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

A solid sphere of diameter 5 mm is rising through an oil at a constant velocity of 1 cm/s. The density and dynamic viscosity of the oil are 850 kg/m³ and 0.7 N-s/m², respectively. Find the density of the material from which the sphere is made.

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

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