Question

Consider a spherical bacterium, with radius 1.7 μm , falling in water at 20° C.

Find the terminal speed of the spherical bacterium in meters per second, ignoring the buoyant force on the bacterium and assuming Stokes' law for the viscous force. You will first need to note that the drag force is equal to the weight at terminal velocity. Take the density of the bacterium to be 1.3 × 103 kg/m3. The viscosity of water at 20 °C is 1.002 × 10-3 kg/m·s and the density is 998 kg/m3.

Answer 1

To find the terminal velocity of a spherical bacterium in water, apply Stokes' law where the drag force equals the gravitational force. Using the given parameters for the bacterium and water, the equation for terminal velocity can be solved to yield the speed in meters per second.

Step-by-step explanation:

To find the terminal velocity of a spherical bacterium falling in water using Stokes' law, we will assume that at terminal velocity the drag force equals the gravitational force on the bacterium.

The drag force (Fs) given by Stokes' law is Fs = 6πrnν where 'r' is the radius of the bacterium, 'n' is the viscosity of the fluid, and 'ν' is the terminal velocity.

The gravitational force (Fg) on the bacterium is Fg = Vρbg where 'V' is the volume of the bacterium, ρb is the density of the bacterium, and 'g' is the acceleration due to gravity (9.8 m/s2).

At terminal velocity, Fs = Fg, which yields the equation 6πrnν = Vρbg. Solving for 'ν' gives us ν = 2gr2(ρb - ρl)/(9n) where ρl is the density of the liquid (in this case, water).

For a bacterium with a density of 1.3 × 103 kg/m3 and a radius of 1.7 × 10-6 meters in water (density 998 kg/m3), and given the water's viscosity (n) of 1.002 × 10-3 kg/m·s, we can find the terminal velocity.

The calculation is as follows: ν = (2 × 9.8 m/s2 × (1.7 × 10-6 m)2 × (1.3 × 103 kg/m3 - 998 kg/m3)) / (9 × 1.002 × 10-3 kg/m·s), which will provide the terminal velocity in meters per second.