$A$ 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).

$A$ tiny spherical oil drop carrying a net charge $q$ is balanced in still air, with a vertical uniform electric field of strength $\frac{81}{7} \pi \times 10^5 \,V / m$. When the field is switched $OFF$, the drop is observed to fall with terminal velocity $2 \times 10^{-3} \,m / s$. Here $g=9.8 \,m / s^2$, viscosity of air is $\eta = 1.8 \times 10^{-5} \,N s / m^2$ and density of oil is $\rho = 900 \,kg / m^3$. The magnitude of $q$ is

$A$ gas bubble of $2 \ cm$ diameter rises through a liquid of density $1.75 \ g \ cm^{-3}$ with a fixed speed of $0.35 \ cm \ s^{-1}$. Neglect the density of the gas. The coefficient of viscosity of the liquid is (in $\text{poise}$)

If the terminal speed of a sphere $A$ [density $\rho_A = 7.5 \ kg \ m^{-3}$] is $0.4 \ ms^{-1}$ in a viscous liquid [density $\rho_L = 1.5 \ kg \ m^{-3}$],the terminal speed of sphere $B$ [density $\rho_B = 3 \ kg \ m^{-3}$] of the same size in the same liquid is: (in $ms^{-1}$)

$A$ spherical ball of radius $1 \times 10^{-4} \,m$ and density $10^4 \,kg \,m^{-3}$ falls freely under gravity through a distance $h$ before entering a tank of water. If the velocity of the ball does not change after entering the water, then the value of $h$ is:

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?