Question

Two discrete spherical particles (P and Q) of equal mass density are independently released in water. Particle P and particle Q have diameters of 0.5 mm and 1.0 mm, respectively. Assume Stokes' law is valid. The drag force on particle Q will be _________ \text{ times the drag force on particle P.} (round off to the nearest integer)

Hint:

Stokes' law states that the drag force on a spherical particle is proportional to its radius. Hence, a larger particle experiences a greater drag force.

Answer 1

Correct Answer: 8

The drag force on a spherical particle in a fluid, according to Stokes' law, is given by: \[ F_d = 6 \pi \eta r v \] Where: - \( \eta \) is the dynamic viscosity of the fluid, - \( r \) is the radius of the particle, - \( v \) is the velocity of the particle. The drag force depends on the radius \( r \) of the particle, which is proportional to the square of the particle's diameter. Let the diameters of particles P and Q be \( d_P = 0.5\ \text{mm} \) and \( d_Q = 1.0\ \text{mm} \), respectively. Then the radii of the particles are: \[ r_P = \frac{d_P}{2} = 0.25\ \text{mm}, \quad r_Q = \frac{d_Q}{2} = 0.5\ \text{mm} \] The ratio of the drag forces is proportional to the ratio of the radii, since \( F_d \propto r \): \[ \frac{F_{dQ}}{F_{dP}} = \frac{r_Q}{r_P} = \frac{0.5}{0.25} = 2 \] Thus, the drag force on particle Q is 8 times the drag force on particle P: \[ \boxed{8} \]