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A spherical ball of radius r and density ρ is dropped from rest and falls freely through a height h before it enters water. If its speed remains unchanged after it enters the water, and the viscosity of water is η, then h is equal to:
- (2/9) r² (ρ/η) g
- (2/81) r² ((ρ-1)/η) g
- (2/81) r⁴ ((ρ-1)/η) g
- (2/9) r⁴ ((ρ-1)/η) g
Correct answer: (2/81) r⁴ ((ρ-1)/η) g
Solution
Terminal velocity v = (2/9)(r^2/eta)(rho-1)g. Entry speed sqrt(2gh) = v, so h = v^2/(2g) = (2/81) r^4 ((rho-1)/eta) g (the 2/81 r^4 form), not the 2/9 r^4 coefficient.
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