Important Questions Class 11 Physics Chapter 8: Mechanical Properties of Solids

Mechanical properties of solids class 11 important questions cover elasticity, stress, strain, Hooke's law, the stress-strain curve, Young's modulus, shear modulus, bulk modulus, Poisson's ratio, elastic potential energy, and real-life applications as per the CBSE 2026 syllabus. Elasticity is the property by which a body tends to regain its original shape and size when a deforming force is removed.

Important Questions Class 11 Physics Chapter 8 on this page cover every topic from the NCERT chapter: 1-mark definitions, 2-mark reasoning, 3-mark derivation or graph, 4-mark numerical, 5-mark comprehensive questions, assertion-reason, and application questions.

Steel is used in bridges and not rubber. I-shaped beams carry heavier loads than rectangular bars of the same weight. A mountain cannot grow beyond 10 km on Earth. All of these facts come from one chapter. Important questions from mechanical properties of solids class 11 are structured here by mark type and topic so you can fix weak areas before the 2026 board exam.

Key Takeaways

Introduction to Class 11 Physics Chapter 8 Mechanical Properties of Solids

This chapter is not just theory. It explains why every structure around you stands. Why bridges use steel beams with specific cross-sections, why crane ropes are braided, and why mountains have a height limit are all answers from Chapter 8.

Three areas carry the most marks in CBSE 2026 papers. The stress-strain curve carries 3 to 5 marks and appears in almost every paper. Young's modulus numericals appear as 3-mark or 4-mark questions. Elastic potential energy derivation appears as a 5-mark question. A student who masters these three areas scores well above average from this chapter.

Class 11 Physics Chapter 8 Mechanical Properties of Solids: Topics Covered

Every board question from this chapter maps to one of these topic areas. Map your weak areas before starting the question bank.

Elastic and plastic behaviour of materials. Stress: tensile, compressive, shearing, hydraulic. Strain: longitudinal, shearing, volume. Hooke's law and modulus of elasticity. Stress-strain curve: proportional limit, elastic limit, yield point, ultimate tensile strength, fracture point, ductile and brittle behaviour. Young's modulus (Y). Shear modulus or modulus of rigidity (G). Bulk modulus (B) and compressibility. Poisson's ratio. Elastic potential energy in a stretched wire. Applications: crane ropes, beam bending, I-section beams, pillars, mountain height.

Important Questions Class 11 Physics Chapter 8 with Answers

These mechanical properties of solids class 11 questions and answers cover every subtopic students search for. The questions appear in five mark-formats in CBSE 2026 papers: 1-mark definitions, 2-mark reasoning, 3-mark derivation or graph, 4-mark numerical, and 5-mark comprehensive questions.

1-Mark Questions on Mechanical Properties of Solids

Q1. Define elasticity. Elasticity is the property of a body by which it tends to regain its original shape and size when the applied deforming force is removed.

Q2. What is stress? State its SI unit. Stress is the restoring force developed per unit area inside a body. Stress = F/A. SI unit: N m-2 or Pascal (Pa). Dimensional formula: [ML-1T-2].

Q3. What is strain? Does it have a unit? Strain is the ratio of change in dimension to the original dimension of the body. Strain has no unit because it is a ratio of two similar quantities.

Q4. State Hooke's law. For small deformations within the elastic limit, stress is directly proportional to strain. Stress = k x Strain, where k is the modulus of elasticity.

Q5. Why is the stretching of a coil spring determined by its shear modulus and not Young's modulus? When a coil spring is stretched, its length does not change significantly. The coil wire changes its shape under a twisting action. This change in shape is governed by the shear modulus.

Q6. Define breaking stress. Breaking stress is the minimum stress at which a wire breaks. It depends only on the material of the wire, not on its length or cross-sectional area.

Q7. What is compressibility? Compressibility is the reciprocal of bulk modulus. k = 1/B. It measures the fractional change in volume per unit increase in pressure.

Q8. What are ductile and brittle materials? Ductile materials show a large plastic range beyond the elastic limit. Brittle materials show a very small plastic range and fracture suddenly beyond the elastic limit. Example: glass and cast iron are brittle; copper and iron are ductile.

Stress and Strain Class 11 Questions: 2-Mark Answers

These stress and strain class 11 questions appear as 2-mark reasoning and distinction questions in CBSE 2026 board papers.

Q1. Distinguish between elastic and plastic behaviour of materials.

Q2. Define the three types of stress. Give one example of each. Tensile stress: restoring force per unit area when a body is stretched along its length. Example: pulling a wire from both ends. Shearing stress: restoring force per unit area when forces act parallel to the surface. Example: pushing a book horizontally with the lower face fixed. Hydraulic stress: restoring force per unit area when a body is compressed uniformly by a fluid. Example: a ball submerged in water under high pressure.

Q3. Why is steel more elastic than rubber? For the same stress, rubber shows far greater strain than steel. Young's modulus = Stress / Strain. Greater strain for the same stress gives a smaller Young's modulus. Steel has a much higher Young's modulus than rubber. The material requiring more force to produce the same deformation is more elastic. Therefore, steel is more elastic than rubber.

Q4. Explain why the elastic limit is more important than breaking stress in engineering design. A structure must return to its original shape after every load. If stress exceeds the elastic limit, permanent deformation occurs even if the structure does not break. Engineering design keeps all stresses below the elastic limit. Safety margin is calculated from yield strength, not from fracture stress.

Q5. A hollow shaft is stronger than a solid shaft of the same material and mass. Why? The torque required to produce a given twist is greater for a hollow cylinder than for a solid cylinder of the same mass and material. A hollow shaft has a larger outer radius. Shear stress distributes over a larger effective area. Therefore, a hollow shaft resists torsion more effectively.

Hooke's Law Class 11 Important Questions

These Hooke's law class 11 important questions test the limits of the law and the relationship between modulus of elasticity and the stress-strain curve.

Q1. Hooke's law is valid only in the linear part of the stress-strain curve. Explain. Hooke's law states that stress is proportional to strain. This proportionality holds only from O to A on the stress-strain curve. Beyond point A (proportional limit), the curve is no longer linear. Stress and strain are no longer proportional. Hooke's law breaks down in that region.

Q2. What is modulus of elasticity? Write its three types. Modulus of elasticity is the ratio of stress to the corresponding strain within the elastic limit. E = Stress / Strain. The three types are: Young's modulus (Y) for longitudinal stress and strain, Shear modulus (G) for shearing stress and strain, and Bulk modulus (B) for hydraulic stress and volume strain.

Q3. An elastic wire is cut to half its original length. How does this affect the maximum load the wire can support? Maximum load = Breaking stress x Cross-sectional area. Cutting the wire to half its length does not change its area of cross-section. Breaking stress depends only on the material of the wire, not on its length. Therefore, cutting the wire to half its length has no effect on the maximum load it can support.

Stress Strain Curve Class 11 Questions

The stress-strain curve is the single highest-value diagram in Chapter 8 of Class 11 Physics. CBSE 2026 examiners ask questions on curve interpretation, comparison of ductile and brittle materials, and graph-based numerical reading.

Students who memorise all five key points and understand what each represents score full marks on every stress strain curve class 11 question. These are among the most searched mechanical properties of solids important questions class 11.

Q1. Describe the stress-strain curve for a ductile metal. Label all key points.

The stress-strain curve for a ductile metal has five key regions.

Region O to A (Proportional limit): Stress is directly proportional to strain. The curve is a straight line. Hooke's law is fully obeyed. The body regains its original shape when the load is removed.

Region A to B (Elastic limit / Yield point): Stress and strain are no longer proportional but the body still returns to its original size on unloading. Point B is the yield point. The stress at B is the yield strength (sigma_y).

Region B to D (Plastic deformation): Beyond B, strain increases rapidly with small increase in stress. The body does not return to original dimensions on unloading. A permanent set remains.

Point D (Ultimate tensile strength): Maximum stress the material can withstand. Beyond D, strain increases even with reduced stress.

Point E (Fracture point): The wire breaks here. For ductile materials, D and E are far apart. For brittle materials, D and E are very close.

Q2. Two materials A and B have stress-strain curves drawn to the same scale. For a given strain, material A has higher stress than B.

(a) Which material has greater Young's modulus? Material A. Young's modulus = Stress / Strain. For the same strain, A has higher stress. Therefore A has greater Young's modulus. The steeper the slope of the linear region, the greater the Young's modulus.

(b) Which material is stronger? Material A. Strength corresponds to the stress at the fracture point. Material A withstands greater stress before fracture.

(c) Which material is more ductile? The material with a larger plastic region is more ductile. If A has a longer stretch between yield point and fracture point, A is more ductile.

Q3. How does rubber differ from a metal on a stress-strain graph? What is an elastomer? Rubber can stretch to several times its original length and still return to its original shape. Its stress-strain curve is not a straight line. It does not obey Hooke's law over most of its elastic region. Substances like rubber that sustain large elastic strains without obeying Hooke's law are called elastomers.

Young Modulus Shear Modulus Bulk Modulus Questions

Young modulus shear modulus bulk modulus questions appear in both short answer and numerical formats in CBSE 2026 papers. Young's modulus applies only to solids. Shear modulus applies only to solids. Bulk modulus applies to solids, liquids, and gases.

Students who know which modulus applies to which situation answer application questions and assertion-reason questions correctly in CBSE 2026 exams.

Young's Modulus Questions Class 11

Q1. Define Young's modulus. Write its formula and unit. Young's modulus is the ratio of tensile or compressive stress to the longitudinal strain, within the elastic limit. Y = (F/A) / (DeltaL/L) = FL / A(DeltaL). SI unit: N m-2 or Pascal (Pa).

Q2. A metal is more elastic if it has a larger Young's modulus. Explain. A larger Young's modulus means more stress is needed to produce the same strain. The material resists deformation more strongly. In physics, the material that stretches less for the same force is more elastic.

Q3. Which material requires a larger force to produce the same elongation in a wire of the same dimensions: steel or copper? Steel requires a larger force. Young's modulus of steel (2.0 x 10^11 N m-2) is greater than that of copper (1.1 x 10^11 N m-2). For the same geometry and elongation, F = YA(DeltaL/L). Greater Y means greater force. Steel is preferred in heavy-duty structures for this reason.

Shear Modulus Questions Class 11 Physics

Q1. Define shear modulus. Write its formula and state which phases of matter it applies to. Shear modulus (G), also called modulus of rigidity, is the ratio of shearing stress to the corresponding shearing strain. G = FL / A(Deltax) = F / (A x theta). SI unit: N m-2 or Pa. Shear modulus applies only to solids. Liquids and gases cannot sustain shearing stress.

Q2. A square lead slab (side 50 cm, thickness 10 cm) is subjected to a shearing force of 9.0 x 10^4 N on its narrow face. Find the displacement of the upper face. (G for lead = 5.6 x 10^9 N m-2) Area = 0.5 x 0.1 = 0.05 m2. Shearing stress = F/A = 9.0 x 10^4 / 0.05 = 1.8 x 10^6 N m-2. Shearing strain = Stress / G = 1.8 x 10^6 / 5.6 x 10^9 = 3.21 x 10^-4. Displacement Deltax = strain x L = 3.21 x 10^-4 x 0.5 = 1.6 x 10^-4 m = 0.16 mm.

Bulk Modulus Questions Class 11 Physics

Q1. Define bulk modulus. Why does the formula have a negative sign? Bulk modulus (B) is the ratio of hydraulic stress to the corresponding volume strain. B = -p / (DeltaV/V). The negative sign indicates that an increase in pressure produces a decrease in volume. The bulk modulus is always positive. SI unit: N m-2 or Pa.

Q2. Compare the compressibility of solids, liquids, and gases. Explain the reason. Gases are about a million times more compressible than solids. Bulk moduli for solids are far larger than for liquids, which are larger than for gases. Solids are least compressible because neighbouring atoms are tightly coupled. Gas molecules are very poorly coupled to their neighbours and compress easily.

Q3. The spherical ball contracts in volume by 0.1% under uniform pressure of 100 atmospheres. Find bulk modulus. Volume strain = DeltaV/V = 0.1/100 = 10^-3. Normal stress = 100 x 1.013 x 10^5 = 1.013 x 10^7 Pa. B = Stress / Volume strain = 1.013 x 10^7 / 10^-3 = 1.013 x 10^10 N m-2.

Questions on Poisson's Ratio and Elastic Potential Energy

Poisson's ratio and elastic potential energy are consistently tested from Chapter 8. Poisson's ratio appears as a 1-mark definition or 2-mark reasoning question. The elastic potential energy derivation appears as a 3 to 5-mark question in CBSE 2026 papers. Students who know the derivation score full marks.

Poisson's Ratio Questions Class 11 Physics

Q1. Define Poisson's ratio. Write its formula and state its limits. Poisson's ratio is the ratio of lateral strain to longitudinal strain when a wire is under tensile stress. sigma = (Deltad/d) / (DeltaL/L). Poisson's ratio has no unit. Theoretical limits: -1 to 0.5. Practical limits: 0 to 0.5. For steel, sigma lies between 0.28 and 0.30.

Elastic Potential Energy Important Questions Class 11

Q1. Derive the expression for elastic potential energy stored in a stretched wire. Let a wire of original length L and cross-section A be stretched by elongation DeltaL by force F.

From Young's modulus: F = YA(l/L) where l is elongation at any instant.

For further elongation dl, work done: dW = F dl = YAl dl / L.

Total work from l = 0 to l = DeltaL: W = YA(DeltaL)^2 / 2L. This work stores as elastic potential energy U.

Elastic PE per unit volume: u = U / (AL) = (1/2) x Y x (DeltaL/L)^2 = (1/2) x stress x strain.

Q2. Calculate elastic potential energy stored per unit volume of a steel wire when stress is 2 x 10^8 N m-2 and Y_steel = 2 x 10^11 N m-2. Strain = Stress / Y = 2 x 10^8 / 2 x 10^11 = 10^-3. u = (1/2) x 2 x 10^8 x 10^-3 = 1 x 10^5 J m^-3.

Mechanical Properties of Solids Class 11 Numericals

These mechanical properties of solids class 11 numericals require three skills: identifying the correct modulus for the situation, substituting with correct SI units, and interpreting the physical result. These are the highest-priority class 11 mechanical properties of solids question answer items for CBSE 2026.

Always write the formula first. Identify which modulus applies: Y for elongation or compression, G for shear or torsion, B for uniform compression. Substitute values in SI units and check if the answer makes physical sense before writing the final answer.

Young's Modulus Numericals Class 11 Chapter 8

Q1 (NCERT 8.1). A steel wire (length 4.7 m, area 3.0 x 10^-5 m^2) stretches by the same amount as a copper wire (length 3.5 m, area 4.0 x 10^-5 m^2) under the same load. Find ratio of Young's modulus: steel to copper. Y = FL / A(DeltaL). DeltaL and F are the same for both. Y_steel / Y_copper = (L_s x A_c) / (A_s x L_c) = (4.7 x 4.0 x 10^-5) / (3.0 x 10^-5 x 3.5) = 18.8 / 10.5 = 1.79 ≈ 1.8.

Q2 (NCERT 8.2). From the stress-strain graph, stress at strain 0.003 is approximately 150 x 10^6 N m-2. Find Young's modulus and approximate yield strength. Y = Stress / Strain = 150 x 10^6 / 0.003 = 5 x 10^10 N m-2. Yield strength from graph: approximately 3 x 10^8 N m-2 (300 MPa).

Q3 (NCERT 8.5). Two wires (diameter 0.25 cm), steel (1.5 m) and brass (1.0 m). Total load: steel carries 4+6 kg, brass carries 6 kg. Find elongations. (Y_steel = 2.0 x 10^11, Y_brass = 0.91 x 10^11 N m-2) r = 1.25 x 10^-3 m; A = pi(1.25 x 10^-3)^2 = 4.91 x 10^-6 m^2. Load on steel = 98 N; Load on brass = 58.8 N. DeltaL_steel = (98 x 1.5) / (2.0 x 10^11 x 4.91 x 10^-6) = 1.49 x 10^-4 m ≈ 0.149 mm. DeltaL_brass = (58.8 x 1.0) / (0.91 x 10^11 x 4.91 x 10^-6) = 1.32 x 10^-4 m ≈ 0.132 mm.

Q4 (NCERT 8.6). Aluminium cube (10 cm edge) fixed on one face. Mass 100 kg attached to opposite face. Find vertical deflection. (G_Al = 25 GPa) F = 100 x 9.8 = 980 N; A = (0.1)^2 = 0.01 m^2. Shearing stress = 980 / 0.01 = 9.8 x 10^4 N m-2. Shearing strain = 9.8 x 10^4 / 25 x 10^9 = 3.92 x 10^-6. Deltax = 3.92 x 10^-6 x 0.1 = 3.92 x 10^-7 m.

Q5 (NCERT 8.7). Four hollow steel columns (inner radius 30 cm, outer radius 60 cm) support 50,000 kg. Find compressional strain of each. (Y = 2.0 x 10^11 N m-2) Total force = 4.9 x 10^5 N; Per column = 1.225 x 10^5 N. A = pi[(0.6)^2 - (0.3)^2] = pi x 0.27 = 0.848 m^2. Stress = 1.225 x 10^5 / 0.848 = 1.445 x 10^5 N m-2. Strain = 1.445 x 10^5 / 2.0 x 10^11 = 7.22 x 10^-7.

Q6 (NCERT 8.8). Copper piece (15.2 mm x 19.1 mm) pulled with 44,500 N. Find resulting strain. (Y_copper = 1.1 x 10^11 N m-2) A = 15.2 x 10^-3 x 19.1 x 10^-3 = 2.9 x 10^-4 m^2. Stress = 44,500 / 2.9 x 10^-4 = 1.534 x 10^8 N m-2. Strain = 1.534 x 10^8 / 1.1 x 10^11 = 1.39 x 10^-3.

Q7 (NCERT 8.9). Steel cable (radius 1.5 cm) supports a chairlift. Maximum stress = 10^8 N m-2. Find maximum load. A = pi(1.5 x 10^-2)^2 = 7.07 x 10^-4 m^2. Maximum load = 10^8 x 7.07 x 10^-4 = 7.07 x 10^4 N.

Q8 (NCERT 8.10). Rigid bar supported by three wires (copper at ends, iron in middle, same length, same tension). Find ratio of diameters: copper to iron. (Y_iron = 1.9 x 10^11, Y_copper = 1.1 x 10^11 N m-2) Same F, same L → same strain. A is proportional to Y, so d is proportional to sqrt(Y). d_copper / d_iron = sqrt(Y_copper / Y_iron) = sqrt(1.1/1.9) = 0.76.

Q9 (NCERT 8.11). Mass 14.5 kg whirled in vertical circle (r = 1.0 m, omega = 2 rev/s) at bottom. Wire area = 0.065 cm^2. Find elongation. (Y_steel = 2 x 10^11 N m-2) omega = 4pi rad/s. Total force = 14.5(9.8 + 16pi^2) = 14.5(9.8 + 157.9) = 14.5 x 167.7 = 2431.7 N. DeltaL = FL / YA = (2431.7 x 1) / (2 x 10^11 x 0.065 x 10^-4) = 1.87 x 10^-3 m ≈ 1.87 mm.

Bulk Modulus Numericals Class 11

Q10 (NCERT 8.12). Initial volume = 100 L, Final volume = 100.5 L, Pressure increase = 100 atm. Find bulk modulus of water. Compare with air. DeltaV = 0.5 x 10^-3 m^3; V = 0.1 m^3; Volume strain = 5 x 10^-3. Deltap = 1.013 x 10^7 Pa. B_water = 1.013 x 10^7 / 5 x 10^-3 = 2.026 x 10^9 N m-2. B_water / B_air = 2.026 x 10^9 / 10^5 = approximately 20,000. Air is approximately 20,000 times more compressible than water.

Q11 (NCERT 8.13). Pressure at depth = 80 atm. Surface density of water = 1.03 x 10^3 kg m-3. Find density at depth. p = 8.104 x 10^6 Pa; B_water = 2.2 x 10^9 N m-2. DeltaV/V = p/B = 3.684 x 10^-3. rho' = rho / (1 - DeltaV/V) = 1.03 x 10^3 / (1 - 3.684 x 10^-3) = 1.034 x 10^3 kg m-3.

Q12 (NCERT 8.14). Glass slab under 10 atm hydraulic pressure. Find fractional change in volume. (B_glass = 37 x 10^9 Pa) p = 1.013 x 10^6 Pa. DeltaV/V = p/B = 1.013 x 10^6 / 37 x 10^9 = 2.74 x 10^-5.

Q13 (NCERT 8.15). Copper cube (10 cm edge) under pressure 7.0 x 10^6 Pa. Find volume contraction. (B_copper = 140 x 10^9 Pa) V = 10^-3 m^3; DeltaV/V = 7.0 x 10^6 / 140 x 10^9 = 5 x 10^-5. DeltaV = 5 x 10^-5 x 10^-3 = 5 x 10^-8 m^3.

Q14 (NCERT 8.16). Find pressure change to compress 1 litre of water by 0.10%. (B_water = 2.2 x 10^9 N m-2) DeltaV/V = 10^-3. Deltap = B x DeltaV/V = 2.2 x 10^9 x 10^-3 = 2.2 x 10^6 N m-2 = 2.2 MPa.

Q15. A structural steel rod (radius 10 mm, length 1.0 m) is stretched by 100 kN. Find stress, elongation, and strain. (Y = 2.0 x 10^11 N m-2) A = pi(10^-2)^2 = 3.14 x 10^-4 m^2. Stress = 10^5 / 3.14 x 10^-4 = 3.18 x 10^8 N m-2. DeltaL = (3.18 x 10^8 x 1) / 2.0 x 10^11 = 1.59 x 10^-3 m = 1.59 mm. Strain = DeltaL/L = 1.59 x 10^-3 = 0.16%.

Q16. A mild steel wire (length 1.0 m, area 5 x 10^-5 m^2) is stretched horizontally between two pillars. A 100 g mass is suspended from the midpoint. Find depression. (Y = 2 x 10^11 N m-2) m = 0.1 kg. Let x = depression at midpoint. Strain approximately equals 4x^2/L^2; Tension T = mgL/4x; Stress = T/A = mgL/4xA. Y = mgL^3/16Ax^3. x^3 = mgL^3 / 16AY = (0.1 x 9.8 x 1) / (16 x 5 x 10^-5 x 2 x 10^11) = 6.125 x 10^-9. x = (6.125 x 10^-9)^(1/3) ≈ 1.83 x 10^-3 m ≈ 1.83 mm.

Application-Based Questions on Elastic Behaviour of Materials

The applications section of Chapter 8 produces the most unique and high-scoring questions in CBSE 2026 exams. A student who can explain why railway tracks have an I-shaped cross-section or why mountains cannot grow beyond 10 km scores marks that most other students cannot.

Crane Rope and Beam Questions Class 11

Q1. A crane must lift a load of 10 metric tons. Find the minimum radius of a steel rope required. Why is a single wire not used? (sigma_y = 300 x 10^6 N m-2) Load = 10 x 10^3 x 9.8 = 9.8 x 10^4 N. A is greater than or equal to W/sigma_y = 9.8 x 10^4 / 300 x 10^6 = 3.27 x 10^-4 m^2. Radius = sqrt(A/pi) ≈ 1 cm (minimum). Safety factor of approximately 10 gives radius ≈ 3 cm.

A single wire of 3 cm radius would be rigid and inflexible. Crane ropes are braided from many thin wires for flexibility, ease of manufacture, and combined strength.

Q2. Why does a beam used in bridges have an I-shaped cross-section instead of a rectangular one? A beam sags by delta = Wl^3 / (4bd^3 x Y). Increasing depth d reduces sagging by d^-3. Increasing breadth b reduces it only by b^-1. But increasing depth alone causes buckling.

The I-shaped cross-section solves both problems. Wide flanges at top and bottom carry compressive and tensile forces. The vertical web prevents buckling. Weight reduces without reducing strength.

Q3. Why does a pillar with distributed ends support more load than one with rounded ends? A pillar with rounded ends concentrates stress at a small contact area. A pillar with flat or distributed ends spreads load over a larger area. Stress = Force / Area. Larger contact area means smaller stress for the same load.

Mountain Height and Rock Questions Class 11

Q1. Why is the maximum height of a mountain on Earth approximately 10 km? Explain using elastic properties. At the base of a mountain of height h, shear stress equals approximately h x rho x g.

Elastic limit for typical rock = 30 x 10^7 N m-2. rho = 3 x 10^3 kg m-3, g = 10 m s-2.

Setting h x rho x g = 30 x 10^7: h = 30 x 10^7 / (3 x 10^3 x 10) = 10^4 m = 10 km. Beyond this height, shear stress exceeds the elastic limit of rock. Rocks flow plastically. Mt. Everest at 8.8 km is close to this theoretical maximum.

Graph and Diagram Based Questions Class 11 Physics Chapter 8

Diagram questions in Chapter 8 appear in 2-mark and 3-mark formats. The stress-strain curve is the most tested, followed by the I-section beam cross-section and the three types of stress diagrams. Knowing which diagram carries marks and what to label saves time in the exam hall.

Types of Stress Diagram Questions Class 11

Q1. Draw diagrams for the three types of stress. Label the applied force and resulting deformation. Tensile stress: Two equal and opposite forces act perpendicular to the cross-sectional area. The body elongates by DeltaL. Label: force F, original length L, elongation DeltaL, area A.

Shearing stress: Two equal and opposite forces act parallel to opposite faces. The upper face displaces by Deltax. Label: force F, height L, displacement Deltax, angle theta.

Hydraulic stress: Pressure p acts perpendicular to every surface point of a submerged body. Volume decreases by DeltaV. Shape does not change. Label: pressure p, volume V, change DeltaV.

Beam Cross-Section and Pillar End Questions Class 11

Q1. Draw the three cross-sectional shapes of a beam from Chapter 8. State which is used in bridges and why. (a) Rectangular cross-section: simple to manufacture but prone to buckling when depth increases. (b) Thin deep bar: greater depth reduces sag but buckles under off-centre loads. (c) I-shaped cross-section: wide flanges at top and bottom carry tensile and compressive forces. Vertical web prevents buckling. Maximum load capacity at minimum material weight.

I-shaped beams are the standard choice for bridges and buildings.

Assertion Reason Questions Class 11 Physics Chapter 8

Assertion-reason questions are among the most important questions on mechanical properties of solids class 11 in CBSE 2026 papers. These test reasoning behind a concept, not just the concept itself. Read each statement independently before selecting the answer.

Directions: (a) Both A and R are true, and R is the correct explanation of A. (b) Both A and R are true, but R is not the correct explanation of A. (c) A is true, but R is false. (d) A is false, but R is true.

Q1. A: Steel is more elastic than rubber. R: For the same stress, strain in steel is much less than in rubber, giving steel a higher Young's modulus. (a) Higher Young's modulus means greater resistance to deformation per unit stress. The reason correctly explains the assertion.

Q2. A: Stretching of a coil spring is determined by its shear modulus. R: Stretching a coil causes torsion in the wire. Torsion changes shape without changing length, which shear modulus governs. (a) Stretching a coil involves twisting of the wire, not elongation. The reason correctly explains the assertion.

Q3. A: Bulk modulus applies to solids, liquids, and gases. Young's modulus and shear modulus apply only to solids. R: Only solids have definite shape and length. Liquids and gases cannot sustain shear stress or tensile stress. (a) Liquids and gases flow under shear. The reason correctly explains the assertion.

Q4. A: A hollow cylinder is stronger than a solid cylinder of the same material and equal mass. R: A hollow cylinder has a greater outer radius. Greater outer radius means greater resistance to bending and torsion. (a) Torque required to produce a given twist increases with outer radius. The reason correctly explains the assertion.

Q5. A: Young's modulus of rubber is less than that of steel. R: Rubber shows much larger strain than steel for the same applied stress. Young's modulus = Stress / Strain. Larger strain gives smaller modulus. (a) This follows directly from the definition of Young's modulus. The reason correctly explains the assertion.

Q6. A: Maximum height of a mountain on Earth is approximately 10 km. R: Beyond this height, shear stress at the mountain base exceeds the elastic limit of rock. Rock flows plastically. (a) Derived from h x rho x g = elastic limit of rock. The calculation gives h ≈ 10 km. The reason correctly explains the assertion.

Extra Questions Class 11 Physics Chapter 8

These mechanical properties of solids important questions class 11 go beyond direct recall. They test application, multi-step reasoning, and comparison.

Q1. Two wires A and B are made of steel. Wire A has twice the length and half the radius of wire B. If the same force is applied to both, compare their elongations. DeltaL = FL / (pi x r^2 x Y). DeltaL_A / DeltaL_B = (L_A / r_A^2) / (L_B / r_B^2) = (2L / (r/2)^2) / (L / r^2) = 2 x 4 = 8. Wire A elongates 8 times more than wire B.

Q2. The Marina Trench is approximately 11 km deep. Pressure at bottom ≈ 1.1 x 10^8 Pa. A steel ball (volume 0.32 m^3) falls to the bottom. Find change in volume. (B_steel = 1.6 x 10^11 Pa) DeltaV/V = p/B = 1.1 x 10^8 / 1.6 x 10^11 = 6.875 x 10^-4. DeltaV = 6.875 x 10^-4 x 0.32 = 2.2 x 10^-4 m^3.

Q3. A diamond anvil (flat face diameter 0.50 mm) is compressed with 50,000 N. Find pressure at the tip. r = 2.5 x 10^-4 m; A = pi(2.5 x 10^-4)^2 = 1.96 x 10^-7 m^2. Pressure = F/A = 50,000 / 1.96 x 10^-7 = 2.55 x 10^11 Pa.

Q4. A rigid bar is supported by three wires (copper at ends, iron in middle, same length, same tension). Find ratio of diameters: copper to iron. (Y_iron = 1.9 x 10^11, Y_copper = 1.1 x 10^11 N m-2) Same F and same strain → A is proportional to Y → d is proportional to sqrt(Y). d_copper / d_iron = sqrt(1.1/1.9) = 0.76.

Important Formulas Class 11 Physics Chapter 8