viscosity (stokes' law)

Question

Sphere radius 1 mm falls steadily in oil (η=0.6 Pa·s), density difference Δρ=900 kg/m³, g=10. Terminal velocity?

RankGuru · Accepted Answer

1. **Stokes' Law Reference**: For a sphere falling through a viscous fluid, terminal velocity ($v_t$) is reached when viscous drag + buoyancy = weight.
2. **Formula**: $v_t = \frac{2 r^2 (ho_s - ho_l) g}{9 \eta} = \frac{2 r^2 \Delta ho g}{9 \eta}$.
3. **Variables**:
   - $r = 1$ mm $= 10^{-3}$ m
   - $\Delta ho = 900 	ext{ kg/m}^3$
   - $\eta = 0.6 	ext{ Pa·s}$
   - $g = 10 	ext{ m/s}^2$
4. **Calculation**:
   - $v_t = \frac{2 (10^{-3})^2 	imes 900 	imes 10}{9 	imes 0.6}$
   - $v_t = \frac{2 	imes 10^{-6} 	imes 9000}{5.4} = \frac{0.018}{5.4} \approx 0.00333$ m/s.
5. **Unit conversion**: $0.00333$ m/s $= 0.333$ cm/s.
6. **Conclusion**: Closest option is **$0.34$ cm/s**.

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viscosity (stokes' law)

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