Answer :

Solution :

Drag force, $\mathrm{D}=({\mathrm{C}}_{\mathrm{D}})(0.5\rho {\mathrm{V}}^2)\left(\frac{\pi {\mathrm{d}}^2}{4}\right),$${\mathrm{C}}_{\mathrm{D}}=\frac{24}{\mathrm{R}\mathrm{e}}$

Dynamic viscosity, μ = 1.1 × 10^-3 kg∙m^-1∙s^-1

Density of bead ρ = 11000 kg/m³, d = 0.1 mm

Buoyant force F_b = 0, W = weight of body

The net force on the body sinking is :

W - F_b = D ⇒ W = D

Weight of the sphere bead W = mg = Volume × density × g

Volume of sphere = $\frac{4}{3}\pi r^3$= $\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3$

W = $\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3\rho g$

For Drag force calculation we need to calculate :

Reynolds number Re = $\frac{\rho Vd}{\mu }$ =$\rho V\frac{0.1\times {10}^{-3}}{1.1\times 0^{-3}}$

Re = ρV/11

$\mathrm{D}=({\mathrm{C}}_{\mathrm{D}})(0.5\rho {\mathrm{V}}^2)\left(\frac{\pi {\mathrm{d}}^2}{4}\right)$, ${\mathrm{C}}_{\mathrm{D}}=\frac{24}{\mathrm{R}\mathrm{e}}$ 

D = $\frac{24\times 11}{\rho V}(0.5\rho V^2)\left(\frac{\pi d^2}{4}\right)$ = 132.V. $\left(\frac{\pi d^2}{4}\right)$

Now, W = D

$\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3\rho g$ = 132 x V $\left(\frac{\pi d^2}{4}\right)$
V = $\frac{\rho gd}{6\times 33}$ = $\frac{11000\times 10\times 0.1\times {10}^{-3}}{6\times 33}$

V = 1/18 m/s