Class 11 Physics Important Chapter 9 Mechanical Properties of Solids

Class 11 Physics Important Chapter 9 Mechanical Properties of Solids Solutions English Medium As Per AHSEC New Syllabus to each chapter is provided in the list so that you can easily browse through different chapters NCERT Class 11 Physics Important Solutions and select need one. AHSEC Class 11 Physics Additional Notes English Medium Download PDF. HS 1st Year Physics Important Solutions in English.

Class 11 Physics Important Chapter 9 Mechanical Properties of Solids

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Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. NCERT Class 11 Physics Additional Question Answer are part of All Subject Solutions. Here we have given HS 1st Year Physics Important Notes in English for All Chapters, You can practice these here.

Chapter: 9

IMPORTANT QUESTION AND ANSWER

Answer the Following Questions:

1. Define stress and give its SI unit.

Ans: Stress is the force per unit area acting normal to the surface. SI unit is Pascal (Pa).

2. What is strain?

Ans: Strain is the ratio of change in length to the original length of the material.

3. State Hooke’s law.

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Ans: For small deformations, stress is directly proportional to strain.

4. What is Young’s modulus? Write its formula.

Ans: Young’s modulus is the ratio of tensile stress to tensile strain, Y = stress/strain = 

5. What is shear stress?

Ans: Shear stress is the restoring force per unit area when forces act parallel to the surface.

6. Define bulk modulus.

Ans: Bulk modulus is the ratio of hydraulic stress to volume strain: B = −p / ΔV/V

7. What is Poisson’s ratio?

Ans: Poisson’s ratio is the ratio of lateral strain to longitudinal strain.

8. What happens if the stress exceeds the elastic limit?

Ans: Permanent deformation occurs, called plastic deformation.

9. Why are ropes made of many thin wires twisted together?

Ans: To combine flexibility and strength while preventing permanent deformation.

10. Why do engineers prefer steel for structural work?

Ans: Steel has a high Young’s modulus and yield strength, making it strong and elastic.

Long Type Question And Answer:

1. Explain the concepts of stress, strain and Hooke’s law. What are the different types of elastic moduli?

Ans: Stress is the internal restoring force per unit area inside a body when an external force tries to deform it. Its unit is Pascal (Pa).

Strain is the fractional change in dimension (like length or volume) caused by stress. It is dimensionless.

Hooke’s Law states that within the elastic limit of a material, the stress is directly proportional to the strain. Mathematically,

Stress = k × strain

where k is the modulus of elasticity.

Types of elastic moduli:

Young’s modulus (Y) – ratio of tensile/compressive stress to longitudinal strain.

Shear modulus (G) – ratio of shear stress to shear strain.

Bulk modulus (B) – ratio of volumetric stress (pressure) to volumetric strain.

These moduli help engineers design materials and structures that can withstand forces without permanent deformation.

2. State and explain Hooke’s law. What are the limitations of Hooke’s law?

Ans: Hooke’s law states that within the elastic limit of a material, the stress applied on it is directly proportional to the strain produced. Mathematically:

Stress = K × Strain

where k is a constant called the modulus of elasticity.

This means if the force is doubled, the deformation (strain) also doubles, but only up to a point called the elastic limit.

Graphically, stress vs. strain is a straight line in the elastic region.

Limitations:

(i) Hooke’s law is valid only up to the elastic limit. Beyond this, the material shows plastic deformation and does not obey Hooke’s law.

(ii) Not all materials obey Hooke’s law. For example, rubber and elastomers show non-linear elasticity.

(iii) It applies only for small deformations and stresses.

3. What is Poisson’s ratio? Explain its significance and typical values for some materials.

Ans: When a wire or rod is stretched, it not only elongates but also contracts in the directions perpendicular to the applied force. This lateral contraction compared to the longitudinal elongation is quantified by Poisson’s ratio.

Poisson’s ratio (ν) is defined as the ratio of lateral strain to longitudinal strain:

v = Lateral strain / Longitudinal strain

where d is diameter and L is length.

Significance:

It provides insight into the dimensional changes of a material under load and is important for engineering designs to predict material behavior in three dimensions.

Typical values:

(i) Steel: 0.28 to 0.30

(ii) Aluminium alloys: about 0.33

(iii) Rubber has much smaller Poisson’s ratio.

(iv) Poisson’s ratio is always less than 0.5 for common materials.

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