Question

The terminal speed (v) of a spherical particle falling in a liquid is given by:

(a) v = (2R^2g)/(9η(rhos - rho₁))
(b) v = (9η(rhos - rho₁))/(2R^2g)
(c) v = (2Rg)/(9η(rhos - rho₁))
(d) v = (9η(rhos - rho₁))/(2Rg)

Answer 1

The terminal speed (v) for a spherical particle falling in a liquid is calculated by balancing gravitational, drag, and buoyant forces, using the equation v = (2R²g)/(9η(ρs - ρ₁)). The correct equation among the options given is (a).

Step-by-step explanation:

The terminal speed (v) of a spherical particle falling in a liquid is derived by balancing the gravitational force with the drag and buoyant forces on the particle. Stokes Law is used to describe the drag force acting on the particle with viscosity η. The formula for the terminal speed (v) is given by v = (2R²g)/(9η(ρs - ρ₁)), where R is the radius of the sphere, ρs is the density of the spherical particle, ρ₁ is the density of the fluid, η is the coefficient of viscosity, and g is the acceleration due to gravity.

The correct expression for the terminal velocity in the given choices is (a) v = (2R²g)/(9η(ρs - ρ₁)).