The terminal velocity $(v_{t})$ of a spherical raindrop depends on the radius $(r)$ of the raindrop as:

$A$ solid ball of volume $V$ is dropped in a viscous liquid. It experiences a viscous force $F$. If a solid ball of volume $2V$ of the same material is dropped in the same fluid,then the viscous force acting on it will be

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6 \ cm \ s^{-1}$. If they coalesce to form one big drop,what will be the terminal speed of the bigger drop (in $cm \ s^{-1}$)? (Neglect the buoyancy of the air)

The terminal velocity of a copper ball of radius $2.0 \; mm$ falling through a tank of oil at $20 \; ^{\circ}C$ is $6.5 \; cm \; s^{-1}$. Compute the viscosity of the oil at $20 \; ^{\circ}C$. Density of oil is $1.5 \times 10^{3} \; kg \; m^{-3}$,density of copper is $8.9 \times 10^{3} \; kg \; m^{-3}$.

$A$ 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).

$A$ metal sphere of radius $R$,density $\rho_1$ moves with terminal velocity $V_1$ through a liquid of density $\sigma$. Another sphere of same radius but density $\rho_2$ moves through the same liquid. Its terminal velocity is $V_2$. The ratio $V_1: V_2$ is