Question

The viscosity of motor oil in which a steel ball falls with a terminal speed is given by:

(a) η = (4.32 cm/s)/(2R^2g/(9(rhos - rho₁)))
(b) η = (2R^2g/(9(rhos - rho₁)))/(4.32 cm/s)
(c) η = (4.32 cm/s)(2R^2g/(9(rhos - rho₁)))
(d) η = (2R^2g/(9(rhos - rho₁)))(4.32 cm/s)

Answer 1

The viscosity of motor oil in which a steel ball falls with a terminal speed can be determined using the equation η = (2R^2g/(9(rhos - rho₁)))/(4.32 cm/s). To find the viscosity, you need to plug in the given values for the radius of the steel ball, terminal speed, density of the ball, and density of the oil.

Step-by-step explanation:

The viscosity of motor oil in which a steel ball falls with a terminal speed is given by the equation:

η = (2R^2g/(9(rhos - rho₁)))/(4.32 cm/s)

Where η is the viscosity, R is the radius of the steel ball, g is the acceleration due to gravity, rhos is the density of the ball, and rho₁ is the density of the oil.

To find the viscosity of the motor oil in this problem, we need to plug in the given values. The radius of the steel ball is 0.8 mm, the terminal speed is 4.32 cm/s, the density of the ball is 7.86 g/mL, and the density of the oil is 0.88 g/mL.

By substituting these values into the equation, we can calculate the viscosity of the motor oil.