Question

A steel ball, with diameter = 3 mm and density = 7600 kg/m3 is dropped in a liquid with density = 1200 kg/m3 . The average time for the ball to travel a distance of 0.50 m is 10.0 s. a. Does the ball rise or fall in the fluid? Explain b. Determine the fluid viscosity.

Answer 1

Answer

given,

diameter of the steel ball = 3 m = 0.003 m

density of the steel = 7600 kg/m³

density of liquid = 1200 kg/m³

distance travel by the ball = 0.5 m

time = t = 10 s

average velocity =$v = \dfrac{0.5}{10}$

  v = 0.05 m/s

a) density of water is less than ball so, ball will fall in the fluid.

gravitational force is equal to buoyancy force and the drag force

$F_g = F_b + F_d$

$F_g = \rho_g Vg$

Density of ball = ρ_s

V is the volume ball

buoyancy force

$F_b = \rho_f Vg$

$F_b = \rho_f V g$

drag force

 $F_d =3 \pi \mu d v$

$F_g = F_b + F_d$

$\rho_g Vg = \rho_f V g+ 3 \pi \mu d v$

$(\rho_g - \rho_f)Vg = 3 \pi \mu d v$

$V= \dfrac{1}{6}\pi d^3$

$(\rho_g - \rho_f). \dfrac{1}{6}\pi d^3.g = 3 \pi \mu d v$

$\mu = \dfrac{(\rho_s-\rho_f)d^2g}{18 v}$

$\mu = \dfrac{(7600 -1200)\times 0.003^2\times 9.8 }{18 \times 0.05}$

μ = 0.63 kg m/s