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

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

e) none of the above answers are correct