Terminal Velocity And Stokes' Law
A small steel ball falling vertically through a tall column of viscous oil eventually moves at constant speed. If a second ball of the same material but twice the radius is dropped through the identical oil, how does its terminal velocity compare?
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About This Question
- Subject
- physics
- Chapter
- properties of solids and liquids
- Topic
- terminal velocity and stokes' law
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
Four times that of the first ball
Terminal velocity is reached when the downward weight of the sphere is balanced by the upward buoyant force and the viscous drag described by Stokes' law, F=6πηrv. Setting net force to zero gives 34πr3(ρ−σ)g=6πηrvt, which rearranges to vt=9η2r2(ρ−σ)g. For balls of the same material in the same oil, every factor except the radius is fixed, so vt∝r2. Doubling the radius therefore multiplies the terminal velocity by 22=4. The option 'same' ignores the radius dependence entirely. The option 'twice' assumes a linear rather than quadratic relation. The option 'half' even gives the wrong direction, since a larger ball falls faster. This quadratic dependence on radius is the central result of the NCERT treatment of viscous fall, and it holds only while the motion stays laminar so that Stokes' linear drag law applies. The weight of the sphere grows as r3 whereas the viscous drag grows only as r, so their balance forces the steady speed to climb as r2. As a check, larger raindrops and hailstones are known to reach higher steady speeds than tiny droplets, which agrees with the strong r2 growth obtained here.
This medium difficulty physics question is from the chapter properties of solids and liquids, covering the topic of terminal velocity and stokes' law. It appeared in the 2025 exam.
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