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,

`"v" = (2"r"^2rho"g")/(9eta)`

`= (2 xx(1.5 xx 10^-5)^2 xx 2 xx10^3 xx 9.8)/(9 xx 1.8 xx 10^-5)`

`= 5.4 xx 10^-2  "m/s"`

`"v" approx 5.4  "cm/s"`