An air bubble of radius $r$ rises steadily through a liquid of density $\rho $ with velocity $v$ . The coefficient of viscosity of liquid is
$\frac{2}{9}\frac{{{r^2}\rho g}}{v}$
$\frac{2}{9}\frac{{{r^2}g}}{{v\rho }}$
$\frac{2}{9}\frac{{v\rho g}}{{{r^2}}}$
$\frac{2}{9}\frac{{\rho g}}{{v{r^2}}}$
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)
An air bubble of $1\, cm$ radius is rising at a steady rate of $2.00\, mm/sec$ through a liquid of density $1.5\, gm$ per $cm^3$. Neglect density of air. If $g$ is $1000\, cm/sec^2$, then the coefficient of viscosity of the liquid is
In an experiment, a small steel ball falls through a Iiquid at a constant speed of $10\, cm/s$. If the steel ball is pulled upward with a force equal to twice its effective weight, how fast will it move upward ? ......... $cm/s$
A small spherical ball of radius $0.1 \,mm$ and density $10^{4} \,kg m ^{-3}$ falls freely under gravity through a a distance $h$ before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of $h$ wil be $m$. (Given $g =10 \,ms ^{-2}$, viscosity of water $=1.0 \times 10^{-5} \,N - sm ^{-2}$ )
Spherical balls of radius $ 'r'$ are falling in a viscous fluid of viscosity '$\eta$' with a velocity $ 'v'. $ The retarding viscous force acting on the spherical ball is