Spherical particles of density 3.0 g/cm³ are shaken in a container of water (viscosity = 1.0 × 10−³ N·s/m³).The water is 8.0 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) 4.5 × 10−⁶ m B) 9.0 × 10−⁶ m C) 2.6 × 10−⁶ m D) 5.6 × 10−⁶ m

Question

Quizplus · Accepted Answer

The terminal velocity of the particles can be found using the Stokes' Law formula: v = (2/9) * (density difference) * g * r^2 / viscosity, where g is the gravitational constant and r is the radius of the particle. At terminal velocity, the buoyant force and the gravitational force are equal. 
Setting v equal to 0.8 cm/s (8.0 cm / 30 minutes) and solving for r gives us r = (9 * viscosity * v / (2 * density difference * g))^1/2. Plugging in the given values, we get r = 2.6 × 10−6 m. Therefore, the correct answer is C.

Spherical Particles of Density 3