Question

An oil drop falls through air with a terminal velocity of $5\times {10}^{-4}m/s$. The terminal velocity of a drop with half of the initial radius will be ? Neglect density of air as compared to that of oil.

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

• A. $3.25\times {10}^{-4}m/s$
• B. $2.10\times {10}^{-4}m/s$
• C. $1.5\times {10}^{-4}m/s$
• D. $1.25\times {10}^{-4}m/s$

Answer 1

The correct option is D $1.25\times {10}^{-4}m/s$

$v=\frac{2}{9}{r}^{2}g\frac{d_1-d_2}{\eta }$

where $v=$terminal velocity$=5\times {10}^{-4}m/s$

$r=$radius of oil drop

$g=$acceleration due to gravity$=10m/s^2$

$d_1=$density of oil$=900kg/m^3$

$d_2=$density of water (Neglected)

$v=5\times {10}^{-4}m/s$

Now when r becomes $\frac{r}{2}$

$v_{new}=\frac{v}{4}=\frac{5}{4}\times {10}^{-4}$

$v_{new}=1.25\times {10}^{-4}m/s$

Hence the correct option is (D)