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A rigid spherical particle undergoes free settling in a liquid of density $750 \: kg \: m^{-3}$ and viscosity $9.81 \times 10^{-3} \: Pa \: s$. Density of the particle is $3000 \: kg \: m^{-3}$ and the particle diameter is $2 \times 10^{-4} \: m$. Acceleration due to gravity is $9.81 \: m \: s^{-2}$. Assuming Stokesâ€™ law to be valid, the terminal settling velocity (in $m \: s^{-1}$) of the particle is

- $2 \times 10^{-3}$
- $3 \times 10^{-3}$
- $4 \times 10^{-3}$
- $5 \times 10^{-3}$