# Specific gravity & Relative Density – are they the same?

A liter of water has a mass of 1.00 kg, but the same volume of methylated spirits has a mass of only 0.79 kg. We can say that the density of methylated spirits is 0.79 that of water. This is called its relative density (RD) or its specific gravity (SG). *The two terms ( relative density and specific gravity ) are interchangeable and although the relative density is more correct, specific gravity is widely used in industry and other sciences.* **Thus, we can say Specific gravity & Relative Density are the same.**

## Specific gravity (SG) definition

Specific gravity (SG) is defined as the ratio of the mass of an object in the air compared with the mass of an equal volume of water.**Specific gravity = mass of object in air/mass of the equal volume of water**

But it can also be written:**specific gravity = weight of an object in air/weight of the equal volume of water**

## formula of specific gravity(SG)

**According to Archimedes’ principle**, the weight of an equal volume of water equals the upthrust (weight loss) when it is submerged.

Thus the formula of specific gravity can be written as follows:

specific gravity = weight in air/weight loss when submerged in water

or specific gravity = weight in air / (weight in air – apparent weight in water)

**SG = W _{A} / ( W_{A} – W_{W} )** ………….. (1) formula of specific gravity

## Numerical problem of specific gravity(SG) or Relative Density(RD)

**Q 1 ) A house brick has a volume of 1900 cm ^{3} and a weight in air of 80 N. What is its apparent weight in water? The density of water is 1.00 g cm^{–3}. What is the specific gravity(SG) of the brick?**

**Solution**• Volume of water displaced = volume of brick = 1900 cm

^{3}.

• Mass of water displaced = density × volume = 1.00 × 1900 = 1900 g = 1.9 kg.

• Weight of water displaced = mg = 1.9 × 10 = 19 N.

• Weight in water (apparent weight) = weight in air – upthrust = 80 N – 19 N = 61 N

In the example above, the SG equation gives a value of 80/19 = 4.2 g cm^{–3}

**Q 2) A crown supposedly made of gold weighs 8.00 N in air. If the SG of gold is 19.3, what should be the crown’s apparent weight in water?**

**Solution:**

SG = weight in air/weight loss when submerged in water

weight loss = weight in air / specific gravity = 8/19.3 = 0.415

The crown’s apparent weight should be 8.00 N – 0.415 N = 7.59 N