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# Specific Gravity Formula

Specific gravity is a concept that we all have seen but do not know its name. Also, the density of the object determines this factor. Moreover, in this topic we will discuss specific gravity, specific gravity formula and its derivation and solved examples. ## Specific Gravity

Specific gravity refers to the ratio of the density of an object and the reference material. Furthermore, the specific gravity can tell us if the object will sink or float in reference material. Besides, the reference material is water that always has a density of 1 gram per cubic centimeter or 1 gram per millimeter.

In simple words, specific gravity defines whether an object will sink or float in water. Moreover, there are many factors that determine whether an object will float or sink.

Get the huge list of Physics Formulas here

### Density

The density of the object refers to how heavy or compact the object is in the given volume. Also, we measure it in mass per unit volume. Furthermore, it is written as grams per cubic centimeter ($$g/cm^{3}$$), grams per millimeter (g/mL), or kilograms per liter (kg/L).

In simple words, density refers to the heaviness or lightness of an object in the given volume. Moreover, the density of the object directly relates to the mass of the object means that the object that has more molecules will have high density and the object that is less molecule will have lower density.

## Specific Gravity Formula

The specific gravity formula is defined with, water as its reference substance and the formula is the ratio of the density of an object to the density of the water. Also, the Greek Symbol Rho $$\rho$$ indicate the density.

The formula looks like this

Specific Gravity = $$\frac{density of the object}{density of the water}$$ = $$\frac{\rho_{object}}{\rho_{ H_{2}O}}$$

Most noteworthy, the specific gravity has no unit for measure because the numerator and the denominator of the formula are the same so they cancel each other out.

### Derivation of the Specific Gravity Formula

$$\rho$$ = refers to the Greek symbol that denotes the density
Object = refers to the density of the object
$$H_{2}O$$ = refers to the density of reference material (water)

Moreover, for specific gravity, it is also important to know the density of the object and also how to calculate the density of the object.

### Density formula

Density = $$\frac{mass}{volume}$$ = $$\frac{m}{v}$$

#### Derivation

m = refers to the mass of the object
v = refers to the volume of the object

Besides, the mass of the object can be in grams, kilograms, and pounds. Also, the density directly relates to the mass of the object. So, we can specify the specific gravity by dividing the mass of an object with the mass of the water.

Specific Gravity = $$\frac{mass of the object}{mass of the water}$$ = $$\frac{m_{object}}{m_{H_{2}O}}$$

Moreover, the mass of the object is also directly related to the density. In addition, the mass is measured in Newtons. Furthermore, we can also find specific gravity with the weight of the object and water

Specific Gravity = $$\frac{weight of the object}{weight of the water}$$ = $$\frac{W_{object}}{W_{H_{2}O}}$$

Most noteworthy, in all these formulas all the units are the same and they cancel each other out.

## Solved Example on Specific Gravity Formula

#### Example 1

A liquid has a mass of 36 grams and the volume of the water (reference material) is 3 mL. Find the specific gravity of the object? Also, specify if the object will sink or float in the water? Besides, the density of the water is 1 g/mL.

#### Solution:

First of all, we need to kind the density of the object. And after that, we will find the specific gravity of the object.

Given:

m = 36 g
v = 3 mL
$$\rho$$ = 1 g/mL

Calculation:

Density of the object = $$\frac{m}{v}$$ = $$\frac{36 g}{3 mL}$$ = 12 g/mL

Now, we know the density of both the elements that is the object and water. So, put the values in the specific gravity equation to know the answer.

Specific Gravity = $$\frac{\rho_{object}}{\rho H_{2}O}$$ = $$\frac{12 g/mL}{1 g/mL}$$ = 12

So the density of the object is 12 g/mL and the specific gravity is 12. Hence, the specific gravity is greater than 1 so the object will sink in the water.

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