$A$ 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:

$A$ sphere of radius $r$ and density $\rho$ is dropped from a height $h$. When it falls into water,it attains a terminal velocity. If the coefficient of viscosity of water is $\eta$,then $h =$

In Millikan's oil 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 air.

$A$ 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 $\rho_w = 10^3 \,kg \,m^{-3}$)

$A$ spherical ball of radius $1 \text{ mm}$ and density $10.5 \text{ g/cc}$ is dropped in glycerine of coefficient of viscosity $9.8 \text{ poise}$ and density $1.5 \text{ g/cc}$. Viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} \text{ N}$. The value of $x$ is (Given, $g = 9.8 \text{ m/s}^2$ and $\pi = 22/7$)

An object falling through a fluid is observed to have acceleration given by $a = g - bv$,where $g$ is the gravitational acceleration and $b$ is a constant. After a long time of release,it is observed to fall with a constant speed. What must be the value of this constant speed?