Spread the love

Relative Density | Specific Gravity

Relative Density

Relative density(R.D.) is defined as the ratio of the density of a given substance to the density of pure water at 4°C.

R.D. = density of a given substance/density of pure water at 4°C

R.D. is a unitless and dimensionless positive scalar physical quantity.

Being a dimensionless/unitless quantity R.D. of a substance is same in SI and CGS system.

Specific Gravity

It is defined as the ratio of the specific weight of a given substance to the specific weight of pure water at 4°C.

Specific gravity = specific weight of a given substance/specific weight of pure water at 4°C

(The specific weight, also known as the unit weight, is the weight per unit volume of a material)

⇒ Specific gravity = ρ_s×g/ρ_w×g

or Specific gravity = ρ_s/ρ_w

But ρ_s/ρ_w is R.D. of a substance

Hence

Specific gravity = R.D. of a substance

Thus specific gravity of a substance is numerically equal to the relative density of that substance and for calculation purposes they are used interchangeably.

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

$\displaystyle V=\left( \frac{4}{3} \right)\pi \left( r_{2}^{3}-r_{1}^{3} \right)=\frac{4}{3}\times 3.14\times \left[ {{\left( \frac{10}{100} \right)}^{3}}-{{\left( \frac{5}{100} \right)}^{3}} \right]=\frac{4}{3}\times 3.14\times \left[ 0.001-0.000125 \right]$

$\displaystyle \Rightarrow V=\frac{4}{3}\times 3.14\times 0.000875\text{ }{{\text{m}}^{\text{3}}}=0.00367\text{ }{{\text{m}}^{\text{3}}}$

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

$\displaystyle \rho =\frac{M}{V}=\frac{2.5}{0.00367}=681.2\text{ kg/}{{\text{m}}^{\text{3}}}$

(b) Relative density of the material of the sphere

$\displaystyle {{\rho }_{r}}=\frac{681.2}{1000}=0.6812$

Example
A body weighs 160 g in air, 130 g in water and 136 g in oil. What is the specific gravity of oil?
Solution.

Weight in air, W_a = 160g

Weight in water, W_w = 130g = W_a – B_w(Buoyant force by water)

Now, B_w = Volume of the body(V) × density of water(ρ_w) × g

⇒ 130g = 160g – Vρ_w g

⇒ Vρ_w g = 160g – 130g

Vρ_w g = 30g …..(1)

Again for oil,

Weight in oil, W_o = 136g = W_a – B_o(Buoyant force by oil)

Now, B_o = Volume of the body(V) × density of oil(ρ_o) × g

⇒ 136g = 160g – Vρ_o g

⇒ Vρ_o g = 160g – 136g

Vρ_o g = 24g …..(2)

Dividing equation (2) by equation (1), we get

$\displaystyle \frac{V{{\rho }_{o}}g}{V{{\rho }_{w}}g}=\frac{24g}{30g}$

$\displaystyle \Rightarrow \text{ Specific gravity of oil = }\frac{{{\rho }_{o}}}{{{\rho }_{w}}}=\frac{24g}{30g}=0.8$

Relative Density | Specific Gravity

Relative Density | Specific Gravity

Relative Density

Like this:

Leave a Reply Cancel reply