CBSE Class 11 Physics NCERT Solutions: Mechanical Properties of Solids

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NCERT Solutions PDF Class 11 PDF

This resource provides detailed NCERT Solutions for Class 11 Physics, Chapter 8: Mechanical Properties of Solids. It covers essential concepts such as elasticity, stress, strain, Young's modulus, modulus of rigidity, and bulk modulus. The solutions explain the behavior of solids under stress and strain, including topics like Hooke's Law, elastic limits, and the properties of materials like steel and copper. This chapter is crucial for understanding how materials deform and respond to applied forces. These solutions are designed to help students grasp the fundamental principles and solve problems effectively, aiding in their preparation for exams by offering clear explanations and step-by-step problem-solving approaches.

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BoardCBSE
ClassClass 11
SubjectPhysics Exemplar
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 8

Chapter summary

Chapter 8, Mechanical Properties of Solids, focuses on the elastic behavior of solid materials. This NCERT Solutions set breaks down key concepts like stress, strain, and elastic moduli (Young's modulus, shear modulus, bulk modulus). It explains how materials respond to applied forces, the elastic limit, and breaking stress. The solutions cover various types of strain and their relation to applied stress, providing a solid foundation for understanding material science and its applications.

Learning outcomes

  • Understand the concepts of stress, strain, and elastic moduli.
  • Differentiate between various types of strain (longitudinal, shear, volumetric).
  • Apply Hooke's Law to solve problems related to elasticity.
  • Analyze the behavior of solids under tensile, compressive, and shear forces.
  • Calculate the change in length, diameter, or volume of materials under stress.
  • Relate the properties of different materials (e.g., copper, iron, steel) to their elastic behavior.

Topics covered

Paper topics

  • Elasticity
  • Stress
  • Strain
  • Young's Modulus
  • Modulus of Rigidity
  • Bulk Modulus
  • Hooke's Law
  • Elastic Limit
  • Breaking Stress
  • Deformation of Solids
  • Properties of Materials (Copper, Iron, Steel)
  • Springs

Important topics

  • Stress and Strain definitions
  • Young's Modulus calculation and application
  • Hooke's Law
  • Relationship between different elastic moduli
  • Effect of temperature on elasticity
  • Applications of elasticity in engineering

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Questions and Solutions

MCQ 1

Q. 1 What is the modulus of rigidity of ideal liquids? Options are: (a) infinity, (b) zero, (c) unity, (d) some finite small non-zero constant value.
Solution: For an ideal liquid, there are no intermolecular forces that resist deformation. This means they cannot sustain any shear stress. Since the modulus of rigidity is defined as the ratio of shear stress to shear strain, and an ideal liquid cannot develop shear stress, its modulus of rigidity is zero. Thus, the correct option is (b).

MCQ 2

Q. 2 Consider a wire. If its length is reduced to half of its original length, what will be the change in the maximum load it can withstand without breaking? Options are: (a) be double, (b) be half, (c) be four times, (d) remain same.
Solution: The breaking stress of a material is a property that depends on the material itself and its cross-sectional area, not its length. The formula for breaking stress is: \text{Breaking stress} = \frac{\text{Breaking force}}{\text{Area of cross-section}} The breaking force is the maximum force the wire can withstand. When the length of the wire is changed, its cross-sectional area remains the same. Therefore, the breaking force, and consequently the maximum load the wire can withstand without breaking, will remain the same. Thus, the correct option is (d).

MCQ 3

Q. 3 If the temperature of a wire is doubled, what happens to its Young's modulus of elasticity? Options are: (a) will also double, (b) will become four times, (c) will remain same, (d) will decrease.
Solution: The Young's modulus of a material is generally dependent on temperature. For most solids, as the temperature increases, the interatomic forces become weaker due to increased thermal vibrations. This leads to a decrease in the material's resistance to deformation. The relationship can be understood by considering thermal expansion. While the provided source shows a proportionality Y \propto \frac{1}{\Delta T}, a more general understanding is that increased temperature reduces the elastic moduli. Therefore, when the temperature of a wire increases (doubling it implies an increase), its Young's modulus will decrease. Thus, the correct option is (d).

MCQ 4

Q. 4 When a spring is stretched by applying a load to its free end, what type of strain is produced in the spring? Options are: (a) volumetric, (b) shear, (c) longitudinal and shear, (d) longitudinal.
Solution: When a load is applied to the free end of a spring, it causes an extension in its length. This change in length per unit original length is known as longitudinal strain. Additionally, the helical structure of the spring undergoes a slight change in shape due to the applied force, which can be associated with shear strain. Therefore, both longitudinal and shear strains are produced in the spring. Thus, the correct option is (c).

MCQ 5

Q. 5 A rigid bar of mass M is supported symmetrically by three wires, each of length l. The wires at each end are made of copper, and the middle one is made of iron. If each wire is to have the same tension, what is the ratio of their diameters? Options are: (a) Y_{copper} / Y_{iron}, (b) \sqrt{\frac{Y_{iron}}{Y_{copper}}}, (c) \frac{Y^2_{\text{iron}}}{Y^2_{\text{copper}}}, (d) \frac{Y_{iron}}{Y_{copper}}.
Solution: We know the formula for Young's modulus is Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}. Here, F is the tension (force), A is the cross-sectional area, L is the original length, and \Delta L is the change in length. The cross-sectional area of a wire is given by A = \pi (D/2)^2 = \frac{\pi D^2}{4}, where D is the diameter. Substituting this into the Young's modulus formula, we get: Y = \frac{F}{(\pi D^2 / 4)} \times \frac{L}{\Delta L} = \frac{4 FL}{\pi D^2 \Delta L} Rearranging this equation to solve for the diameter D, we get: D^2 = \frac{4 FL}{\pi \Delta L Y} \quad \Rightarrow \quad D = \sqrt{\frac{4 FL}{\pi \Delta L Y}} Since the bar is supported symmetrically and each wire has the same tension (F) and the same original length (L), and assuming the vertical displacement (\Delta L) is the same for all wires under equal tension, we can see that the diameter D is inversely proportional to the square root of Young's modulus Y: D \propto \sqrt{\frac{1}{Y}} Therefore, the ratio of the diameter of the copper wire (D_{copper}) to the diameter of the iron wire (D_{iron}) is: \frac{D_{copper}}{D_{iron}} = \sqrt{\frac{Y_{iron}}{Y_{copper}}} Thus, the correct option is (b).

MCQ 6

Q. 6 A mild steel wire of length 2L and cross-sectional area A is stretched horizontally between two pillars. A mass m is suspended from the mid-point of the wire. Assuming the vertical displacement x is very small compared to L, what is the strain in the wire? Options are: (a) \frac{x^2}{2L^2}, (b) \frac{x}{L}, (c) x^2/L, (d) x^2 / 2L.
Solution: Let the initial length of the wire be 2L. When a mass m is suspended from the mid-point, the wire sags by a vertical distance x. Consider one half of the wire, which has an initial length L. After sagging, the length of this half becomes the hypotenuse of a right-angled triangle with sides L (horizontal distance from pillar to midpoint) and x (vertical sag). Using the Pythagorean theorem, the new length of this half-wire is \sqrt{L^2 + x^2}. The total length of the wire is now 2\sqrt{L^2 + x^2}. The change in length (\Delta L) is the final length minus the initial length: \Delta L = 2\sqrt{L^2 + x^2} - 2L. Since x is very small compared to L (x << L), we can use the binomial approximation \sqrt{L^2 + x^2} = L\sqrt{1 + x^2/L^2} \approx L(1 + \frac{1}{2}\frac{x^2}{L^2}). Substituting this approximation into the expression for \Delta L: \Delta L \approx 2 \left( L(1 + \frac{x^2}{2L^2}) \right) - 2L = 2L + \frac{x^2}{L} - 2L = \frac{x^2}{L}. The strain is defined as the ratio of the change in length to the original length: \text{Strain} = \frac{\Delta L}{\text{Original Length}} = \frac{x^2/L}{2L} = \frac{x^2}{2L^2}. Thus, the correct option is (a).

Common mistakes

  • Confusing breaking stress with tensile strength.
  • Incorrectly applying formulas for strain when both length and shape change.
  • Not considering the effect of temperature on Young's modulus.
  • Misinterpreting the relationship between diameter and Young's modulus in wire stretching problems.

Revision tips

  • Focus on understanding the definitions of stress, strain, and the three elastic moduli.
  • Practice solving problems involving Hooke's Law and calculating changes in dimensions.
  • Review the relationship between temperature and Young's modulus.
  • Pay attention to the units and dimensions used in calculations.
  • Visualize the deformation of solids under different types of stress.

Practice MCQs

Q1. What is the modulus of rigidity for ideal liquids?

Q2. If a wire's length is reduced to half, how does its maximum load withstand capacity change?

Q3. When the temperature of a wire is doubled, what happens to its Young's modulus of elasticity?

Q4. What type of strain is produced in a spring when it is stretched by a load at its free end?

Q5. For a rigid bar supported by three wires (two copper, one iron) of the same length and tension, what is the ratio of their diameters?

Frequently asked questions

What are the main concepts covered in CBSE Class 11 Physics Chapter 8?

Chapter 8 covers the mechanical properties of solids, including elasticity, stress, strain, Young's modulus, modulus of rigidity, bulk modulus, Hooke's Law, elastic limit, and breaking stress.

How do these NCERT Solutions help students?

These solutions provide clear, step-by-step explanations for each question, helping students understand complex concepts and problem-solving techniques for the Mechanical Properties of Solids chapter.

What is the difference between stress and strain?

Stress is the internal restoring force per unit area within a deformed body, while strain is the measure of deformation, defined as the ratio of change in dimension to the original dimension.

What is Young's modulus and when is it used?

Young's modulus (Y) is a measure of a solid's resistance to elastic deformation under tensile or compressive stress. It is used for calculating changes in length of wires or rods.

Why is the modulus of rigidity zero for ideal liquids?

Ideal liquids cannot sustain shear stress; they flow without resistance. Therefore, their modulus of rigidity, which measures resistance to shear deformation, is zero.

Does temperature affect the elastic properties of solids?

Yes, generally, the elastic moduli of solids decrease as temperature increases. This is because increased thermal vibrations weaken the interatomic forces.

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