Mechanical Properties of Solids: Comprehensive NEET Physics Notes

1. Introduction to Mechanical Properties of Solids

1.1 Elasticity and Plasticity

In everyday life, we observe that materials can be deformed by applying forces. A solid object has a definite shape and size, but it can be stretched, compressed, or bent under an external force. The property of a material by which it returns to its original shape after the deforming force is removed is called elasticity. If a material does not return to its original shape, it is said to exhibit plasticity.

Did You Know? The concept of elasticity is crucial in designing structures such as bridges, buildings, and vehicles to ensure they can withstand external forces without permanent deformation.

2. Stress and Strain

2.1 Types of Stress

When a force is applied to a solid body, it deforms. The restoring force developed in the body per unit area is called stress. There are different types of stress:

• Tensile Stress: Occurs when a body is stretched.
• Compressive Stress: Occurs when a body is compressed.
• Shearing Stress: Occurs when forces are applied parallel to the cross-sectional area.

2.2 Types of Strain

Strain is the fractional change in the dimensions of a body. It is dimensionless and categorized as:

• Longitudinal Strain: Change in length divided by the original length.
• Shearing Strain: Relative displacement of layers divided by the original length.
• Volumetric Strain: Change in volume divided by the original volume.

NEET Tip: Stress is measured in pascals (Pa) while strain is dimensionless. Remembering these units can help avoid confusion during exams.

3. Hooke’s Law

3.1 Linear Relationship

For small deformations, stress is directly proportional to strain. This is known as Hooke’s Law, expressed as:

$\text{stress}=k\times \text{strain}$

Where kkk is the modulus of elasticity, which differs for various materials.

3.2 Elastic Moduli

Different types of moduli measure the stiffness of materials:

• Young’s Modulus ($Y$): Ratio of tensile stress to longitudinal strain.
• Shear Modulus ($G$): Ratio of shearing stress to shearing strain.
• Bulk Modulus ($B$): Ratio of hydraulic stress to volumetric strain.

Mnemonic: "Yummy Sheep Baa" - Young's, Shear, and Bulk modulus help remember the three types of elastic moduli.

4. Stress-Strain Curve

4.1 Elastic and Plastic Deformation

The stress-strain curve for a material shows how it deforms under stress. Key points on the curve include:

• Proportional Limit: The end of the linear region where Hooke's Law is valid.
• Elastic Limit: Maximum stress a material can withstand without permanent deformation.
• Yield Point: Point beyond which the material undergoes permanent deformation.
• Ultimate Strength: Maximum stress the material can withstand.
• Fracture Point: The stress at which the material breaks.

Common Misconception: Many students confuse the yield point with the elastic limit. The yield point is where permanent deformation begins, while the elastic limit is the maximum stress a material can handle and still return to its original shape.

5. Applications of Elastic Behaviour

5.1 Engineering and Structural Design

Knowledge of the elastic properties of materials is essential in designing buildings, bridges, automobiles, and other structures. Engineers must ensure materials can withstand loads without undergoing significant deformation.

Real-life Application: In designing skyscrapers, engineers use materials with high Young’s modulus to ensure the building remains stable under various loads such as wind and seismic activities.

6. Mechanical Properties of Fluids

6.1 Pressure in Fluids

Fluid pressure is defined as the force exerted by a fluid per unit area. In a fluid at rest, pressure is the same in all directions and increases with depth due to the weight of the fluid above.

$P=\rho gh$

Where:

• $P$is the pressure,
• $\rho$ is the fluid density,
• $g$ is the acceleration due to gravity,
• $h$ is the height of the fluid column.

6.2 Buoyancy and Archimedes’ Principle

A fluid exerts an upward buoyant force on an object submerged in it. According to Archimedes’ Principle, the buoyant force is equal to the weight of the fluid displaced by the object.

$F_b=\rho Vg$

Where:

• $F_b$​ is the buoyant force,
• $V$ is the volume of the fluid displaced.

NEET Problem-Solving Strategy: To solve buoyancy problems, always start by calculating the volume of the displaced fluid and then use Archimedes’ Principle to find the buoyant force.

Quick Recap

• Stress and Strain: Fundamental concepts in understanding material deformation.
• Hooke's Law: Defines the proportional relationship between stress and strain.
• Elastic Moduli: Young's, Shear, and Bulk moduli are key to material stiffness.
• Stress-Strain Curve: Illustrates elastic and plastic deformation stages.
• Fluid Pressure: Increases with depth and affects buoyancy.

Concept Connection

Understanding mechanical properties of solids and fluids is crucial in both Chemistry (e.g., behavior of materials under different conditions) and Biology (e.g., how blood flows through vessels).

Practice Questions

1. Question: Calculate the stress and strain on a steel rod of length 2 m and cross-sectional area 0.01 m² when a force of 2000 N is applied.
   Solution: $\text{Stress}=\frac{F}{A}=\frac{2000}{0.01}=2\times 10^5\text{Pa}$
   $\text{Strain}=\frac{\Delta L}{L}=\frac{2\times 10^5}{Y}(\text{Assuming }Y=2\times 10^{11}\text{Pa})$
   $\text{Strain}=\frac{2\times 10^5}{2\times 10^{11}}=1\times 10^{-6}$
2. Question: A copper wire and a steel wire of the same length and cross-sectional area are connected end to end and stretched. The Young's modulus of copper is $1.1\times 10^{11}\text{Pa}$, and that of steel is $2.0\times 10^{11}\text{Pa}$. If the net elongation is 0.5 mm, find the individual elongations.
   Solution: Using the relationship: $\frac{\Delta L_{\text{Cu}}}{\Delta L_{\text{Steel}}}=\frac{Y_{\text{Steel}}}{Y_{\text{Cu}}}$
   $\Delta L_{\text{Cu}}=2.5\times \Delta L_{\text{Steel}}$
   $\Delta L_{\text{Cu}}+\Delta L_{\text{Steel}}=0.5\text{mm}$
   Solving these, we get:
   $\Delta L_{\text{Cu}}=0.35\text{mm},\Delta L_{\text{Steel}}=0.15\text{mm}$
3. Question: Calculate the buoyant force on a 10 cm³ iron block submerged in water. The density of iron is $7.8\times 10^3{\text{kg/m}}^3$ and that of water is $1.0\times 10^3{\text{kg/m}}^3$.
   Solution: $F_b={\rho }_{\text{water}}Vg$
   $F_b=1.0\times 10^3\times 10\times 10^{-6}\times 9.8$
   $F_b=9.8\times 10^{-2}\text{N}$
4. Question: A cylinder with a base area of 0.1 m² and height 1 m is filled with water. Calculate the pressure at the bottom of the cylinder.
   Solution:$P=\rho gh$
   $P=1.0\times 10^3\times 9.8\times 1$
   $P=9.8\times 10^3\text{Pa}$
5. Question: What is the strain energy stored in a steel wire of length 2 m and cross-sectional area 0.01 m² when it is stretched by 1 mm? Young’s modulus of steel is $2\times 10^{11}\text{Pa}$.
   Solution: $U=$