For every material, the bulk modulus is defined as the proportion of volumetric stress to volumetric strain. The bulk modulus, in simpler terms, is a numerical constant that is used to quantify and explain the elastic characteristics of a solid or fluid when pressure is applied. We’ll go over the bulk modulus formula with examples in this article. Let’s have a look at this Bulk modulus property.

Bulk modulus

The volume strain modulus is known as the bulk modulus. Volume strain is defined as the difference in volume divided by the original volume. 

Bulk modulus is the change in the relative volume of an object when a unit compressive or tensile stress is uniformly applied to the object’s surface. 

The bulk modulus is measured in pascals. The Bulk modulus explains how an object reacts when crushed evenly from all sides. And it is denoted by ‘K’ or ‘B’.

Bulk modulus Formula

The formula of Bulk modulus is,

K = ΔP × V / ΔV

Where,

• K = Bulk modulus (Pascal)
• V = Actual volume of object (m³)
• ΔP = Change in pressure (Pascal)
• ΔV = Change in volume (m³)

Note: Dimension of Bulk modulus is, L^-1M¹T^-2

Derivation for bulk modules formula

By Hooke’s Law,

Stress is directly proportional to Strain

Hydraulic stress α Volume strain

Hydraulic stress = B × Volume strain

B is the bulk modulus of elasticity, and the proportionality constant is B.

p = B × ΔV / V

∴ B = ΔP × V / ΔV

Sample Problems 

Question 1: What is the bulk modulus of elasticity of a liquid compressed in a cylinder from 0.0125 m³ volume at 80 N/cm² pressure to 0.0124 m³ volume at 150 N/cm² pressure?

Solution:

Given: ΔP = 150 – 80 = 70 N/cm², ΔV = 0.0124 – 0.0125 = -0.0001 m³, V = 0.0125 m³

Since,

K = ΔP × V / ΔV

∴ K = 70 × 0.0125 / 0.0001

∴ K = 8750 N/cm²

Question 2: If the pressure of the liquid is increased from 70 N/cm² to 130 N/cm², calculate the bulk modulus of elasticity. The amount of liquid in the container shrinks by 0.15 %.

Solution:

Given: ΔV = V × 0.15 /100 = 15V × 10^-4 m³, Volume of liquid = V m3, ΔP = 130 – 70 = 60 N/cm²

Volumetric Strain = 15 × 10^-4 

K = ΔP × V / ΔV

∴ K = 60 / 15 × 10^-4

∴ K = 4 × 10⁴ N/cm²

Question 3: Assume that the change in pressure is 80 N/Cm², that the object’s actual volume is 0.128 m³, and that the bulk modulus is 7390 N/cm². Calculate the volume change.

Solution:

Given: ΔP = 80 N/cm², K = 7390 N/cm², V = 0.128 m³

Since,

K = ΔP × V / ΔV

∴ ΔV = ΔP × V / K

∴ ΔV = 80 × 0.128 / 7390

∴ ΔV = 10.24 / 7390

∴ ΔV = 0.0013 m³

Question 4: A rubber ball’s volume falls by 0.1 percent when transported to a depth of 200 meters in a pool. The volume elasticity in N/m² will be if the density of water is 1 × 10³ kg/m³ and g = 10 m/s².

Solution:

Given: ΔP = ρgh = 10³ × 10 × 200 = 20000000, ΔV = 0.1, V = 100

Since,

K = ΔP × V / ΔV

∴ K = 20000000 × 100 / 0.1

∴ K = 2 × 10⁹ N/m²

Question 5: If the pressure change is 78 N/Cm², the volume change is 0.128 m³, and the bulk modulus is 7390 N/cm². Calculate the object’s actual volume?

Solution:

Given: ΔP = 78 N/cm², K = 6390 N/cm², ΔV = 0.1 m³

Since,

K = ΔP × V / ΔV

∴ V = K × ΔV / ΔP

∴ V = 6390 × 0.1 / 78

∴ V = 8.192 m³

Question 6: When a liquid is compressed in a cylinder from 0.0225 m³ volume at 50 N/cm² pressure to 0.0124 m³ volume at 110 N/cm² pressure, what is the bulk modulus of elasticity?

Solution:

Given: ΔP = 110 – 50 = 60 N/cm², ΔV = 0.0124 – 0.0225 = -0.0101 m³, V = 0.0225 m³

Since,

K = ΔP × V / ΔV

∴ K = 60 × 0.0225 / 0.0101

∴ K = 133.6 N/cm²