A small drop of water falls from rest through a large height h in air, the final velocity is,
A. Proportional to \[\sqrt h \]
B. Proportional to h
C. inversely Proportional to h
D. independent of h

Hint: When a particle is moving in a fluid then the fluid applies opposing force in the direction opposite to the direction of motion of the particle, this is called viscous force. When the drag force is balanced by the force of gravity then the particle starts moving with constant speed.

Complete step by step solution:
When the particle size is small then we assume it be spherical in shape. Let a small water drop of radius r fall from rest through a large height h in air. The density of water is \[\rho \] and the density of air is \[\sigma \]. Taking the drag force of air on a water droplet, it acts opposite to the motion of the water droplet.

When the water droplet is falling then the direction of motion is downward, so the direction of drag force is upward and the magnitude of the drag force is proportional to the square root of velocity of the water droplet.The weight, i.e. force of gravity is acting downward. With fall in height, the speed of the water droplet keeps increasing and so the drag force.

There comes a point when the upward drag force becomes equal in magnitude to the downward force of gravity. After this point the drag force ceases to increase and hence the net force acting on the droplet is zero. So the water droplet starts motion with constant velocity. The velocity is called the terminal velocity.

The terminal velocity is given as,
\[v = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9\eta }}\]
As there is no term regarding the height h, so the final velocity of the droplet is independent of the height.

Therefore, the correct option is D.

Note: While describing the motion of the small water droplet, we assume the motion of the water droplet is vertical and there is no horizontal force acting on it due to air drag.