What is the velocity $v$ of a metallic ball of radius $r$ falling in a tank of liquid at the instant when its acceleration is one-half that of a freely falling body ? (The densities of metal and of liquid are $\rho$ and $\sigma$ respectively, and the viscosity of the liquid is $\eta$).
$\frac{{{r^2}g}}{{9\eta }}(\rho - 2\sigma )$
$\frac{{{r^2}g}}{{9\eta }}(2\rho - \sigma )$
$\frac{{{r^2}g}}{{9\eta }}(\rho - \sigma )$
$\frac{{2{r^2}g}}{{9\eta }}(\rho - \sigma )$
The terminal velocity of a copper ball of radius $5\,mm$ falling through a tank of oil at room temperature is $10\,cm\,s ^{-1}$. If the viscosity of oil at room temperature is $0.9\,kg\,m ^{-1} s ^{-1}$, the viscous drag force is :
The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to
The graph between terminal velocity (along $y-axis$ ) and square of radius (along $x-axis$ ) of spherical body of density $\rho $ allowed to fall through a fluid of density $\rho $ is $a$
A raindrop with radius $R=0.2\, {mm}$ fells from a cloud at a height $h=2000\, {m}$ above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attainde by the raindrop is : (In ${ms}^{-1}$)
[Density of water $f_{{w}}=1000\;{kg} {m}^{-3}$ and density of air $f_{{a}}=1.2\; {kg} {m}^{-3}, {g}=10 \;{m} / {s}^{2}$ Coefficient of viscosity of air $=18 \times 10^{-5} \;{Nsm}^{-2}$ ]
A liquid drop of mass $m$ and radius $r$ is falling from great height. Its velocity is proportional to ............