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Mechanical Properties of Solids NCERT Solutions, Complete PDF Content and Important Questions
NCERT Solutions for Class 11 Physics Chapter 8 PDF Download (2026) + Important Questions: Mechanical Properties of Solids Welcome to the fascinating world of materials! Have you ever wondered why bridges are made of steel instead of aluminium? Or why a rubber band snaps back when you pull it, but a piece of clay just stays squished? All these real-world mysteries are solved in Class 11 Physics Chapter 8: Mechanical Properties of Solids . For students gearing up for the CBSE Class 11 Exams in 2026 , as well as competitive entrance tests like JEE Main and NEET, this chapter is an absolute goldmine. It bridges the gap between theoretical physics and real-world engineering. You will learn how materials behave under stress, which is a fundamental concept for anyone aspiring to build the future (literally!). In this comprehensive guide, we will break down the entire chapter into simple, easy-to-digest notes.
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ChapterMechanical Properties of Solids
SubjectPhysics
Class11
BoardCBSE
DifficultyModerate
Exam Weightage4 to 5 Marks
Learning Objectives
- Understand the concepts of elasticity, plasticity, and the restoring forces in solids.
- Calculate different types of stress (longitudinal, shear, hydraulic) and corresponding strains.
- Solve numerical problems.
- Analyze the Stress-Strain Curve and.
- Identify regions of elastic limit, yield point, and breaking point.
- Calculate Young's Modulus, Shear Modulus, and Bulk Modulus for various materials.
- Understand elastic potential energy stored in a stretched wire.
Key Concepts, Definitions and Formulas
- Elasticity: The property of a body by virtue of which it tends to regain its original size and shape after the applied force is removed
- Plasticity: The property by which a body permanently retains its altered shape even after the deforming force is removed
- Stress: The internal restoring force set up per unit area of a deformed body
- Pascal (Pa). Strain: The ratio of change in dimension to the original dimension
- It has no unit (dimensionless). Longitudinal Strain: $\epsilon = \frac{\Delta L}{L}$
- Volumetric Strain: $\frac{\Delta V}{V}$
- Hooke's Law: For small deformations, stress is directly proportional to strain
- Energy stored in a stretched wire. Formula: $U = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$ To truly understand how different materials react to stress, let's explore Hooke's Law and Young's Modulus interactively
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Full NCERT Solutions
NCERT Q8.1: A steel wire of length 4.7 m and cross-sectional area $3.0 \times 10^{-5} \text{ m}^2$ stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of $4.0 \times 10^{-5} \text{ m}^2$ under a given load. What is the ratio of the Young's modulus of steel to that of copper?
Answer: Given: For Steel: $L_s = 4.7 \text{ m}$, $A_s = 3.0 \times 10^{-5} \text{ m}^2$ For Copper: $L_c = 3.5 \text{ m}$, $A_c = 4.0 \times 10^{-5} \text{ m}^2$ Both wires stretch by the same amount ($\Delta L_s = \Delta L_c = \Delta L$) under the same load ($F_s = F_c = F$). Formula: Young's Modulus $Y = \frac{F \cdot L}{A \cdot \Delta L}$ Calculation: Ratio $\frac{Y_s}{Y_c} = \frac{\frac{F \cdot L_s}{A_s \cdot \Delta L}}{\frac{F \cdot L_c}{A_c \cdot \Delta L}}$ $\frac{Y_s}{Y_c} = \left(\frac{L_s}{L_c}\right) \times \left(\frac{A_c}{A_s}\right)$ $\frac{Y_s}{Y_c} = \left(\frac{4.7}{3.5}\right) \times \left(\frac{4.0 \times 10^{-5}}{3.0 \times 10^{-5}}\right)$ $\frac{Y_s}{Y_c} = \frac{4.7}{3.5} \times \frac{4}{3} = \frac{18.8}{10.5} \approx 1.79$ Result: The ratio of the Young's modulus of steel to that of copper is approximately 1.79 : 1 .
NCERT Q8.2: Figure shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material? (Assume standard textbook graph where at strain 0.002, stress is $150 \times 10^6 \text{ N/m}^2$, and yield point is at $300 \times 10^6 \text{ N/m}^2$) .
Answer: (a) Young's Modulus ($Y$): From the linear portion of the graph (which obeys Hooke's Law): Stress = $150 \times 10^6 \text{ N/m}^2$ Strain = $0.002$ $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{150 \times 10^6}{0.002} = 75 \times 10^9 \text{ N/m}^2 = 75 \text{ GPa}$. (b) Approximate Yield Strength: The yield strength is the maximum stress the material can sustain without crossing the elastic limit. From the graph, the curve begins to deviate from the straight line (elastic limit) at approximately $300 \times 10^6 \text{ N/m}^2$. Result: Yield strength $\approx 3 \times 10^8 \text{ N/m}^2$.
NCERT Q8.3: The stress-strain graphs for materials A and B are shown in Figure. The graphs are drawn to the same scale. (a) Which of the materials has the greater Young's modulus? (b) Which of the two is the stronger material?
Answer: (a) Young's modulus is determined by the slope of the stress-strain curve in the linear region ($Y = \text{Stress} / \text{Strain}$). The graph for material A has a steeper slope than material B. Therefore, Material A has a greater Young's modulus. (b) The strength of a material is determined by its ultimate tensile strength (the maximum stress it can handle before breaking). The peak of the curve for material A is higher than that for material B. Hence, Material A is stronger.
NCERT Q8.4: Read the following two statements below carefully and state, with reasons, if it is true or false. (a) The Young's modulus of rubber is greater than that of steel; (b) The stretching of a coil is determined by its shear modulus.
Answer: (a) False. For a given applied force, rubber stretches much more than steel. Since Strain is in the denominator for Young's Modulus ($Y = \text{Stress}/\text{Strain}$), a larger strain means a smaller Young's Modulus. Therefore, steel is actually more elastic and has a much greater Young's modulus than rubber. (b) True. When a coil spring is stretched, the wire of the spring itself gets twisted (it undergoes shear deformation) rather than simply lengthening. Therefore, the stretching of a coil heavily depends on the shear modulus of the material.
NCERT Q8.5: Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires. ($Y_{steel} = 2.0 \times 10^{11} \text{ Pa}$, $Y_{brass} = 0.91 \times 10^{11} \text{ Pa}$).
Answer: Given: Diameter $D = 0.25 \text{ cm} = 2.5 \times 10^{-3} \text{ m}$ Radius $r = 1.25 \times 10^{-3} \text{ m}$ Area $A = \pi r^2 = 3.14 \times (1.25 \times 10^{-3})^2 = 4.91 \times 10^{-6} \text{ m}^2$ 1. Elongation of Steel Wire: The steel wire supports the weight of both masses (e.g., 4 kg and 6 kg). Let total mass $M_s = 4 + 6 = 10 \text{ kg}$. Force $F_s = M_s g = 10 \times 9.8 = 98 \text{ N}$. Length $L_s = 1.5 \text{ m}$. $\Delta L_s = \frac{F_s L_s}{A \cdot Y_s} = \frac{98 \times 1.5}{(4.91 \times 10^{-6}) \times (2.0 \times 10^{11})}$ $\Delta L_s = \frac{147}{9.82 \times 10^5} \approx 1.49 \times 10^{-4} \text{ m}$ 2. Elongation of Brass Wire: The brass wire only supports the lower mass (6 kg). Force $F_b = 6 \times 9.8 = 58.8 \text{ N}$. Length $L_b = 1.0 \text{ m}$. $\Delta L_b = \frac{F_b L_b}{A \cdot Y_b} = \frac{58.8 \times 1.0}{(4.91 \times 10^{-6}) \times (0.91 \times 10^{11})}$ $\Delta L_b = \frac{58.8}{4.468 \times 10^5} \approx 1.3 \times 10^{-4} \text{ m}$
Extra Important Questions for 2026 Exams
Important Q1: Which of the following is the most elastic material? A) Rubber B) Plastic C) Steel D) Glass
Answer: C) Steel Explanation: Steel requires the maximum force to produce a given strain, meaning it has the highest Young's modulus among the options.
Important Q2: The restoring force per unit area is known as: A) Strain B) Stress C) Elasticity D) Pressure
Answer: B) Stress Explanation: By definition, stress is the internal restoring force developed per unit cross-sectional area.
Important Q3: If the length of a wire is reduced to half, its Young's modulus will: A) Become half B) Double C) Remain the same D) Become four times
Answer: C) Remain the same Explanation: Young's modulus is an intrinsic property of the material. It depends on the nature of the material, not on its physical dimensions like length or radius.
Important Q4: Differentiate between elastic and plastic behavior of materials.
Answer: Elastic behavior is when a material fully regains its original shape and size after the deforming force is removed (e.g., a steel spring). Plastic behavior is when a material retains the deformed shape and does not return to its original state even after the force is removed (e.g., putty or wet clay).
Important Q5: What is the yield point on a stress-strain curve?
Answer: The yield point (or elastic limit) is the maximum stress a material can withstand without undergoing permanent deformation. If stretched beyond this point, the material behaves plastically.
Important Q6: Define Poisson's ratio.
Answer: Within the elastic limit, the ratio of lateral strain to the longitudinal strain is constant for a given material. This constant is called Poisson's ratio ($\sigma$). It has no units.
Important Q7: Why do springs have a spiral shape?
Answer: A spiral shape ensures that when a force is applied, the wire undergoes shearing (twisting) rather than just linear stretching. Materials usually have a lower shear modulus than Young's modulus, making it easier to stretch or compress a spiral spring.
Important Q8: Explain the stress-strain curve for a metallic wire under increasing load. Mark the elastic limit, yield point, and fracture point.
Answer: Proportional Region (O to A): Stress is directly proportional to strain (Hooke's Law obeys here). The graph is a straight line. Elastic Limit / Yield Point (B): Beyond A, stress is not proportional to strain, but the body still regains its original shape if the load is removed. Point B is the yield point. Plastic Region (B to D): If load increases beyond B, the strain increases rapidly even for small stress changes. The wire gets permanently deformed. Ultimate Tensile Strength (D): The maximum stress the wire can handle. Fracture Point (E): The point where the wire eventually snaps or breaks.
Important Q9: Derive an expression for the elastic potential energy stored in a stretched wire.
Answer: Let a wire of length $L$ and area $A$ be stretched by distance $l$ by applying force $F$. Young's Modulus $Y = \frac{F \cdot L}{A \cdot l} \implies F = \frac{Y A l}{L}$ Small work done $dW$ to stretch it by $dl$ is $dW = F \cdot dl = \left(\frac{Y A l}{L}\right) dl$. Total work done $W = \int_{0}^{\Delta L} \frac{Y A l}{L} dl = \frac{Y A}{L} \left[ \frac{l^2}{2} \right]_{0}^{\Delta L} = \frac{1}{2} \frac{Y A (\Delta L)^2}{L}$. This work is stored as potential energy ($U$). Rewriting: $U = \frac{1}{2} \times \left(\frac{Y A \Delta L}{L}\right) \times \Delta L = \frac{1}{2} \times F \times \Delta L$. Energy density (Energy per unit volume) = $\frac{U}{AL} = \frac{1}{2} \times \frac{F}{A} \times \frac{\Delta L}{L} = \frac{1}{2} \times \text{Stress} \times \text{Strain}$.
Important Q10: A solid sphere of radius 10 cm is subjected to a uniform pressure of $5 \times 10^8 \text{ Pa}$. If the bulk modulus of the material is $3.14 \times 10^{11} \text{ Pa}$, calculate the decrease in its volume.
Answer: Radius $r = 10 \text{ cm} = 0.1 \text{ m}$. Original Volume $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (0.1)^3 = 4.186 \times 10^{-3} \text{ m}^3$. Pressure $P = 5 \times 10^8 \text{ Pa}$. Bulk Modulus $B = 3.14 \times 10^{11} \text{ Pa}$. Formula: $B = \frac{P V}{\Delta V} \implies \Delta V = \frac{P V}{B}$ $\Delta V = \frac{(5 \times 10^8) \times (4.186 \times 10^{-3})}{3.14 \times 10^{11}}$ $\Delta V \approx \frac{20.93 \times 10^5}{3.14 \times 10^{11}} \approx 6.66 \times 10^{-6} \text{ m}^3$. Decrease in volume is $6.66 \times 10^{-6} \text{ m}^3$. A construction company is selecting materials to build suspension bridge cables. They test two materials, X and Y. Material X has a high Young's Modulus but a low breaking stress. Material Y has a moderate Young's Modulus and a very high breaking stress.
Important Q11: Why is Young's Modulus an important factor in building a bridge?
Answer: A high Young's Modulus means the material will not stretch or sag significantly under the heavy load of cars and trucks, keeping the bridge stable and safe.
Important Q12: Which material (X or Y) should the engineers prefer for the suspension cables and why?
Answer: They should prefer Material Y. While X prevents stretching, its low breaking stress means it could snap suddenly under a heavy load or high wind. Material Y can handle much heavier loads before breaking, making it safer for critical support structures.
Important Q13: What is the term for the maximum stress a material can withstand before breaking?
Answer: Ultimate Tensile Strength. (Directions: Select A if both are true and Reason explains Assertion. Select B if both are true but Reason does not explain Assertion. Select C if Assertion is true, Reason is false. Select D if Assertion is false, Reason is true.)
Important Q14: Assertion (A): Steel is more elastic than rubber. Reason (R): For a given load, the strain produced in steel is much less than the strain produced in rubber.
Answer: A . Both A and R are true, and R is the correct explanation. Because steel resists deformation more strongly (less strain), it has a higher Young's modulus and is thus more elastic in physics terms.
Important Q15: Assertion (A): The stress-strain graph for all elastic materials is always a straight line. Reason (R): Hooke's Law states that stress is directly proportional to strain.
Answer: D . Assertion is false. The graph is only a straight line within the proportional limit. Beyond that, the material may still be elastic up to the yield point, but the graph is curved (Hooke's Law is not obeyed there).
FAQ Section
Is Mechanical Properties of Solids important for Class 11 boards?
Yes, it carries a weightage of about 4-5 marks. Questions are usually straightforward, making it an easy chapter to score full marks in.
Where can I download the Class 11 Physics NCERT PDF for this chapter?
You can download the official, updated NCERT textbooks for free directly from the official NCERT website (ncert.nic.in) under the 'Publications' section.
What is the difference between Bulk Modulus and Young's Modulus?
Young's Modulus measures a solid's resistance to changes in length (linear stress). Bulk Modulus measures a material's resistance to changes in volume under uniform hydraulic pressure.
Why is a negative sign used in the Bulk Modulus formula?
The negative sign indicates that as the pressure increases (positive change), the volume of the material decreases (negative change). The negative sign ensures the Bulk Modulus value remains positive.
Are Elastomers covered in the 2026 CBSE syllabus?
Yes, elastomers like rubber and tissue (aorta) are important. They do not obey Hooke's Law well and have large elastic regions without permanent deformation
More Class 11 Physics Chapters
Ch 1: Units and MeasurementsCh 2: Motion in a Straight LineCh 3: Motion in a PlaneCh 4: Laws of MotionCh 5: Work, Energy and PowerCh 6: System of Particles and Rotational MotionCh 7: GravitationCh 8: Mechanical Properties of SolidsCh 9: Mechanical Properties of FluidsCh 10: Thermal Properties of MatterCh 11: ThermodynamicsCh 12: Kinetic TheoryCh 13: OscillationsCh 14: Waves