Viscosity (stokes' Law)
Hardphysics
Sphere radius 1 mm falls steadily in oil (η=0.6 Pa·s), density difference Δρ=900 kg/m³, g=10. Terminal velocity?
Select the correct option:
Solution
Incorrect! Answer:
0.34 cm/s
- Stokes' Law Reference: For a sphere falling through a viscous fluid, terminal velocity (vt) is reached when viscous drag + buoyancy = weight.
- Formula: vt=9η2r2(ρs−ρl)g=9η2r2Δρg.
- Variables:
- r=1 mm =10−3 m
- Δρ=900 kg/m3
- η=0.6 Pa\cdotps
- g=10 m/s2
- Calculation:
- vt=9×0.62(10−3)2×900×10
- vt=5.42×10−6×9000=5.40.018≈0.00333 m/s.
- Unit conversion: 0.00333 m/s =0.333 cm/s.
- Conclusion: Closest option is 0.34 cm/s.
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About This Question
- Subject
- physics
- Chapter
- properties of solids and liquids
- Topic
- viscosity (stokes' law)
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
0.34 cm/s
- Stokes' Law Reference: For a sphere falling through a viscous fluid, terminal velocity (vt) is reached when viscous drag + buoyancy = weight.
- Formula: vt=9η2r2(ρs−ρl)g=9η2r2Δρg.
- Variables:
- r=1 mm =10−3 m
- Δρ=900 kg/m3
- η=0.6 Pa\cdotps
- g=10 m/s2
- Calculation:
- vt=9×0.62(10−3)2×900×10
- vt=5.42×10−6×9000=5.40.018≈0.00333 m/s.
- Unit conversion: 0.00333 m/s =0.333 cm/s.
- Conclusion: Closest option is 0.34 cm/s.
This hard difficulty physics question is from the chapter properties of solids and liquids, covering the topic of viscosity (stokes' law). It appeared in the 2025 exam.
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