A tiny droplet of oil of density 'ρ' and radius 'r' falls through air under force of gravity. If viscosity of air is 'η', the terminal velocity acquired by the oil drop is given by:
Correct answer: C. vt = 2gr2ρ / 9η
- A. vt = 4gr2ρ / 9η
- B. vt = 9ηr2ρ / 4g
- C. vt = 2gr2ρ / 9η
- D. vt = 9ηr2ρ / 2g
Explanation
Explanation:Using strokes law, 6πηrvt = mg Where, m = 4/3 πr³ρComparing both equations, we get: r² = 9ηvt /2ρg vt = 2gr²ρ / 9η
This question appeared on the UHS MDCAT 2009 paper, which you can sit online with every answer explained.
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About Terminal Velocity
Terminal velocity is the constant speed reached when gravitational force is balanced by fluid resistance. The topic covers drag, buoyant force, the conditions for zero acceleration, and how viscosity, object size, shape and density affect the limiting speed, with a distinction between free fall before and after terminal velocity.
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