Determine the density of a material:

Archimedes' principle enables us to determine the density of a material and it also helps us know whether or not a given object will float in a liquid. In order to prove that the floatation depends on the relative densities of the object and the liquid, let us consider a (solid) body of volume V and density ρS which is immersed in a liquid having density ρL. Let W_a is the weight of the body in air and WL is the weight of the body when it is fully immersed in the liquid. Hence, the apparent loss of weight of the body = W_a - W_L . Now, according to the Archimedes' Principle, the weight (= V ρ g ) of the liquid displaced by the immersed body is equal to the apparent loss in the weight of the body; that is,

W_a  - W_L  = ρ_L g V

or,  ρs gV - W_L  = ρ_L g V

or, W_L  = (ρs  - ρ_L ) V g

From Eq. you may note that if ρS < ρ_L, W_L will be negative. Since effective weight of the body cannot be negative, the body will rise to the surface on the other hand of going to the bottom and a part of it will come out of the liquid. The body will stop rising when its effective weight becomes zero. This is the condition of floatation.

Further, the Archimedes' principle can also be used for the determination of density of a solid. You must have remembered that when a piece of wood or plastic or cork is put on the surface of water, it's not immersing fully; a part of the body remains outside above the free surface of water. Let V be the volume of the body placed on the surface of water so that its volume outside the water is V - Vsub, where Vsub is the volume of the submerged portion of the body. The loss in weight of the body will be equal to the weight of the water displaced, that is, Vsub ρL g.

Therefore, we can write :

W_a - W_L = V_sub ρ_L g

or,

 W_L      = (ρ_S V- ρ_LV_sub ) g

Since the body is floating, we have ρ_S < ρ_L and V > V_sub , we can find a value of V_sub (which is not fixed) such that WL = 0 i.e., the apparent weight of the body within the liquid (or gas) is zero. Within such as condition; we have from Eq.

V_sub/V  = ρS/ρL

 Eq. can be used to determine the density of a liquid. To do so, we can take a block of wood which has a graduated scale on one side and let it float in the liquid whose density is to be determined. We note down the mark up to which the block dips in the liquid and determine V_sub. Thus, knowing V, V_sub and ρs, we can determine ρ_L using Eq.

Till now, you studied the properties of fluids at rest and some of their applications.