mercury on the bottom of the tube is. . . . . . $\mathrm{N}$.

(Given, atmospheric pressure $=10^5 \mathrm{Nm}^{-2}$, density of mercury $=1.36 \times 10^4 \mathrm{~kg} \mathrm{~m}^3, \mathrm{~g}=10 \mathrm{~ms}^2$, $\left.\pi=\frac{22}{7}\right)$

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 

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?

In Millikan's oll drop experiment, what is the terminal speed of an uncharged drop of radius $2.0 \times 10^{-5} \;m$ and density $1.2 \times 10^{3} \;kg m ^{-3} .$ Take the viscosity of air at the temperature of the experiment to be $1.8 \times 10^{-5}\; Pa\; s$. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to atr.

A sphere is dropped under gravity through a fluid of viscosity $\eta$ . If the average acceleration is half of the initial acceleration, the time to attain the terminal velocity is ($\rho$ = density of sphere ; $r$ = radius)

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