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?

If the terminal velocity of a metal sphere of mass $8 \ g$ falling through a liquid is $3 \ cm s^{-1}$,then the terminal velocity of another sphere of mass $64 \ g$ made of the same metal falling through the same liquid is (in $cm s^{-1}$)

Consider two solid spheres $P$ and $Q$ each of density $8 \ g \ cm^{-3}$ and diameters $1 \ cm$ and $0.5 \ cm$,respectively. Sphere $P$ is dropped into a liquid of density $0.8 \ g \ cm^{-3}$ and viscosity $\eta = 3 \ \text{poiseuille}$. Sphere $Q$ is dropped into a liquid of density $1.6 \ g \ cm^{-3}$ and viscosity $\eta = 2 \ \text{poiseuille}$. The ratio of the terminal velocities of $P$ and $Q$ is:

$A$ spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta$. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v_t)$ is called the time constant $(\tau)$. Dimensionally,$\tau$ can be represented by:

$A$ spherical ball of radius $1 \times 10^{-4} \,m$ and density $10^5 \,kg/m^3$ 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, then the value of $h$ is approximately: (The coefficient of viscosity of water is $9.8 \times 10^{-6} \,N s/m^2$) (in $\,m$)

$A$ 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 $h =$