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 = `\v_t` = 3 cm h⁻¹ = 3 ｘ `\frac {10⁻²}{3600}`m 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 = `\frac {2⍴gr^2}{9η}`

or

r = `\sqrt frac {9ηv_t}{2⍴g}`


By putting the corresponding values

r = `\sqrt frac {9 ｘ8 ｘ 10⁻⁴ N m⁻² s ｘ 8.33 m s⁻¹}{2 ｘ 1246 kg m⁻³ｘ 9.8 m s⁻²}`

r = `\sqrt frac {599.76 ｘ 10⁻⁴ N m⁻¹}{24,421.6 kg m⁻² s⁻²}`

r = `\sqrt frac {599.76 ｘ 10⁻⁴ N m⁻¹}{24,446.52 N m⁻³}`

r = `\sqrt {0.0245335532419338 ｘ 10⁻⁴ m²}`

r = `\sqrt {2.453 ｘ 10⁻⁶ m²}`

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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