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State Stoke's law derive an expression of terminal velocity. |
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Answer» Let us understand the use of Stokes law to derive the terminal velocity of a body falling through a viscous liquid,under gravity Weight of the body = mg = Vρg W = 4/3 πr3ρg where r is the radius of the body, r is density, g is the gravity due to upward viscous drag Fv = 6phvr (Stokes law). where h is coefficient of viscosity, v is the velocity of body, r is radius of the body. Upthrust or Buoyant force FT = weight of displaced liquid = Volume of body x density of liquid x acceleration due to gravity FT = Vρg = 4/3 πr3σg When the body moves with terminal velocity, that is, V = VT, total upward force = downward force 6πη VT r + 4/3 πr3σg = 4/3 πr3ρg 6πη VT r = = 4/3 πr3 (ρ - σ)g VT = [2r2(ρ - σ)g]/9η |
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