Question

Find the terminal velocity of a free falling water drop of radius $0.04\text{mm}$:-
The coefficient of viscosity of air is $1.9\times {10}^{-5}{\text{Ns/m}}^2$ and its density is $1.2{\text{kg/m}}^3$. Density of water is $1000{\text{kg/m}}^3$. Take $g=10{\text{m/s}}^2$.

• A. $15\text{cm/s}$
• B. $19\text{cm/s}$
• C. $2.5\text{cm/s}$
• D. $13\text{cm/s}$

Answer 1

The correct option is B $19\text{cm/s}$

The forces on the drop are
(i) Weight acting downwards
$W=\frac{4}{3}\pi r^3\rho g$
$\left[\begin{array}{c}r=\text{radius of drop}\\ \rho =\text{density of drop}\end{array}\right]$


(ii) Buoyant force acting upwards on water droplet
$F_B=\frac{4}{3}\pi r^3{\rho }_{a}g$
[${\rho }_a=\text{density of air}$]
(iii) Drag force on the drop in upward direction
$F_v=6\pi \eta rv$
$\left[\begin{array}{c}\eta =\text{viscosity of air}\\ v=\text{terminal velocity}\end{array}\right]$

$\because {\rho }_a<<\rho$, $W>>F_B$
So, force of buoyancy may be neglected.
Hence, force equation for equilibrium becomes
$Fv=W$
$\Rightarrow 6\pi \eta rv=\frac{4}{3}\pi r^3\rho g$
$\Rightarrow v=\frac{2}{9}\frac{r^2\rho g}{\eta }$
$=\frac{2\times (0.04\times {10}^{-3}{)}^2\times 1000\times 10}{9\times 1.9\times {10}^{-5}}=0.187\text{m/s}$$=18.7\text{cm/s}\approx 19\text{cm/s}$