Question

A drop of radius 1.5 x 10^-5 and density 2 x 10³ kg/m³ falls through air. The viscosity of air is 1.8 x 10^-5 N s/m². Neglecting buoyancy due to air, the terminal speed of the drop is ____________.

Options

• 2.8 cm/s

• 3.8 cm/s

• 4.8 cm/s

• 5.4 cm/s

Answer 1

A drop of radius 1.5 x 10^-5 and density 2 x 10³ kg/m³ falls through air. The viscosity of air is 1.8 x 10^-5 N s/m². Neglecting buoyancy due to air, the terminal speed of the drop is 5.4 cm/s.

Explanation:

Neglecting buoyancy due to air,

${\displaystyle \text{v}=\frac{2{\text{r}}^2\rho \text{g}}{9\eta }}$

${\displaystyle =\frac{2\times {(1.5\times {10}^{-5})}^2\times 2\times {10}^3\times 9.8}{9\times 1.8\times {10}^{-5}}}$

${\displaystyle =5.4\times {10}^{-2} \text{m/s}}$

${\displaystyle \text{v}\approx 5.4 \text{cm/s}}$