A raindrop with radius R = 0.2 mm falls 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 buoyancy may be neglected, then the terminal speed attained by the raindrop is:
$[Density of water f_{w} = 1000 {kgm}^{- 3} and Density of air f_{a} = 1.2 {kgm}^{- 3}, g = 10 m / s^{2} . Coefficient of viscosity of air = 1.8 \times 10^{- 5} {Nsm}^{- 2}]$

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43.56 ms^-1

14.4 ms^-1

250.6 ms^-1

4.94 ms^-1

answer is B.

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Detailed Solution

Given,

$Radius r = 0.2 mm, height h = 2000 m$

The velocity will be constant when the plummet reaches its terminal speed (v), hence a=0.

$F_{net} = 0$

$mg = F_{v} = 6 πηRv$

$v = \frac{mg}{6 πηR} = \frac{ρ_{w} (\frac{4}{3} {πR}^{3}) g}{6 πηR} = \frac{2 ρ_{w} R^{2} g}{9 η}$

$v = \frac{2 \times 1000 \times {(0.2 \times 10^{- 3})}^{2} \times 10}{9 \times 1.8 \times 10^{- 5}} = 4.94 m / s$

Hence the correct answer is 4.94m/s.

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