The terminal velocity of a spherical ball of radius r falling through a viscous liquid is proportional to

Concept:

Terminal Velocity:

• it is the maximum constant velocity acquired by the body while falling freely in a viscous medium.
• When a body falls freely through a viscous medium, three forces act on it.

1. Weight of the body acting vertically downwards.
2. Upward thrust due to buoyancy equal to the weight of viscous medium displaced.
3. Viscous drag acting in the direction opposite to the motion of the body.

• Mathematically, the terminal velocity for a spherical body is given by the formula:

\({\rm{v}} = \frac{{\left( {\frac{2}{9}} \right){{\rm{r}}^2}{\rm{g}}\left( {{\rm{ρ }} - {\rm{σ }}} \right)}}{{\rm{η }}}\)

Where, v = Terminal velocity, r = Radius of the spherical body, g = Acceleration due to gravity (constant for big and smaller spheres), ρ = Density of the body (constant for big and smaller spheres), σ = Density of the medium through which the body is falling (constant for big and smaller spheres), η = Coefficient of the viscosity of the medium through which the body is falling (constant for big and smaller spheres).

Explanation:

• From the above discussion, it's clear that the terminal velocity is directly proportional to the square of the radius of the spherical body.

i.e v α r²

So, the correct option is r².