91Ó°ÊÓ

Drag force is given as,$F_{\text{drag}}=bv$.

Drag constant,$b=6\mathrm{π}\mathrm{η}R.$

Newton's second law $F=mg.$

The force on the spherical particle due to gravity is given by Newton's second law$F=mg$.....(1)

Here,

$m$is the mass of the spherical particle.

$g$is the acceleration due to gravity.

The drag force on the particle is given as

$F_{\text{drag}}=b{v}_{\text{terminal}}……\left(2\right)$

$=6\mathrm{π}\mathrm{η}R{v}_{\text{terminal}}$.

In the condition of equilibrium, the drag force on the particle must balance the force of gravity

$\begin{array}{c}mg=6\mathrm{π}\mathrm{η}R{v}_{\text{tcrminal}}\\ v_{\text{tcrminial}}=\frac{mg}{6\text{ise}R}\end{array}$.

Diameter of the particle, $D=50\mathrm{μ}\mathrm{m}.$

Height,$h=300\mathrm{m}.$

Density of the dust, $\mathrm{ÒÏ}=2700\mathrm{kg}/{\mathrm{m}}^3.$

Viscosity,$\mathrm{η}=2×{10}^{−5}\mathrm{Ns}/{\mathrm{m}}^2.$

The terminal speed of the spherical particle is given by:

$v_{\text{lerminal}}=\frac{mg}{6mpR}$.

Here,

$m$is the mass.

$g$is the acceleration due to gravity.

$\mathrm{η}$is the viscosity.

$R$is the radius of the spherical particle.

The mass of the particle is calculated as

$\begin{array}{c}m=\mathrm{ÒÏ}×\frac{4}{3}×\mathrm{Ï€}×{\left(\frac{D}{2}\right)}^3\\ =2700×\frac{4}{3}×\mathrm{Ï€}×{\left(\frac{s1×{10}^{−6}}{2}\right)}^3\\ =1.76×{10}^{−10}\mathrm{kg}.\end{array}$

The terminal velocity of the particle is calculated as:

$v_{\text{lerminal}}=\frac{mg}{6mpR}.$

Plugging the values in the above equation:

role="math" localid="1647775897792" $\begin{array}{rl}{v}_{\text{terminal}}& =\frac{1.76×{10}^{−10}×9.81}{6×\mathrm{m}×2×{10}^{−3}×25×{10}^{−6}}\\ & =0.183\mathrm{m}/\mathrm{s}.\end{array}$

The time required by the particle to settle down is calculated as

$\begin{array}{c}t=\frac{h}{v_{\text{lerminal}}}\\ =\frac{300}{0.183}\\ =1640\sec \\ =27.32\min \\ ≈27\min .\end{array}$

Part (a): Expression of the terminal speed of the spherical particle is $V_{terminal}=\frac{mg}{6\mathrm{Ï€³¾¸é}}.$

Part (b): The time required by the particle to settle back to the ground is$27\min .$