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9-1.Fluid Mechanics
normal
Water drop whose radius is $0.0015\, mm$ is falling through the air. If the coefficient of viscosity of air is $1.8 \times 10^{-5}\, kg/m-s$, then assuming buoyancy force as negligible the terminal velocity of the dorp will be
A
$2.72 \times {10^{ - 4}}\,m/s$
B
$2.72 \times {10^{ - 3}}\,m/s$
C
$2.72 \times {10^{ - 2}}\,m/s$
D
$2.72 \times {10^{ - 1}}\,m/s$
Solution
From Stoke's law the terminal velocity of water drop assuming buoyancy froce negligible, is
$\mathrm{v}=\frac{2}{9} \frac{\mathrm{r}^{2} \rho \mathrm{g}}{\eta}$
$=\frac{2}{9} \times \frac{\left(1.5 \times 10^{-6}\right)^{2}\left(1.0 \times 10^{3}\right) \times 9.8}{1.8 \times 10^{-5}}$
$=2.72 \times 10^{-4} \mathrm{m} / \mathrm{s}$
Standard 11
Physics
