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If a ball of steel (density ρ = 7.8 g cm⁻³) attains a terminal velocity of 10 cm s⁻¹ when falling in water (Coefficient of viscosity of water = 8.5 × 10⁻⁴ Pa s), then, its terminal velocity in glycerine (ρ = 1.2 g cm⁻³, η = 13.2 Pa s) would be, nearly
- 6.25 × 10⁻⁴ cm s⁻¹
- 6.45 × 10⁻⁴ cm s⁻¹
- 1.5 × 10⁻⁵ cm s⁻¹
- 1.6 × 10⁻⁵ cm s⁻¹
Correct answer: 6.25 × 10⁻⁴ cm s⁻¹
Solution
v2/v1 = [(7.8-1.2)/13.2] / [(7.8-1.0)/8.5e-4] = (0.5)/(8000) = 6.25e-5. v2 = 10*6.25e-5 = 6.25e-4 cm/s.
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