The terminal velocity $\left( v _{ t }\right)$ of the spherical rain drop depends on the radius ( $r$ ) of the spherical rain drop as
$r^{1 / 2}$
$r$
$r^{2}$
$r ^{3}$
The diameter of an air bubble which was initially $2\,mm$, rises steadily through a solution of density $1750\,kg\,m\,m ^{-3}$ at the rate of $0.35\,cms ^{-1}$. The coefficient of viscosity of the solution is poise (in nearest integer). (the density of air is negligible).
A Spherical ball of radius $1 mm$ and density $10.5 g / cc$ is dropped in glycerine of coefficient of viscosity $9.8$ poise and density $1.5 g / cc$. Viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} N$. The value of $x$ is $\text { (Given, } g =9.8 m / s ^2 \text { and } \pi=\frac{22}{7} \text { ) }$
If the terminal speed of a sphere of gold ( density $= 19.5 kg/m^3$) is $0.2\ m/s$ in a viscous liquid (density $= 1.5\ kg/m^3$ ), find the terminal speed (in $m/s$) of a sphere of silver (density $= 10.5\ kg/m^3$) of the same size in the same liquid ...... $m/s$
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)