Question

An oil drop falls through air with a terminal velocity $5\times {10}^{-4}\text{m/s}$. Find the radius of the drop.
Neglect the density of air as compared to that of oil. (Take viscosity of air $=3.6\times {10}^{-5}{\text{N-s/m}}^2,g=10{\text{m/s}}^2$, density of oil ${\rho }_o=900{\text{kg/m}}^3)$

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

Answer 1

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

Given viscosity of air $\left(\mu \right)=3.6\times {10}^{-5}{\text{N-s/m}}^2$
Terminal velocity $\left(V_T\right)=5\times {10}^{-4}\text{m/s}$ Now, terminal velocity is given by
$V_T=\frac{2({\rho }_o-{\rho }_a)r^{2}g}{9\mu }$
where $r=$ radius of drop
$\because {\rho }_a<<<{\rho }_o$
So, $V_T=\frac{2{\rho }_o{r}^{2}g}{9\mu }$
$\Rightarrow 5\times {10}^{-4}=\frac{2\times 900\times r^2\times 10}{9\times 3.6\times {10}^{-5}}$
$\Rightarrow r^2=9\times {10}^{-12}$
$\Rightarrow r=3\times {10}^{-6}\text{m}$