Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce the terminal velocity would be
Correct answer: C. 7.94 cm/s
- A. 10 cm/s
- B. 2.5 cm/s
- C. 7.94 cm/s
- D. 5 x √2 cm/s
Explanation
When two drops of the same radius coalesce, the volume of the new drop is twice that of a single drop. If the original radius is 'r', the new radius 'R' can be found using the volume relation: 4/3πR³ = 2(4/3πr³), giving R = r * 2^(1/3). According to Stokes' Law, terminal velocity 'v' is proportional to the square of the radius (v ∝ R²). Therefore, the new terminal velocity v' = 5 * (2^(2/3)) ≈ 7.94 cm/s. Thus, Option C is correct, while Options A, B, and D incorrectly apply relationships or assumptions about linearity or scaling.
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About Fluid Dynamics
Fluid flow is described through speed, pressure and density, including steady flow and the equation of continuity for conserving mass. Bernoulli's equation relates pressure, speed and height, while fluid drag and terminal velocity explain why a falling object eventually moves at constant speed when drag balances its weight.
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