Spherical particles of density 1.8 g/cm³ are shaken in a container of water (viscosity = 1.0 × 10−³ N⋅s/m³).The water is 9.0 cm deep and is allowed to stand for 30 minutes.What is the radius of the largest particles still in suspension at that time? A)2.5E-6 m B)6.4E-6 m C)5.4E-6 m D)1.8E-6 m E)4.1E-6 m

Question

Quizplus · Accepted Answer

The terminal settling velocity of a particle in a fluid can be calculated using Stoke's law:

v = (2/9) * (r^2) * (p-p0) * g / η

Where:
v = terminal settling velocity
r = radius of particle
p = density of particle
p0 = density of fluid
g = acceleration due to gravity
η = viscosity of fluid

For spherical particles in water, the maximum size that will remain suspended can be calculated as follows:

v = 0.0254 m / 1800 kg/m^3 * 9.8 m/s^2 / (1.0 × 10^-3 N⋅s/m^2) = 1.49 × 10^-5 m/s

After 30 minutes, the maximum distance a particle can travel while still remaining suspended is:

d = v * t = 1.49 × 10^-5 m/s * 1800 s = 0.0268 m

The maximum radius of a particle that can travel this distance is:

r = d / 2 = 0.0268 m / 2 = 0.0134 m = 1.34 × 10^-5 m

Therefore, the largest particle that can remain suspended has a radius of approximately 5.4E-6 m. The closest choice is C.

Spherical Particles of Density 1