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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 buoyance may be neglected, then the terminal speed attained by the raindrop is : [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 $= 1.8 \times 10^{-5} Nsm^{-2}$ ]
Option: 1 $250.6\: ms^{-1}$
Option: 2 $43.56\: ms^{-1}$
Option: 3 $4.94\: ms^{-1}$
Option: 4 $14.4\: ms^{-1}$

Answers (1)

best_answer

$F \rightarrow Frictional\: \: drag \: \: face \: \: (stokes\: \: law)$

$B \rightarrow Buo yancy\: \: force\: \: B \simeq 0 ( neglected)$

$At\: terminal \: \: velocity,$

$\begin{aligned} &F=m g\\ \end{aligned}$

$\begin{aligned} &6 \pi \eta R{V_T}=m g=\left(\rho _{w}\right) \times \frac{4}{3} \pi R^{3} g\\ \end{aligned}$

$V_{T}=\frac{\rho _{w\times 4\pi R^{3}}g}{3\times 6\pi \eta R}$

$\begin{aligned} &V=\frac{45 \rho w}{18 \eta } R^{2} g\\ \end{aligned}$

$\begin{aligned} &V_{T}=\frac{2}{9} \times \frac{10^{3} \times\left(2 \times 10^{-4}\right)^{2} \times 10}{1.8 \times 10^{-5}}\\ \end{aligned}$

$\begin{aligned} &h=\frac{8 \times 10^{-4} \times 10^{5}}{9 \times 1.8}=\frac{40}{8.1}\\ \end{aligned}$

$\begin{aligned} &V_{T}=4.94 \mathrm{~m} / \mathrm{s} \end{aligned}$

The correct option is (3)

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vishal kumar

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