Two spherical water drops of equal radius r are falling through air at a steady terminal velocity v cm/s each. If the two drops merge to form a single larger drop, what is the new terminal velocity?

(A) 4v
(B) 4^(1/3) * v
(C) 2v
(D) 64v

Correct answer: (B) 4^(1/3) * v

Solution

When two drops of radius r merge, the new radius R satisfies (4/3)pi*R³ = 2*(4/3)pi*r³, giving R = 2^(1/3)*r. Since terminal velocity by Stokes' law is proportional to r², the new velocity is v*(R/r)² = v*(2^(1/3))² = v*2^(2/3) = v*4^(1/3).

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