Consider a metallic sphere of radius ‘a’ and density p to fall under gravity in a liquid of density. The viscous force F acting on the metallic sphere increases as its velocity increases. A stage is reached when the weight W of the sphere becomes equal to the sum of the upward viscous force F and the upward thrust U due to buoyancy (Figure). Now, there is no net force acting on the sphere and it moves down with a constant velocity v called **terminal velocity.**

**W – F – U = O … (1)**

Terminal velocity of a body is defined as the constant velocity acquired by a body while falling through a viscous liquid.

From (1), W = F + U … (2)

According to Stoke’s law, the viscous force F is given by **F = 6πη av**

The buoyant force U = Weight of liquid displaced by the sphere = **4/3 πa ^{3 }σg**

The weight of the sphere, W = **4/3 πa ^{3 }ρg**

Substituting in equation (2),

**4/3 πa ^{3 }ρg = 6πη av + 4/3 πa^{3 }σg**

**So, V = 2/9 [a ^{2} (ρ-σ)g]/η**