A rigid sphere is falling from a height of 1m in a fluid of density 1000 m3/kg. The time it took to fall is 500 seconds. If the density of the sphere is 1100 m3/kg and the viscosity of the fluid is 100 cP, what is the diameter of the sphere? (a. 6.1 m , b. 0.61 m , c. 0.061 m , d. 0.0061 m)

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Tringa0 Tringa0 · Answer 1 · 2019-07-12T09:32:46+02:00

Answer:

0.00191 meter is the diameter of the sphere.

Explanation:

Velocity of the rigid sphere = [tex]v=\frac{1m}{500 s}[/tex]

Density of the fluid = [tex]\rho _f=1000 kg/m^3[/tex]

Density of the particle= [tex]\rho _p=1100 kg/m^3[/tex]

Viscosity of the fluid =[tex]\eta =100 centiPoise = 1 Poise=0.1 Kg/(m s)[/tex]

Radius of the sphere = r

[tex]v=\frac{2}{9}\times \frac{\rho _p-\rho _f}{\eta }gr^2[/tex]

(Terminal velocity)

[tex]\frac{1 m}{500 s}=\frac{2}{9}\times \frac{1100 kg/m^3-1000 kg/m^3}{0.1 kg/(m s)}\times 9.8 m/s^2\times r^2[/tex]

r = 0.00191 m

0.00191 meter is the diameter of the sphere.