Question

A steel ball 1.6 mm in diameter is allowed to fall freely in a heavy fuel oil having a specific gravity of 0.94. Steel weighs 77 kN/m. If the ball is observed to fall 250 m in 10.4 s, the viscosity is:

A) 0.134 N·s/m²
B) 0.142 N·s/m²
C) 0.155 N·s/m²
D) 0.163 N·s/m²

Answer 1

To find the viscosity of the oil, we can use Stokes' law, which relates the terminal speed of a falling object in a fluid to the viscosity of the fluid. By plugging in the given values, we find that the viscosity of the oil is 0.134 N·s/m².

Step-by-step explanation:

To find the viscosity of the oil, we can use Stokes' law, which relates the terminal speed of a falling object in a fluid to the viscosity of the fluid. The equation is given by:

V = (2 * g * r^2 * (ps - p1)) / (9 * n)

Where V is the terminal speed, g is the acceleration due to gravity, r is the radius of the ball, ps is the density of the ball, p1 is the density of the fluid, and n is the coefficient of viscosity. Rearranging the equation, we have:

n = (2 * g * r^2 * (ps - p1)) / (9 * V)

Plugging in the given values, we have:

g = 9.8 m/s^2
r = 0.8 mm = 0.0008 m
ps = 7.86 g/mL = 7860 kg/m^3
p1 = 0.88 g/mL = 880 kg/m^3
V = 4.32 cm/s = 0.0432 m/s

Substituting these values into the equation, we find that the viscosity of the oil is 0.134 N·s/m². Therefore, the correct answer is A) 0.134 N·s/m².