A small plastic sphere of diameter 4 mm is observed to fall with a constant terminal velocity of 5 cm s ⁻¹ through a tall jar containing a viscous liquid. If the coefficient of viscosity of the liquid is 0.2 N s m ⁻² , what is the magnitude of the viscous drag force acting on the sphere? (Take = 3.14 )

Step-by-step solution

According to Stokes' law, the viscous drag force F acting on a spherical body of radius r falling with terminal velocity v through a fluid of viscosity is given by: F = 6 r v Given: Radius of the sphere, r = 4 mm 2 = 2 10⁻³ m Terminal velocity, v = 5 cm s ⁻¹ = 5 10⁻² m s ⁻¹ Coefficient of viscosity, = 0.2 N s m ⁻² Substituting the values into the formula: F = 6 3.14 0.2 (2 10⁻³) (5 10⁻²) F = 6 3.14 0.2 10 10⁻⁵ F = 12 3.14 10⁻⁵ F = 37.68 10⁻⁵ N F = 3.768 10⁻⁴ N 3.77 10⁻⁴ N Answer: 3.77 10⁻⁴ N