$A$ small metallic sphere of diameter $2 \ mm$ and density $10.5 \ g/cm^3$ is dropped in glycerine having viscosity $10 \ \text{Poise}$ and density $1.5 \ g/cm^3$. The terminal velocity attained by the sphere is . . . . . . $cm/s$. $(\pi = \frac{22}{7}$ and $g = 10 \ m/s^2)$

State Stokes' law. By using it,deduce the expression for:
$(i)$ Initial acceleration of a smooth sphere.
$(ii)$ Equation of terminal velocity of a sphere falling freely through a viscous medium.
$(iii)$ Explain: Upward motion of bubbles produced in a fluid.

$A$ small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If after entering the water the velocity of the ball does not change and it continues to fall with the same constant velocity inside the water,then the value of $h$ will be $m$. (Given $g = 10 \,m \,s^{-2}$,viscosity of water $\eta = 1.0 \times 10^{-5} \,N \,s \,m^{-2}$,density of water $\rho_w = 10^3 \,kg \,m^{-3}$)

$A$ viscous fluid is flowing through a cylindrical tube. The velocity distribution of the fluid is best represented by the diagram

$A$ solid sphere falls with a terminal velocity of $10 \, cm/s$ in the Earth's gravitational field. If it is allowed to fall freely in a region outside the gravitational field of the Earth,the terminal velocity will be:

$A$ copper ball of radius $3.0 \,mm$ falls in an oil tank of viscosity $1 \,kg / ms$. Then, the terminal velocity of the copper ball will be (Density of oil $= 1.5 \times 10^3 \,kg / m^3$, Density of copper $= 9 \times 10^3 \,kg / m^3$ and $g = 10 \,m / s^2$.)