Question

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

• A. $49\, ms^{-1}$
• B. $98\, ms^{-1}$
• C. $392 \,ms^{-1} $
• D. $980 \,ms^{-1} $

Answer 1

Answer: (B)

Terminal velocity, $v=\frac{2}{9} r^{2} \frac{(p-\sigma)}{\eta} g$
Neglecting buoyancy effect of the fluid,
$v=\frac{2}{9} \cdot \frac{\rho}{\eta} r^{2} g$
Putting the given values, we get
$v=\frac{2}{9} \times \frac{10^{3} \times\left(0.9 \times 10^{-3}\right)^{2}}{1.8 \times 10^{-5}} \times 9.8=98 \,ms ^{-1}$