# Problems on Volumetric Stress and Volumetric Strain, Bulk Modulus

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#### Example – 1:

A solid rubber ball has its volume reduced by 14.5% when subjected to a uniform stress of 1.45 × 104 N/m². Find the bulk modulus for rubber.

• Solution:
• Given: Volumetric strain = 14.5 % = 14.5 × 10-2 , Volumetric stress =  1.45 × 104 N/m²,
• To Find: Bulk modulus of elasticity = ?

Bulk modulus of elasticity = K = Volumetric stress / Volumetric strain

∴  K = (1.45 × 104) / (14.5 × 10-2) = 105 N/m²

Ans: Bulk modulus of elasticity of rubber is 105 N/m²

#### Example -2:

• What pressure should be applied to a lead block to reduce its volume by 10% Bulk modulus for lead = 6 × 109 N/m².
• Solution:
• Given: Volumetric strain = 10 % = 10 × 10-2 , Bulk modulus of elasticity =  6 × 109 N/m².
• To Find: Pressure intensity =?

Bulk modulus of elasticity = K = Volumetric stress / Volumetric strain

∴  Volumetric stress = K ×Volumetric strain

∴  Pressure intensity = K ×Volumetric strain

∴  Pressure intensity = 6 × 109 ×10 × 10-2

∴  Pressure intensity = 6 × 108 N/m²

Ans: Pressure intensity is 6 × 108 N/m²

#### Example – 3:

• A volume of 5 litres of water is compressed by a pressure of 20 atmospheres. If the bulk modulus of water is 20 × 108 N/m². , find the change produced in the volume of water. Density of Mercury = 13,600 kg/m³; g = 9.8 m/s². Normal atmospheric pressure  = 75 cm of mercury.
• Solution:
• Given: Original Volume = 5 L = 5 × 10-3 m³, Pressure = dP = 20 atm = 20 × 75 × 10-2 × 13600 × 9.8 N/m², Bulk modulus of elasticity of water =  20 × 108 N/m².
• To Find: Change in volume = dV =?

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴  Change in volume = dV   = (dP × V)/ K

∴    dV  =  5 × 10-6  m³ = 5 cc

Ans: The change produced in the volume is 5 cc.

#### Example – 4:

• A volume of 10-3 m³ of water is subjected to a pressure of 10 atmospheres. The change in volume is 10-6 m³. Find the bulk modulus of water. Atm. pressure = 10 N/m².
• Solution:
• Given: Original Volume = 10-3 m³, Pressure = dP = 10 atm = 10 × 76 × 10-2 × 13600 × 9.8 N/m², Change in volume = dV =10-6 m³,
• To Find: Bulk modulus of elasticity of water =?

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴  K = (10 × 76 × 10-2 × 13600 × 9.8 × 10-3)/ 10-6

∴  K = 1.01 × 109 N/m²

Ans: Bulk modulus of elasticity of water is 1.01 × 109 N/m²

#### Example – 5:

• Two litres of water, when subjected to a pressure of 10 atmospheres, are compressed by 1.013 cc. Find the compressibility of water.
• Solution:
• Given: Original Volume = 2 L = 2 × 10-3 m³, Pressure = dP = 10 atm = 10 × 76 × 10-2 × 13600 × 9.8 N/m², Change in volume = dV =1.013 cc = 1.013 × 10-6 m³,
• To Find: Compressibility of water =?

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴  K = (10 × 76 × 10-2 × 13600 × 9.8 × 2 × 10-3)/(1.013 × 10-6)

∴  K = 2 × 109 N/m²

Compressibility = 1/K = 1/(2 × 109)

Compressibility =  5 × 10-10 m²/N

Ans: Compressibility of water is 5 × 10-10 m²/N

#### Example – 6:

• Bulk modulus of water is 2.05 × 109 N/m². What change of pressure will compress a given quantity of water by 0.5%?
• Solution:
• Given: Bulk modulus of water = K = 2.05 × 109 N/m², Volumetric strain = 0.5 % = 0.5 × 10-2  = 5 × 10-3
• To Find: Change in pressure = dP =?

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = dP / Volumetric strain

∴  dP = K  × Volumetric strain

∴  dP = 2.05 × 109  × 5 × 10-3

∴  dP = 1.025 × 107  N/m²

Ans : Change in pressure is 1.025 × 107  N/m²

#### Example – 7:

• Calculate the change in volume of a lead block of volume 1 m³ subjected to pressure of 10 atmospheres. Also calculate compressibility of lead. 1 atm = 1.013 × 105 N/m², K = 8 × 105 N/m².
• Solution:
• Given: Original Volume = 1 m³, Pressure = dP = 10 atm = 10 × 1.013 × 105 N/m², Bulk modulus of elasticity = K = 8 × 109 N/m².
• To Find: Change in volume = dV =?, Compressibility = ?

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴  Change in volume = dV  =  (dP × V)/ K

∴  Change in volume = dV  =  (10 × 1.013 × 105 × 1)/ 8 × 109

∴    dV  =  1.27  × 10-4

Compressibility = 1/K = 1/(8 × 109)

Compressibility =  1.25 × 10-10 m²/N

Ans: Change in volume is  1.27  × 10-4 m³ and compressibility of lead is 1.25 × 10-10 m²/N

#### Example – 8:

• Find the increase in the pressure required to decrease volume of mercury by 0.001%. Bulk modulus of mercury = 2.8 × 1010 N/m².
• Solution:
• Given: Volumetric strain = 0.001% = 0.001 × 10-2 = 10-5, Bulk modulus of elasticity = 2.8 × 1010 N/m².
• To Find: Pressure intensity =?

Bulk modulus of elasticity = K = Volumetric stress / Volumetric strain

∴  Volumetric stress = K ×Volumetric strain

∴  Pressure intensity = K ×Volumetric strain

∴  Pressure intensity = 2.8 × 1010 × 10-5

∴  Pressure intensity = 2.8 × 105 N/m²

Ans: Pressure intensity is 2.8 × 105 N/m²

#### Example – 9:

• A solid brass sphere of volume 0.305 m³ is dropped in an ocean. where water pressure is 2 × 107 N/m². The bulk modulus of water is 6.1 × 1010 N/m². What is the change in volume of the sphere?
• Solution:
• Given: Original Volume = 0.305 m³, Pressure = dP = 2 × 107 N/m²², Bulk modulus of elasticity = K =6.1 × 1010 N/m²
• To Find: Change in volume = dV =?

Volumetric Stress = Pressure intensity = dP

Bulk modulus of elasticity = K = (dP × V)/ dV

∴  Change in volume = dV  =  (dP × V)/ K

∴  Change in volume = dV  =  (2 × 107 × 0.305)/ (6.1 × 1010)

∴    dV  =  10-4

Ans: Change in volume = 10-4

 Science > You are Here

1. Najibu Tanimu
• Hemant More