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A raindrop with radius $R=0.2\, {mm}$ fells from a cloud at a height $h=2000\, {m}$ above the ground. Assume that the drop is spherical throughout its fall and the force of buoyance may be neglected, then the terminal speed attainde by the raindrop is : (In ${ms}^{-1}$)
[Density of water $f_{{w}}=1000\;{kg} {m}^{-3}$ and density of air $f_{{a}}=1.2\; {kg} {m}^{-3}, {g}=10 \;{m} / {s}^{2}$ Coefficient of viscosity of air $=18 \times 10^{-5} \;{Nsm}^{-2}$ ]
$14.4$
$2.47$
$43.56$
$4.94$
Solution
At terminal speed
${F}_{\text {net }} =0$
${Mg} ={F}_{{v}}=6 \pi {\eta} {Rv}$
${V} =\frac{{mg}}{6 \pi \eta {Rv}}$
${V} =\frac{\rho_{{w}} \frac{4 \pi}{3} {R}^{3} {g}}{6 \pi \eta {R}}$
$=\frac{2 p_{{w}} {R}^{2} {g}}{9 {\eta}}$
$=\frac{400}{81}\, {m} / {s}$
$=4.94\, {m} / {s}$
