**Density , Relative Density of Substance:**

Matter exists in three states solid, liquid and gas. Any state of matter that can flow is a fluid. Liquids and gases are therefore, referred to as fluids.

### What is Density ?

Density of a substance is the mass per unit volume . Homogenous bodies have uniform density .

$\large Density = \frac{Mass}{Volume}$

The unit of density is kg/m^{3} in S.I. system. Dimension of density is given by [ρ] = = [M] [L^{-3}].

### What is the Relation between Density & Relative Density or Specific Gravity ?

Relative density of a substance is the ratio of the density of the substance to the density of pure water (ρ_{w} = 1000 kg/m^{3}) at 4°C.

$ \displaystyle \rho_r = \frac{\rho}{\rho_w} $

“Relative density is dimensionless ”

Solved Example : A hollow metallic sphere has inner and outer radii respectively as 5 cm and 10 cm. If the mass of the sphere is 2.5 kg, find (a) density of the material (b) relative density of the material of the sphere.

Solution: The volume of the material of the sphere is

$ \displaystyle V = \frac{4}{3}\pi( r_2^3 – r_1^3)$

$ \displaystyle V = \frac{4}{3}\times 3.14( (0.1)^3 – (0.05)^3)$

= 0.00367 m^{3}

(a) Therefore, density of the material of the sphere is

$ \displaystyle \rho = \frac{M}{V} $

$ \displaystyle \rho = \frac{2.5}{0.00367} $

= 681.2 kg/m^{3}

(b) $ \displaystyle \rho_r = \frac{\rho}{\rho_w} $

$ \displaystyle \rho_r = \frac{681.2}{1000} $

= 0.6812

__How to Calculate Density of a mixture ?__

Let a number of substances of masses M1, M2, M3 etc., and densities ρ_{1}, ρ_{2}, ρ_{3}, etc. respectively be mixed together.

The total mass of the mixture is = M_{1} + M_{2} + M_{3} + ….

The total volume $ \displaystyle = \frac{M_1}{\rho_1} + \frac{M_2}{\rho_2} + \frac{M_3}{\rho_3} + …..$ ;

provided that the substances retain their individual states within the mixture.

Therefore, the density of the mixture is

$ \displaystyle \rho_{mix} = \frac{M_1 + M_2 + M_3 + ….}{\frac{M_1}{\rho_1} + \frac{M_2}{\rho_2} + \frac{M_3}{\rho_3} +….} $

#### The same expression may written in terms of the volumes:

The density of the mixture is

$ \displaystyle \rho_{mix} = \frac{\rho_1 V_1 + \rho_2 V_2 + ….}{V_1 + V_2 + ….} $

where V_{1}, V_{2}, V_{3} … represent the volumes of substances of densities ρ_{1}, ρ_{2}, ρ_{3} . . . in the mixture

Solved Example : Two liquids of densities 2.5 gm/cm^{3} and 0.8 gm/cm^{3} are taken in the ratio of their masses as 2:3 respectively. Find the average density of the liquid combination.

Solution: Let the masses be ‘ 2 m ‘ gm and ‘ 3 m ‘ gm respectively.

Therefore, the volume of the first liquid of density 2.5 gm/cm^{3} is V_{1} = 2m/2.5 cm^{3}

That of the second liquid is V_{2} = 3m/0.8 cm^{3}

Total volume $ \displaystyle V = V_1 + V_2 $

$ \displaystyle V = \frac{2m}{2.5} + \frac{3m}{0.8} $

Total mass = 2m + 3m = 5 m gm

Therefore, the average density

$ \displaystyle \rho_{av} = \frac{5m}{V} $

$ \displaystyle \rho_{av} = \frac{5m}{\frac{2m}{2.5} + \frac{3m}{0.8}} $

$ \displaystyle \rho_{av} = \frac{10}{9.1} $

= 1.09 gm/cm^{3}

Solved Example: Two miscible liquids of densities 1.2 gm/cc and 1.4 gm/cc are mixed with a proportion ratio of their volumes equal to 3:5. What is the density of resulting liquid?

Solution: density of resulting liquid is

$\large \rho = \frac{\rho_1 V_1 + \rho_2 V_2}{V_1 + V_2} $

$\large \rho = \frac{\rho_1 \frac{V_1}{V_2} + \rho_2 }{\frac{V_1}{V_2}+ 1} $

$\large \rho = \frac{1.2 \times \frac{3}{5} + 1.4 }{\frac{3}{5}+ 1} $

$\large \rho = \frac{3.6 + 7}{8} = \frac{10.6}{8}$

ρ ≅ 1.32

### Also Read :

How to calculate Pressure & Variation of Pressure with depth ? What is Pascal’s law & Its Application ? How to Measure Pressure with Barometer & Manometer ? What is Archimede’s Principle & Buoyancy ? |