Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of 6 cm s–1. If they coalesce to form one big drop, what will be its terminal speed? Neglect the buoyancy due to air

Engineering

Physics

Physical Quantities

Acceleration

Viscosity and Terminal Velocity New

Question

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of 6 cm s^–1. If they coalesce to form one big drop, what will be its terminal speed? Neglect the buoyancy due to air

24 cms^–1

1.5 cms^–1

6 cms^–1

32 cms^–1

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Terminal velocity v_t for a sphere is given by $v_{t} = \frac{2 r^{2} (ρ - ρ_{0}) g}{9 η}$ . Since buoyancy is neglected (ρ₀=0), v_t ∝ r².

Mass is conserved. The mass of the big drop is 8 times the mass of a small drop. Since mass ∝ volume ∝ r³, the radius of the big drop R = (8)^1/3r = 2r.

Therefore, the terminal velocity of the big drop v_{t_big} ∝ R² = (2r)² = 4r². This is 4 times the original v_t ∝ r².

Original terminal speed was 6 cm/s, so the new terminal speed is 4 × 6 = 24 cm/s.

Final Answer: 24 cms^–1