Question

What will be the approximate terminal velocity of a rain drop of diameter $1.8\times {10}^{-3}$m, when density of rain water $\approx .{10}^{3}kg{m}^{-3}$ and the co=efficient of viscosity of air $\approx .1.8\times {10}^{-5}Ns{m}^{-2}$? (Neglect buoyancy of air).

• A. $49m{s}^{-1}$
• B. $98m{s}^{-1}$
• C. $392m{s}^{-1}$
• D. $980m{s}^{-1}$

Answer 1

Answer: (B)

$98m{s}^{-1}$

$6\pi \eta rv=\frac{4}{3}\pi r^{3}pg$
$v=\frac{2}{9\eta }{r}^{2}pg$
$v=\frac{2\times 0.9\times 0.9\times {10}^{-6}\times {10}^3\times 9.8}{9\times 1.8\times {10}^{-5}}$
$=9.8\times {10}^{-6}\times {10}^7=98m{s}^{-1}$.