Today we will look at the interesting topics in physics that are bulk modulus. To begin with, the bulk modulus is defined as the proportion of volumetric stress related to the volumetric strain of specified material, while the material deformation is within the elastic limit. To put in more simple words, the bulk modulus is nothing but a numerical constant that is used to measure and describe the elastic properties of a solid or fluid when pressure is applied on all the surfaces.

The bulk modulus of elasticity is one of the measures of the mechanical properties of solids. Other elastic modules include Young’s modulus and Shear modulus. In any case, the bulk elastic properties of a material are used to determine how much it will compress under a given amount of external pressure. Here it is important to find and note the ratio of the change in pressure to the fractional volume compression.

The value is denoted with a symbol of K and it has the dimension of force per unit area. It is expressed in the units per square inch (psi) in the English system and newtons per square meter (N/m^{2}) in the metric system.

Following is the table explaining other **related concepts**:

## Bulk Modulus Of Elasticity Formula

It is given by the ratio of pressure applied to the corresponding relative decrease in the volume of the material.

Mathematically, it is represented as follows:

B = ΔP /(ΔV/V)

Where:

**B**: Bulk modulus

**ΔP**: change of the pressure or force applied per unit area on the material

**ΔV**: change of the volume of the material due to the compression

**V**: Initial volume of the material in the units of in the English system and N/m^{2} in the metric system.

### Solved Example

**What is the bulk modulus of a body that experienced a change of pressure of 5*10 ^{4}N/m^{2} and its volume goes from 4 cm^{3} to 3.9 cm^{3}?.**

The bulk modulus is calculated using the formula,

B = ΔP /(ΔV/V)

B = (5*10^{4} N/m^{2})/((4 cm^{3} – 3.9 cm^{3})/4 cm^{3}) = 0.125 *10^{4} N/m^{2}

B = 1.25 *10^{4} N/m^{2}

### Uses

Bulk modulus is used to measure how incompressible a solid is. Besides, the more the value of K for a material, the higher is its nature to be incompressible. For example, the **value of K for steel is 1.6×10 ^{11 }N/m^{2}** and the value of K for glass is 4×10

^{10}N/m

^{2}. Here, K for steel is more than three times the value of K for glass. This implies that glass is more compressible than steel.

Now, a quick exercise for you. Try finding the value for diamond and comparing it with the value of steel and glass.

Consider this situation. You go deep-sea diving into the Mariana Trenches. This is the following information you have in hand.

Bulk Modulus of Bone = 1.5×10^{10 }N/m^{2}

Atmospheric Pressure = 1.01×10^{5} N/m^{2}

The pressure at deep point = 1.09×10^{8} N/m^{2}

You have two tasks;

- Find out what the value of dVV will be for your bones
- Appreciate atmospheric pressure because you don’t die here

While in solids, Young’s modulus is commonly used, the value of K varies in gases, as they are extremely compressible. The concept of Bulk Modulus is also used in liquids. Temperatures of fluid and entrained air content are the two factors highly controlled by the bulk modulus.

Explore more about physics formulas and calculators with our expert faculty, here at BYJU’S.

what is the difference between bulk modulus and density of liquid?

Bulk Modulus is a constant that describes how resistant a substance is to compression. It is defined as the ratio between pressure increase and the resulting decrease in a material’s volume. The density of a substance describes the degree of compactness of the substance. It is defined as mass per unit volume.

THANK YOU VERY MUCH

Which molecule possess high bulk modulus?

Osmium has extremely high bulk modulus scaling between 395 and 462 N/m

^{2}.