Question

An oil drop falls through air with a terminal velocity of $5\times {10}^{-4}m/s$. Viscosity of oil is $1.8\times {10}^{-5}$ $Ns/m^2$ and density of oil is $900kg/m^3$. Neglecting density of air as compared to that of the oil, choose the correct statements.

• A. Radius of the drop is $6.20\times {10}^{-2}m$
• B. Radius of the drop is $2.14\times {10}^{-6}m$
• C. Terminal velocity of the drop at half of this radius is $1.25\times {10}^{-4}m/s$
• D. Terminal velocity of the drop at half of this radius is $2.5\times {10}^{-4}m/s$

Answer 1

Answer: (B), (C)

The correct options are
B Radius of the drop is $2.14\times {10}^{-6}m$
C Terminal velocity of the drop at half of this radius is $1.25\times {10}^{-4}m/s$

Terminal velocity of oil drop $V_T=\frac{2}{9}(\rho -\sigma )\frac{g{r}^2}{\eta }(\text{As}\sigma =0)V_T=\frac{2}{9}\left(\rho \right)\frac{g{r}^2}{\eta }5\times {10}^{-4}=\frac{2}{9}\times 900\times \frac{10\times r^2}{1.8\times {10}^{-5}}\Rightarrow r^2=4.5\times {10}^{-12}\Rightarrow r=2.14\times {10}^{-6}m\text{Now if}{r}_1=\frac{r}{2},\text{Let new terminal velocity be}{V}_T^{\prime }\text{As}{V}_T\propto r^2\Rightarrow \frac{V_T^{\prime }}{V_T}=\frac{r_1^2}{r^2}\Rightarrow \frac{V_T^{\prime }}{V_T}=\frac{1}{4}\Rightarrow V_T^{\prime }=\frac{5\times {10}^{-4}}{4}\Rightarrow V_T^{\prime }=1.25\times {10}^{-4}m{s}^{-1}$