In a fluid, at a point, density ρ is defined as: ρ = lim_{ΔV → 0} Δm/ΔV = dm/dV.**(1)** In case of homogenous isotropic substance, it has no directional properties, so is a scalar.

**(2)** It has dimensions [ML^{-3}] and S.I. unit kg/m^{3} while C.G.S. unit g/cc with 1g /cc = 10^{3}kg/m^{3}

**(3)** Density of substance means the ratio of mass of substance to the volume occupied by the substance while density of a body means the ratio of mass of a body to the volume of the body. So for a solid body.

Density of body = Density of substance

While for a hollow body, density of body is lesser than that of substance [As V_{body} > V_{sub}]

**(4)** When immiscible liquids of different densities are poured in a container the liquid of highest

density will be at the bottom while that of lowest density at the top and interfaces will be plane.

**(5)** Sometimes instead of density we use the term relative density or specific gravity which is defined as:

RD = Density of body / Density of water

**(6)** If m₁ mass of liquid of density ρ₁ and m₂ mass of density ρ₂ are mixed, then as

m = m₁ + m₂ and V = (m₁/ ρ₁) + (m₂/ ρ₂) [As V = m/ρ]

ρ = m/V = (m₁ + m₂)/ (m₁/ ρ₁) + (m₂/ ρ₂) = ∑mᵢ/ ∑ (mᵢ/ρᵢ)

If m₁ = m₂ ρ = 2ρ₁ρ₂/ (ρ₁ + ρ₂) = Harmonic mean

**(7)** If V₁ volume of liquid of density ρ₁ and V₂ volume of liquid of density ρ₂ are mixed, then as:

m = ρ₁ V₁ + ρ₂ V₂ and V = V₁ + V₂ [As ρ = m/V]

If V₁ = V₂ = V ρ = (ρ₁ + ρ₂)/ 2 = Arithmetic Mean

**(8)** With rise in temperature due to thermal expansion of a given body, volume will increase while mass will remain unchanged, so density will decrease, i.e.

ρ/ ρₒ = (m/ V)/ (m/ Vₒ) = Vₒ/V = Vₒ/ Vₒ (1 + γΔθ) [As V = Vₒ (1 + γΔθ)]

Or ρ = ρₒ/ (1 + γΔθ) ≈ ρₒ (1 – γΔθ)

**(9)** With increase in pressure due to decrease in volume, density will increase, i.e.,

ρ / ρₒ = (m/ V)/ (m/ Vₒ) = Vₒ/V [As ρ = m/ V]

But as by definition of bulk-modulus

B = – Vₒ Δp/ ΔV i.e., V = Vₒ [1 – (Δp/B)]

So, ρ = ρₒ [1 – (Δp/B)¯¹] ≈ ρₒ [1 + (Δp/B)]

**Problem: **A homogeneous solid cylinder of length (L<H/ 2). Cross-sectional area A/5 is immersed such that it floats with its axis vertical at the liquid-liquid interface with length L/4 in the denser liquid as shown in the fig. The lower density liquid is open to atmosphere having pressure Pₒ. Then density D of solid.(a) 5/4 d

(b) 4/5 d

(c) Ad

(d) d/5

**Solution: (a)**:

Weight of cylinder = up thrust due to both liquids

V x D x g = (A/5 x ¾ L) x d x g + (A/5 x L/4) x 2d x g

Þ (A/5 x L) D x g = (A x L x d x g)/ 4

Þ D/5 = d/4

∴ D = 5/4 d