A spherical liquid drop of radius R falls through a gas of viscosity η with terminal speed v₁. The drop is then broken into 64 identical droplets, and each smaller droplet falls through the same gas with terminal speed v₂. Neglecting changes in density and using Stokes' law, what is v₁ ÷ v₂?
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Question

A spherical liquid drop of radius R falls through a gas of viscosity η with terminal speed v₁. The drop is then broken into 64 identical droplets, and each smaller droplet falls through the same gas with terminal speed v₂. Neglecting changes in density and using Stokes' law, what is v₁ ÷ v₂?

Options

G1 Physics · Accepted Answer

Conservation of volume gives (4 ÷ 3)πR³ = 64(4 ÷ 3)πr³, where r is the radius of each smaller droplet. Therefore, R³ = 64r³. Since 64 = 4³, R = 4r. Thus, r = R ÷ 4. For Stokes' terminal velocity, v(t) ∝ radius², when the fluid, density difference and gravity are unchanged. Therefore, v₁ ÷ v₂ = R² ÷ r² = [R ÷ (R ÷ 4)]² = 4² = 16. Correct option: D.