A small glass marble of radius 2 ~mm falls with a constant velocity of 5 ~cm ~s ⁻¹ through a tall cylinder filled with a liquid. If the viscous drag force acting on the marble is 3.768 10⁻³ ~N , the coefficient of viscosity of the liquid is: (Take = 3.14 )

Step-by-step solution

According to Stokes' law, the viscous drag force is given by: F = 6 r v Rearranging the formula to solve for the coefficient of viscosity : = F 6 r v Given values: F = 3.768 10⁻³ ~N r = 2 ~mm = 2 10⁻³ ~m v = 5 ~cm ~s ⁻¹ = 5 10⁻² ~m ~s ⁻¹ Substituting the values: = 3.768 10⁻³ 6 3.14 2 10⁻³ 5 10⁻² = 3.768 10⁻³ 18.84 10 10⁻⁵ = 3.768 10⁻³ 1.884 10⁻³ = 2.0 ~Pa s Answer: 2.0 ~Pa s