Terminal Velocity Calculation

Question

A spherical raindrop of radius 0.2 mm falls through air; if its radius were doubled to 0.4 mm under identical conditions, the terminal velocity of the larger drop would become:

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NCERT Class 11, Chapter 10 (Mechanical Properties of Fluids) gives the terminal velocity of a sphere in a viscous medium as $v_t = \dfrac{2r^2(\rho - \sigma)g}{9\eta}$. The key feature is that terminal velocity is proportional to the square of the radius, $v_t \propto r^2$, since density, gravity and viscosity stay fixed. Doubling the radius multiplies the velocity by $2^2 = 4$. Therefore the larger drop falls four times faster at its terminal speed. The option 'two times' wrongly assumes a linear dependence on radius. The option 'half' inverts the relationship entirely. The option 'eight times' incorrectly applies a cubic dependence, which actually governs the drop's mass and volume, not its terminal velocity. The physical reason for the square-law is that the gravitational driving force grows with volume (as $r^3$) while the viscous drag grows only with radius and speed (as $r v$); balancing these gives $v \propto r^2$. As a plausibility check, larger raindrops are observed to fall noticeably faster than fine mist droplets, which lingers almost suspended in air, and this agrees with the strong square-law dependence and confirms the factor of four for a doubled radius.

Terminal Velocity Calculation

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