If you were to measure the viscosity of a fluid you would do the following experiment https://www.wikihow.com/Measure-Viscosity
The formula for dynamic viscosity is the following one:
$n=\frac{2gr^{2}(p_s-p_l)}{9v}=\frac{ms^{-2}m^{2}(kgm^{-3})}{ms^{-1}}=kgm^{-1}s^{-1}= Pa·s$
where n is dynamic viscosity of the liquid in Pa s, g is gravity ($9.81{ms^-}^2$), r the radius of the ball in meters, $p_s$ density of sphere, $p_l$ density of liquid and v the velocity at which the ball travels through the fluid.
This source and many more state that the dynamic viscosity of water is $8.90*{10^-}^4$ Pa s https://www.engineersedge.com/physics/water__density_viscosity_specific_weight_13146.htm
If we plug this value for viscosity and try to find a value for the velocity at witch the sphere travels through the water (in this case). Then we come up with the following.
$8.90*{10^-}^4=\frac{2(9.81)r^{2}(p_s-1000)}{9v}$ (Density of water is 1000kg per meter cube)
Let's say that we drop an aluminum ball of 0.01m radius. Aluminum has a density of 2700kg/$m^3$
$8.90*{10^-}^4=\frac{2(9.81)(0.01)^{2}(2700-1000)}{9v}$
The final value for $v$ is 416.4 ${ms^-}^1$
Is the sphere really traveling through water at that high speed. Is the formula wrong? Or is it my math or what? I'm so confused.