Question

A raindrop with radius $R=0.2\text{mm}$ falls from a cloud at a height $h=2000\text{m}$ 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 ${\rho }_w=1000{\text{kg m}}^{-3}$ and density of air ${\rho }_a=1.2{\text{kg m}}^{-3},g=10{\text{m/s}}^2$
Coefficient of viscosity of air$=1.8\times {10}^{-5}{\text{Nsm}}^{-2}$

• A. $250.6{\text{ms}}^{-1}$
• B. $43.56{\text{ms}}^{-1}$
• C. $4.94{\text{ms}}^{-1}$
• D. $14.4{\text{ms}}^{-1}$

Answer 1

The correct option is C $4.94{\text{ms}}^{-1}$

When the drop will attain terminal speed $\left(v\right)$, the velocity will be constant, thus $a=0$
$\Rightarrow F_{net}=0$

From the FBD, we have
$mg=F_v=6\pi \eta Rv$
$\Rightarrow v=\frac{mg}{6\pi \eta Rv}$
$\Rightarrow v=\frac{{\rho }_w\frac{4\pi }{3}{R}^{3}g}{6\pi \eta R}$
$=\frac{2{\rho }_w{R}^{2}g}{9\eta }$
$=\frac{2\times 1000\times (0.2\times {10}^{-3}{)}^2\times 10}{9\times 1.8\times {10}^{-5}}$
$=\frac{400}{81}\text{m/s}$
$=4.94\text{m/s}$