Question 15.12: A solid sphere of diameter 5 mm is rising through an oil at ......

Given data:
Diameter of sphere                           D = 5 mm = 0.005 m
Velocity                                               U= 1 cm/s = 0.01 m/s

Density of oil                                  \rho_{o}=850\,{\mathrm{kg}}/{\mathrm{m}}^{3}

Viscosity of oil                                  μ = 0.7 N-s/m²

Reynolds number is found to be

R e=\frac{\rho U D}{\mu}=\frac{850 \times 0.01 \times 0.005}{0.7}=0.06<1

The total drag force is found from Stokes’ formula (Eq. (15.15))

F_{D}=3\pi\mu D U_{\infty}  (15.15)

F_D=3 \pi \mu D U

=3\pi\times0.7\times0.005\times0.01=0.0033\,\mathrm{N}

Let {\boldsymbol{\rho}}_{s} be the density of sphere.

\therefore \text { Weight of sphere }                     W = Density of sphere x g x Volume of sphere

=\rho_s \times g \times \frac{\pi}{6} D^3

=\rho_{s}\times9.81\times{\frac{\pi}{6}}(0.005)^{3}=0.00000064\rho_{s}\,\mathrm{N}

\therefore  Buoyant force      {\boldsymbol{F}}_{B} = Density of oil × g × Volume of sphere

=\rho_{o}\times g\times{\frac{\pi}{6}}D^{3}=850\times9.81\times{\frac{\pi}{6}}(0.005)^{3}=0.000546~\mathrm{N}

Using Eq. (15.20), we have

W=F_D+F_B

or                                                0.00000064 \rho_s=0.00033+0.000546

or                                                  \rho_s=\frac{0.000876}{0.00000064}=1368.75 \mathrm{~kg} / \mathrm{m}^3