Given:
Radius of the drops r = 0.02 cm = 2 × 10-4 m
Viscosity of air η = 1.8 × 10-4 poise = 1.8 × 10-5 decapoise
Acceleration due to gravity g = 9.9 m/s2
Density of water ρ = 1000 kg/m3
Let v be the terminal velocity of a drop.
The forces acting on the drops are
(i) The weight `` mg`` acting downwards
(ii) The force of buoyance, i.e., `` \left(\frac{4}{3}\right)\pi {r}^{3}\rho g`` acting upwards
(iii) The force of viscosity, i.e., 6πηrv acting upwards
Because the density of air is very small, the force of buoyance can be neglected.
From the free body diagram:
`` 6\pi \eta rv=mg``
`` 6\pi \eta rv=\frac{4}{3}\pi {r}^{3}\rho g``
`` v=2{r}^{2}\frac{\,\mathrm{\,\rho \,}g}{9\eta }``
`` =2\times {\left(0.02\times {10}^{-2}\right)}^{2}\times 1000\times \frac{\left(9.9\right)}{9}\times \left(1.8\times {10}^{-5}\right)``
`` =5\,\mathrm{\,m\,}/\,\mathrm{\,s\,}.``
Hence, the required vertical speed of the falling raindrops is 5 m/s.
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