Question

If Stokes' law is considered valid below a Reynolds number of unity, then the largest raindrop that will fall in accordance with this law will have a diameter of approximately
$(assuming{\rho }_w=1000kg/m^3,{\rho }_a=1.3kg/m^3,v_w={10}^{-6}{m}^2/s,v_a=1.5\times {10}^{-5}{m}^2/s$

• A. 0.75 mm
• B. 0.3mm
• C. 0.15
• D. 0.075mm

Answer 1

The correct option is D 0.075mm

The weight of rain drop,

$W={\rho }_w\left(\frac{\pi D^3}{6}\right)g$

The drag force,

$F_D=3\pi {\mu }_{air}DV$

where V is the velocity
$W=F_D$

$\therefore V=\frac{{\rho }_w{D}^{2}g}{18{\mu }_{air}}=\frac{{\rho }_w{D}^{2}g}{18{\rho }_a{v}_a}$

Reynolds number,

$Re=\frac{VD}{V_a}=\frac{{\rho }_w{D}^{2}g}{18{\rho }_a(v_a{)}^2}$

$\therefore \frac{{\rho }_w{D}^{2}g}{18{\rho }_a(v_a{)}^2}=1.0$

$\Rightarrow D={\left[\frac{18{\rho }_a(v_a{)}^2}{{\rho }_{w}g}\right]}^{1/3}$

$={\left[\frac{18\times 13\times (1.5\times {10}^{-5}{)}^2}{1000\times 10}\right]}^{1/3}$

= 0.081 mm