Certain globular protein particle has a density of 1246 Kg m⁻³ it falls through pure water (η = 8 ｘ 10⁻⁴ N m⁻² s) with a terminal speed of 3 cm h⁻¹. Find the radius of the particle. (Ans: 1.6 ｘ 10⁻⁶ m )



Data Given:


Density of particle = ⍴ = 1246 kg m⁻³

Viscosity = η = 8 ｘ 10⁻⁴ N m⁻² s

Terminal speed = vt = 3 cm h⁻¹ = 3 ｘ 10⁻²3600m s⁻¹ = 8.33 m s⁻¹

Value of g = 9.8 m s⁻²



To Find:


Radius of the particle  = r = ?


Solution:


By using the formula for terminal speed

vt = 2⍴gr29η

or

r = `\sqrt frac {9ηv_t}{2⍴g}`


By putting the corresponding values

r = √9ｘ8ｘ10⁻⁴Nm⁻²sｘ8.33ms⁻¹2ｘ1246kgm⁻³ｘ9.8ms⁻²

r = √599.76ｘ10⁻⁴Nm⁻¹24,421.6kgm⁻²s⁻²

r = √599.76ｘ10⁻⁴Nm⁻¹24,446.52Nm⁻³

r = √0.0245335532419338ｘ10⁻⁴m²

r = √2.453ｘ10⁻⁶m²

or

r = 1.566 ｘ 10⁻³ m  -----------Ans.

Hence the radius of the particle is 1.566 ｘ 10⁻³ m.




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