Bhaskar Das Arts and Board Exam Specialist
Arts and Board Exam Specialist | Updated On - Dec 11, 2025
Solids possess definite shape and volume due to strong intermolecular forces. When external forces are applied, solids deform, and their mechanical properties describe how they resist or recover from such deformations. Mechanical properties of Solids is an important chapter in Class 11 Physics that explains how solid materials respond to external forces.
When an external force (push or pull) is applied to something like metal, plastic, or rubber, its shape changes. This helps us understand two important concepts in physics: stress and strain.

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Introduction to Mechanical Properties of Solids
The study of Mechanical Properties of Solids, Physics Class 11, also helps engineers and technicians prevent accidents from happening. Some mechanical properties include plasticity, stress, elasticity, strain, Young’s modulus, etc. These characteristics help determine a suitable solid for a given specific work.
- Elasticity: The property by which a body regains its original shape and size after removing the deforming force.
- Plasticity: The property by which a body does not regain its original shape and size and undergoes permanent deformation.
- Stress: Restoring force per unit area developed in a body when deformed. Mathematically, the Stress formula can be expressed as: σ = F / A. In Physics, the SI unit of Stress is N/m2 or Pascal (Pa). The three types of stress studied in the Mechanical Properties of Solids Class 11 are, namely: Tensile stress, Compressive stress, and Shear (tangential) stress.
- Strain: Change in dimension per unit original dimension. It is dimensionless. Three types of strain studied in the Mechanical Properties of Solids Class 11 are, namely: Longitudinal, Volumetric, and Shear strain.
Mechanical Properties of Solids: Questions Weightage in JEE & NEET
The overall weightage of this Chapter in JEE and NEET Exams is given below:
| Exam | Approximate Weightage | Number of Questions (Tentative) |
|---|---|---|
| JEE Mains | 2–3% | 1–2 questions |
| JEE Advanced | 3–4% | 1–2 questions (may include multi-concept) |
| NEET | 2–3% | 1–2 questions (often combined with fluids) |
Important Questions on Mechanical Properties of Solids
Very Short Answer Questions (1 Mark Questions)
Ques: Why is the negative sign used in the definition of bulk modulus?
Ans: Pressure is taken as positive when it compresses the material, but volume decreases. The negative sign ensures that the bulk modulus is positive.
Ques: Can Poisson’s ratio be negative?
Ans: Yes, in some auxetic materials (e.g., certain foams), Poisson's ratio is negative – they expand laterally when stretched.
Ques: Which material has the highest Young’s modulus?
Ans: Diamond has the highest Young's modulus (≈ 1200 GPa) among known materials.
Ques: Why do rubber and steel have almost the same Poisson’s ratio but vastly different Young’s modulus?
Ans: Poisson’s ratio depends on molecular structure and lateral contraction behaviour, whereas Young’s modulus depends on bond strength. Rubber has weak intermolecular forces (low Young's modulus, Y) but a similar lateral response to steel.
Ques: What happens to elastic moduli with an increase in temperature?
Ans: Generally, elastic moduli decrease with a rise in temperature because thermal vibrations weaken interatomic forces.
Ques: Is strain a tensor quantity?
Ans: Yes, strain is a second-rank tensor because it has both normal and shear components in three dimensions.
Short Answer Questions (2 Mark Questions)
Ques. Define longitudinal strain.
Ans. It is defined to be the ratio of change in length to the original length of an object when it gets deformed by an external force, i.e. Longitudinal strain = l /L= change in length/ original length
Ques. Define volumetric strain.
Ans. It is defined as the ratio of change in volume per unit of the original volume of the body when deformed by an external force change in volume AV i.e. Volumetric strain = change in volume/ original volume =ΔV/V
Ques. Define longitudinal or tensile stress.
Ans. It is defined as the restoring force developed per unit area of cross-section of a body when the length of the body increases in the direction of the deforming force.
Read More: Stress
Ques. What are the factors due to three states of matter to differ from one another?
Ans. Three states of matter differ from each other due to the following two factors:
(a) The different amounts of tester atomic and intermolecular forces.
(b) The degree of haphazard thermal motion of the atoms and molecules of a substance depends upon the temperature.
Ques. When we stretch a wire, we have to perform work. Why? What happens to the energy given to the wire in this process?
Ans. In a normal situation/circumstances, the atoms of a solid are at the locations of minimum potential energy. When we intentionally stretch a wire, the work has to be done against interatomic forces. This work is stored in the wire in the form of elastic potential energy for the time being.
Very Long Answer Questions (5 Marks Questions)
Ques. (a) What are the factors affecting elasticity?
(b) Define Poisson’s ratio.
(c) Define breaking load
Ans. (a) The following factors affect the elasticity of a material:
- Effect of hammering and rolling on a material: It causes a decrease in the plasticity of the material due to the break-up of crystal grains into smaller units and thus, the elasticity of the material increases.
- Effect of Annealing on a material: Annealing results in an increase in the plasticity of the material due to, the formation of large crystal grains. Henceforth, the elasticity of the material decreases.
- Effect of the presence of impurities on a material: The effect of the presence of impurities in a material can be both ways i.e. it can increase as well as decrease the elasticity of the material. The type of effect depends upon the nature of the impurity present in the material.
- Effect of temperature on a material: In most cases, an increase in the temperature of the material causes a decrease in the elasticity of the material. The elasticity remains constant with temperature.
(b) Poisson’s Ratio (σ): Within elastic limits, it is defined as the ratio of lateral strain (β) to the linear strain, i.e. σ = βα
(c) Breaking Load: It is defined as the product of the breaking stress and area of cross-section of the given object. It is also called the maximum load a body (cable/ wire) can support or hold i.e. it is given as breaking load = Breaking stress × area of cross-section. It should be noted that breaking stress is always constant for the given material.
Ques. Why are the bridges declared unsafe after long use?
Ans. A bridge, during its use, undertakes alternative strains a large number of times each day, depending upon the movement of vehicles on it. When a bridge is used for a long time, it exhausts its elastic strength, resulting in a significant increase in the number of strains in the bridge for a given stress. Ultimately, the bridge may collapse. Thus, to avoid this, the bridges are stated to be unsafe after long use.
Ques. Give reasons why the springs are made of steel and not of copper.
Ans. Spring should have a large restoring force, which is set up in it when it is deformed, which in turn depends upon the elasticity of the material of the spring. Since Young’s modulus of elasticity of steel is more than that of copper, steel is preferred in making such springs.
Read More: Interaction of Forces & Effects of Force Interaction
Important MCQs on Mechanical Properties of Solids
- The SI unit of stress is:
- Pascal
- Newton
- Joule
- Watt
Answer: a) Pascal
Stress = Force/Area = N/m2 = Pascal.
2. Within the elastic limit, the stress is proportional to strain. This was stated by:
- Newton
- Hooke
- Young
- Poisson
Answer: b) Hooke
3. Toisson’s ratio cannot have a value:
- 0.1
- 0.5
- 0.8
- -o.1
Answer: c) 0.8
For isotropic materials, −1 < ν < 0.5. Value > 0.5 violates
Thermodynamic stability.
4. Energy stored in a stretched wire of length L, across-section A, stretched by ΔL is:
- (Y A (ΔL)²)/(2L)
- (Y A ΔL)/(2L)
- (Y A (ΔL)²)/L
- Y A ΔL / L
Answer: a) (Y A (ΔL)²)/(2L)
U = (½) x stress x strain x volume = (½) x (Y strain) x strain x
(AL) = (Y A (ΔL)²)/(2L)




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