Final answer:
The terminal velocity of a sphere in a viscous fluid is proportional to the square of the radius of the sphere, the difference in density between the sphere and the fluid, and inversely proportional to the fluid viscosity. Therefore, if all other factors are constant, terminal velocity is inversely proportional to the radius of the sphere, making the correct answer (b) 1/r.
Step-by-step explanation:
The question asks about the terminal velocity of a sphere falling in a viscous fluid. Terminal velocity is reached when the gravitational force acting downwards on the body is balanced by the drag force acting upwards, plus any buoyant force acting upwards. The drag force experienced by a spherical object in a viscous fluid at low Reynolds numbers can be described by Stokes' Law, which states that the drag force, Fs, is proportional to the velocity v, the fluid's viscosity n, and the radius r of the sphere: Fs = 6πrηv. At terminal velocity, the net force on the sphere is zero, hence gravitational force minus buoyant force equals the drag force.
When solving for terminal velocity, the sphere's mass m and radius r are taken into account to find gravitational force and buoyant force, respectively. For a sphere of density ps and radius R, falling through a fluid of density p1 with viscosity n, the terminal velocity v can be shown to be proportional to R2(ps - p1)/n. This relationship indicates that the terminal velocity is dependent on the square of the radius, the difference in density between the sphere and the fluid, and inversely proportional to the viscosity.
Therefore, the terminal velocity is not directly proportional to just any single factor such as 1/√r, 1/r, 1/√m, or 1/m, rather it depends on a combination of these factors as indicated above. The correct answer from the provided options would then be (b) 1/r, considering that if we keep all other factors constant, an increase in radius r would cause a decrease in terminal velocity, assuming a proportional relationship.