Question

An oil drop falls through air with a terminal velocity of $5\times {10}^{-4}\text{m/s}$. The radius of the drop will be:

Neglect density of air compared to that of oil. (Viscosity of air $=\frac{18\times {10}^{-5}}{5}{\text{N.s/m}}^2,g=10{\text{m/s}}^2$, density of oil $=900{\text{kg/m}}^3$

• A. $2.5\times {10}^{-6}\text{m}$
• B. $2\times {10}^{-6}\text{m}$
• C. $3\times {10}^{-6}\text{m}$
• D. $4\times {10}^{-6}\text{m}$

Answer 1

The correct option is C $3\times {10}^{-6}\text{m}$

Viscosity of air,
$\eta =\frac{18\times {10}^{-5}}{5}{\text{N.s m}}^{-2}$
Terminal velocity is, $v_T=5\times {10}^{-4}{\text{ms}}^{-1}$
We know, $v_T=\frac{2r^2(\sigma -\rho )g}{9\eta }$
Here, $\sigma =900{\text{kg/m}}^3$, is the density of an oil droplet moving through the viscous medium of air with density $\rho \approx 0$
$\Rightarrow 5\times {10}^{-4}=\frac{2r^2\left(900\right)\left(10\right)}{9\left(\frac{18\times {10}^{-5}}{5}\right)}$
$\Rightarrow r^2=9\times {10}^{-12}{\text{m}}^2$
$\Rightarrow r=3\times {10}^{-6}\text{m}$
This gives the value of the radius of the oil droplet.