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Q.Derive an expression for terminal velocity of a spherical object falling under gravity through a viscous medium.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022Subjective· 3mImportance★★★★★
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At terminal velocity the net force is zero: weight = viscous drag (Stokes' law) + upthrust.

At terminal velocity vtv_t, the sphere (radius rr, density ρ\rho) falling through a viscous fluid (density σ\sigma, coefficient of viscosity η\eta) moves with zero acceleration, so:

Weight=Viscous force (Stokes’)+Buoyant force\text{Weight} = \text{Viscous force (Stokes')} + \text{Buoyant force}

43πr3ρg=6πηrvt+43πr3σg\frac{4}{3}\pi r^3 \rho g = 6\pi\eta r v_t + \frac{4}{3}\pi r^3\sigma g

Solving for vtv_t: …

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