The largest and the smallest balls used in the experiment are with diameter 9.52 mm, and 2.38 mm respectively. For a glycerin with viscosity 1.0 Pa. s, what is the time necessary for each ball to reach a velocity 95% of the terminal velocity? Density of the ball material is 1.42 g/cm^3. Round the result to three decimal places.

The answer is = 0.0080

Explanation:

From the stokes law,

v = g * D^2 * (d p - d m) / (18 V)

Here, v is the terminal Velocity, D is the diameter Of a Particle, V is the Viscosity of Medium, dp is the density Of Particle, dm is the density Of Medium.

The density of ball is 1.42 gm/cc

v = 9.81 x 9.52 x 10^-3 ^2 (1420 - 1300) / 18 x 1.0

=5.92x10^-4

v (t) = 0.99 x V trm

=0.99 x 0.0059

= 0.00059 m/s

v (t) / Vterm = 1-e^ (-t/r)

0.99 = 1 - e ^ (-t / 9.52 x 10^-3)

e ^ (-t / 9.52 x 10^-3) = 0.01

Take natural log

-t / 9.52 x 10^-3 = - 4.6

t = 0.0438 s

For the smaller ball

v = 9.81 x 2.38 x 10^-3 ^2 (1420 - 1300) / 18 x 1.0

= 0.000037 m/s

v (t) / Vterm=1-e^ (-t/r)

0.99 = 1 - e ^ (-t / 2.38 x 10^-3)

e ^ (-t / 2.38 x 10^-3) = 0.01

Take natural log

-t / 2.38 x 10^-3 = - 4.6

t = 0.0109 s

Reynolds number is,

R = rho vd / mu

= 1420 x 0.00059 x 9.52 x 10^-3 x / 1.0

= 0.0080