Step 1: Understanding the Concept:
A dust particle thrown up by a blast falls back down through calm air. Because it is so small and light, it very quickly reaches a steady falling speed and keeps that speed the rest of the way down. That steady speed is the Stokes' law terminal velocity.
Step 2: Key Formula or Approach:
Stokes' law for a sphere settling in a viscous fluid, with the fluid (air) density ignored as instructed, gives $v_t = \dfrac{\rho_p g d^2}{18\mu}$. Once $v_t$ is known, the settling time is simply the drop height divided by this constant speed, $t = h / v_t$.
Step 3: Detailed Explanation:
Work in convenient sub units first. The diameter is $d = 2.5\ \mu m$, which is $2.5 \times 10^{-3}\ mm = 2.5 \times 10^{-6}\ m$, so $d^2 = 6.25 \times 10^{-12}\ m^2$.
The numerator of the terminal velocity formula is $\rho_p g d^2 = 3600 \times 9.81 \times 6.25 \times 10^{-12} = 2.207 \times 10^{-7}$ (SI units).
The denominator is $18\mu = 18 \times 1.80 \times 10^{-5} = 3.24 \times 10^{-4}$.
Dividing gives $v_t = 2.207 \times 10^{-7} / 3.24 \times 10^{-4} = 6.81 \times 10^{-4}\ m/s$, which is only about 0.68 mm every second, showing how slowly a fine dust particle falls.
The fall height is 500 m, so the time taken is $t = 500 / (6.81 \times 10^{-4}) = 733945\ s$.
There are $24 \times 3600 = 86400$ seconds in a day, so $t = 733945 / 86400 = 8.49$ days.
Step 4: Final Answer:
The dust particle needs close to 8.49 days to settle out of the atmosphere and reach the ground.