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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 × 10^{4} N/m². Find the bulk modulus for rubber.

**Solution:****Given:**Volumetric strain = 14.5 % = 14.5 × 10^{-2}, Volumetric stress = 1.45 × 10^{4}N/m²,**To Find:**Bulk modulus of elasticity = ?

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

∴ K = (1.45 × 10^{4}) / (14.5 × 10^{-2}) = 10^{5} N/m²

**Ans: **Bulk modulus of elasticity of rubber is 10^{5} N/m²

**Example -2:**

- What pressure should be applied to a lead block to reduce its volume by 10% Bulk modulus for lead = 6 × 10
^{9}N/m².

**Solution:****Given:**Volumetric strain = 10 % = 10 × 10^{-2}, Bulk modulus of elasticity = 6 × 10^{9}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 × 10^{9} ×10 × 10^{-2}

∴ Pressure intensity = 6 × 10^{8} N/m²

**Ans: **Pressure intensity is 6 × 10^{8} 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 × 10
^{8}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 × 10^{8}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^{5 }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 × 10^{9} N/m²

**Ans:** Bulk modulus of elasticity of water is 1.01 × 10^{9} 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 × 10^{9} N/m²

Compressibility = 1/K = 1/(2 × 10^{9})

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 × 10
^{9}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 × 10^{9}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 × 10^{9} × 5 × 10^{-3 }

∴ dP = 1.025 × 10^{7} N/m²

**Ans : **Change in pressure is 1.025 × 10^{7} 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 × 10
^{5}N/m², K = 8 × 10^{5}N/m².

**Solution:****Given:**Original Volume = 1 m³, Pressure = dP = 10 atm = 10 × 1.013 × 10^{5}N/m², Bulk modulus of elasticity = K = 8 × 10^{9}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 × 10^{5} × 1)/ 8 × 10^{9}

∴ dV = 1.27 × 10^{-4 }m³

Compressibility = 1/K = 1/(8 × 10^{9})

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 × 10
^{10}N/m².

**Solution:****Given:**Volumetric strain = 0.001% = 0.001 × 10^{-2}= 10^{-5}, Bulk modulus of elasticity = 2.8 × 10^{10}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 × 10^{10} × 10^{-5}

∴ Pressure intensity = 2.8 × 10^{5} N/m²

**Ans: **Pressure intensity is 2.8 × 10^{5} N/m²

#### Example – 9:

- A solid brass sphere of volume 0.305 m³ is dropped in an ocean. where water pressure is 2 × 10
^{7}N/m². The bulk modulus of water is 6.1 × 10^{10}N/m². What is the change in volume of the sphere?

**Solution:****Given:**Original Volume = 0.305 m³, Pressure = dP = 2 × 10^{7}N/m²², Bulk modulus of elasticity = K =6.1 × 10^{10}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 × 10^{7} × 0.305)/ (6.1 × 10^{10})

∴ dV = 10^{-4 }m³

**Ans:** Change in volume = 10^{-4 }m³

Science > Physics > Elasticity > You are Here |

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The problems are treated step by step and the required solution of unknown is well achieved. But along the way there’s typing error in some questions , like in Question 7 and 9.

However I have a question, why the Bulk modulus of water is not constant? Foristance in Question 4 finally the Bulk modulus of water is 1.013×10^9 N/m2 and rest of problems 2×10^9 N/m2. WHY???

THANK YOU SO MUCH FOR YOUR HIGHEST PARTICIPATION IN HELPING STUDENTS LIKE ME TO ACHIEVE BETTER ACADEMIC PERFORMANCE.

Thank you Najibu Tanimu. The actual value of water is Bulk modulus of water is 2.2 ×10^9 N/m2. The answers obtained in those problems is due to given data.

Najibu we have started this site about one year back. Our aim is to make knowledge available to each deserving and inquisitive student. Be part of mission to spread the website in student community. It will take about next five years to add articles on science, commerce, law and management. I don't know your nationality. Just you can help us by suggesting topics.