Question

According to the Stokes’ law, the rate of settling of the particles depends on the terminal settling velocity v_t which is

Where

d_p = Particle diameter

ρ_p = Density of particle

ρ_a = Density of air

μ_a = Velocity of air

P = Air pressure

C = Constant 

1. \(\frac{{gd_p^2}}{{18{\mu _a}}}\left( {{\rho _p} - {\rho _a}} \right)\left[ {1 + \frac{{2C}}{{{d_p}P}}} \right]\)
2. \(\frac{{gd_p^2}}{{18{\mu _a}}}\left( {{\rho _p} + {\rho _a}} \right)\left[ {1 - \frac{{2C}}{{{d_p}P}}} \right]\)
3. \(\frac{{gd_p^2}}{{18{\mu _a}}}\left( {{\rho _p} - {\rho _a}} \right)\left[ {1 - \frac{{2C}}{{{d_p}P}}} \right]\)
4. \(\frac{{gd_p^2}}{{18{\mu _a}}}\left( {{\rho _p} + {\rho _a}} \right)\left[ {1 + \frac{{2C}}{{{d_p}P}}} \right]\)

Answer 1

Correct Answer - Option 1 : \(\frac{{gd_p^2}}{{18{\mu _a}}}\left( {{\rho _p} - {\rho _a}} \right)\left[ {1 + \frac{{2C}}{{{d_p}P}}} \right]\)

Stokes’s law expresses the settling velocities of small spherical particles in a fluid medium.

Terminal velocity is the constant velocity the particle attains when acceleration becomes 0.

According to the Stokes’ law, the terminal settling velocity vt is given by

\(\frac{{gd_p^2}}{{18{\mu _a}}}\left( {{\rho _p} - {\rho _a}} \right)\left[ {1 + \frac{{2C}}{{{d_p}P}}} \right]\)

Where

dp = Particle diameter

ρp = Density of particle

ρa = Density of air

μa = Velocity of air

P = Air pressure

C = Constant