A ball of radius R falls through oil with terminal velocity 5 cm/sec. Another ball of same material, has radius 3R. Its terminal velocity in the same oil will be [Use Stoke’s law but neglect buoyant force]

Engineering

Physics

Acceleration

Viscosity and Terminal Velocity New

Forces and Types of Forces

Question

A ball of radius R falls through oil with terminal velocity 5 cm/sec. Another ball of same material, has radius 3R. Its terminal velocity in the same oil will be [Use Stoke’s law but neglect buoyant force]

15 cm/sec

4.5 cm/sec

0.45 m/s

4.5 m/sec

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Terminal velocity v_t for a sphere in viscous fluid under Stoke's law (neglecting buoyancy) is given by balancing weight and viscous drag:

$m g = 6 π η R v_{t}$

Mass m ∝ R³ (same material). Thus v_t ∝ m/R ∝ R³/R ∝ R².

For radius 3R, v_t increases by factor (3)² = 9. So new terminal velocity is 9 × 5 cm/sec = 45 cm/sec = 0.45 m/s.

Final Answer: 0.45 m/s