Question

A large tank is filled with water (density \(=1\ \mathrm{g\,cm^{-3}}\)) up to a height of \(5\ \mathrm{m}\). A \(100\ \mu\mathrm{m}\) diameter solid spherical particle (density \(=0.8\ \mathrm{g\,cm^{-3}}\)) is released at the bottom of the tank. The particle attains its terminal velocity \((v_t)\) after traveling to a certain height in the tank. Use acceleration due to gravity as \(10\ \mathrm{m\,s^{-2}}\) and water viscosity as \(10^{-3}\ \mathrm{Pa\,s}\). Neglect wall effects on the particle. If Stokes’ law is applicable, the absolute value of \(v_t\) (in \(\mathrm{mm\,s^{-1}}\)) is ________________ (rounded off to two decimal places).

Hint:

Stokes’ law holds for very small Reynolds number \(Re=\rho_f v_t d/\mu \ll 1\).
Use absolute density difference \(|\rho_p-\rho_f|\); sign only indicates direction (here, upward since \(\rho_p<\rho_f\)).

Answer 1

Correct Answer: 1

Step 1: Apply Stokes' terminal velocity for a sphere
For creeping flow, \[ v_t=\frac{(\rho_p-\rho_f) g d^2}{18\mu}. \] Given \(\rho_p=0.8\ \mathrm{g\,cm^{-3}}=800\ \mathrm{kg\,m^{-3}}\), \(\rho_f=1000\ \mathrm{kg\,m^{-3}}\), \(g=10\ \mathrm{m\,s^{-2}}\), \(d=100\ \mu\mathrm{m}=1\times 10^{-4}\ \mathrm{m}\), \(\mu=10^{-3}\ \mathrm{Pa\,s}\). Step 2: Compute magnitude
\[ |v_t|=\frac{|800-1000| \times 10 \times (1\times 10^{-4})^2}{18\times 10^{-3}} =\frac{200\times 10\times 10^{-8}}{18\times 10^{-3}} =\frac{2\times 10^{-5}}{1.8\times 10^{-2}} =1.111\dots \times 10^{-3}\ \mathrm{m\,s^{-1}}. \] Step 3: Convert to \(\mathrm{mm\,s^{-1}}\) and round
\[ |v_t|=1.111\dots\ \mathrm{mm\,s^{-1}} \approx \boxed{1.11\ \mathrm{mm\,s^{-1}}}. \]