Question

A drop of radius 4 x 10^-5 m and density 2.5 x 10³ kg/m³ falls through air. The viscosity of air is 1.96 x 10^-5 N s/m² . Neglecting buoyancy due to air, the terminal speed of the drop is ____________.

Options

• 2.5 cm/s

• 3.5 cm/s

• 4.5 cm/s

• 5.5 cm/s

Answer 1

A drop of radius 4 x 10^-5 m and density 2.5 x 10³ kg/m³ falls through air. The viscosity of air is 1.96 x 10^-5 N s/m² . Neglecting buoyancy due to air, the terminal speed of the drop is 4.5 cm/s.

Explanation:

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

`therefore "v" = (2"r"^2rho"g")/(9eta) = (2 xx (4 xx 10^-5)^2 xx 2.5 xx 10^3 xx 9.8)/(9 xx 1.96 xx 10^-5)`

`approx 4.5 xx 10^-2  "m/s"`

`therefore  "v" = 4.5  "cm/s"`