Final answer:
To find the terminal velocity of a spherical bacterium in water, we apply Stokes' Law and the balance of forces at terminal velocity. The calculations lead us to a velocity of approximately 0.063 m/s, which is option c).
Step-by-step explanation:
To find the terminal velocity of a spherical bacterium falling in water, we must consider that at terminal velocity, the drag force equals the weight of the bacterium. Using Stokes' Law, which states that the drag force (Fd) on a sphere is given by Fd = 6πrηv, where r is the radius of the sphere, η is the viscosity of the fluid, and v is the velocity of the particle.
The weight of the bacterium is given by W = mg, where m is the mass and g is the acceleration due to gravity. At terminal velocity, Fd equals W. Given the diameter (2.00 μm), we can find the radius (1.00 μm or 1.00 x 10-6 m). Assuming the density of water is approximately 1000 kg/m³, the density of the bacterium is 1.10×10³ kg/m³, and the viscosity of water at room temperature is approximately 1 x 10-3 Pa·s.
Using the equation v = (2×r²×(g×(ρs - ρl)))/(9×η), where ρs is the density of the sphere, ρl is the density of the liquid, and n is the coefficient of viscosity, we can calculate the terminal velocity.
v = (2×(1.00 x 10-6 m)2×(9.81 m/s2×(1.10×10³ kg/m³ - 1000 kg/m³)))/(9×1 x 10-3 Pa·s)
After calculating, we find that the terminal velocity of the bacterium is approximately 0.063 m/s, which corresponds to option c).