Mechanical Properties of Solids – Class 11 Physics Notes, Formulas, MCQs & Numericals

NCERT Chapter 8 made simple: stress and strain, Hooke's law, the stress–strain curve, Young's modulus, shear modulus, bulk modulus, Poisson's ratio and elastic potential energy, with 8 diagrams, a calculator, 12 solved numericals, 20 MCQs, true/false and exam-style questions.

Physics 07 October, 2026 30 min read

1. Introduction: Elasticity and Plasticity

In the chapter on rotation we treated bodies as rigid, meaning a hard solid object with a definite shape and size. In reality no body is perfectly rigid: even a steel bar bends, stretches or compresses when a large enough force acts on it. The mechanical properties of solids tell us how much, and in what way, real solids deform.

Elasticity

The property by which a body regains its original size and shape when the deforming force is removed. A steel spring is a good example, and the deformation is called elastic deformation.

Plasticity

The property by which a body stays permanently deformed after the force is removed. Putty and mud are close to ideal plastics, and the deformation is plastic deformation.

Engineers use these ideas to design buildings, bridges, cranes, cars and aeroplanes. They answer questions such as: why does a railway track have an I-shape, why is glass brittle while brass is not, and how light can an aeroplane be and still be strong? This chapter is Chapter 8 in the current NCERT Class 11 Physics book (it was Chapter 9 in older editions), and it builds on Hooke's law and Young's modulus.

2. Stress and Strain (Types, Formulas, Units)

When forces act on a body that stays in static equilibrium, the body deforms a little. It develops an internal restoring force equal in magnitude and opposite in direction to the applied force. The restoring force per unit area is the stress.

Stress

Stress σ = F / A

F = force applied normal to the cross-section (N) | A = area of cross-section (m²) | SI unit: N m⁻² = pascal (Pa) | Dimensions: [M L⁻¹ T⁻²]

Three Types of Stress and Strain

Figure 1: (a) A cylinder stretched by equal forces normal to its faces. (b) A block under a tangential (shearing) force. (c) A solid sphere compressed uniformly by a fluid. Dashed outlines show the original shape; deformations are exaggerated.

Type of stress	How it acts	Strain produced	Formula
Tensile / compressive (longitudinal)	Equal, opposite forces normal to the cross-section	Longitudinal strain (change in length)	ΔL / L
Shearing (tangential)	Equal, opposite forces parallel to opposite faces	Shearing strain (change in shape)	Δx / L = tan θ ≈ θ
Hydraulic (volume)	Fluid pressure acting normally at every point of the surface	Volume strain (change in volume only)	ΔV / V

Strain Has No Unit

Strain is the ratio of a change in dimension to the original dimension, so it has no unit and no dimensional formula. Also, stress is not a vector: unlike force, it cannot be given one specific direction. And in a wire hanging from a ceiling, the tension at any cross-section is F, not 2F, so the tensile stress is F/A.

Hindi Terms for Revision (हिंदी शब्दावली)

Stress = प्रतिबल · Strain = विकृति · Elasticity = प्रत्यास्थता · Plasticity = सुघट्यता · Hooke's law = हुक का नियम · Young's modulus = यंग गुणांक · Bulk modulus = आयतन प्रत्यास्थता गुणांक · Shear modulus (modulus of rigidity) = अपरूपण / दृढ़ता गुणांक · Poisson's ratio = प्वासों अनुपात

3. Hooke's Law

For small deformations, stress and strain are proportional to each other. This is Hooke's law:

Hooke's Law

Stress = k × Strain

k = modulus of elasticity (a property of the material) | valid only in the linear region of the stress–strain curve

Hooke's law is an empirical law (found by experiment) and holds for most materials, but some materials, such as rubber and aortic tissue, do not show this linear relationship. For the spring version F = −kx, the force–extension graph and the experiment are explained in our detailed Hooke's law guide.

4. Stress–Strain Curve

To get the stress–strain curve, a test wire or cylinder is stretched in small steps. The applied force (stress) and the fractional change in length (strain) are recorded at each step and plotted. The shape of the curve differs from material to material and shows how a material deforms as the load increases.

Figure 2: O–A is linear (Hooke's law, slope = Young's modulus). B is the yield point with yield strength σy. D is the ultimate tensile strength σu and E is the fracture point. Unloading from C leaves a permanent set.

Reading the Curve Region by Region

Region	What happens	On removing the load
O → A	Linear. Stress ∝ strain. Hooke's law is obeyed.	Body regains original dimensions (elastic).
A → B	Stress and strain are not proportional.	Body still returns to original size.
B (yield point)	Elastic limit. Stress here = yield strength σy.	Beyond B, deformation is no longer fully recoverable.
B → D	Strain increases rapidly for a small increase in stress (plastic deformation).	At C, a permanent set remains even at zero stress.
D	Maximum stress = ultimate tensile strength σu.	—
D → E	Strain grows even though the force falls. Fracture occurs at E.	Wire breaks.

Ductile vs Brittle

If the ultimate strength point D and the fracture point E are far apart, the material is ductile (copper, mild steel: they can be drawn into wires). If they are close, the material is brittle (glass, cast iron: they snap with almost no plastic stretching).

Figure 3: Typical stress–strain behaviour of a brittle solid, a ductile metal and an elastomer. For an elastomer the loading and unloading paths differ slightly, and there is no well-defined plastic region.

Elastomers such as rubber and the elastic tissue of the aorta (the large blood vessel leaving the heart) can be stretched to several times their length and still return to their shape. Their elastic region is very large but does not obey Hooke's law over most of it.

5. Elastic Moduli: Young's, Shear and Bulk Modulus

The ratio of stress to strain within the elastic limit (region OA) is a characteristic of the material, called its modulus of elasticity. It is the region of greatest importance in structural and manufacturing design. There are three moduli, one for each kind of stress.

5.1 Young's Modulus (Y)

For a given material the strain is the same whether the stress is tensile or compressive. The ratio of tensile (or compressive) stress σ to longitudinal strain ε is Young's modulus.

Young's Modulus

Y = σ / ε = (F/A) / (ΔL/L) = F L / (A ΔL)

SI unit: N m⁻² (Pa) | Dimensions: [M L⁻¹ T⁻²] | Because strain has no unit, Y has the same unit as stress

Figure 4: Metals have large Young's moduli, so a large force produces only a small change in length. Wood, bone and glass have comparatively small values. Data as in NCERT Table 8.1 (1 GPa = 10⁹ N m⁻²).

To stretch a thin steel wire of cross-section 0.1 cm² by 0.1 %, a force of 2000 N is needed. For aluminium, brass and copper wires of the same area, the forces are 690 N, 900 N and 1100 N. So steel is more elastic than copper, brass and aluminium, which is why it is preferred in heavy-duty machines and structural design.

Material	Young's modulus Y (10¹¹ N m⁻²)	Young's modulus (GPa)
Tungsten	3.6	360
Steel	2.0	200
Iron (wrought)	1.9	190
Copper	1.1	110
Brass	0.91	91
Aluminium	0.70	70
Glass	0.65	65
Lead	0.16	16
Bone	0.094	9.4

Wire Extension and Young's Modulus Calculator

Enter the load, original length and diameter of a wire and pick a material. The calculator gives stress, strain, extension and the elastic energy stored.

Material

Young's modulus Y (GPa)

Load F (N)

Original length L (m)

Diameter d (mm)

Press Calculate to see the results.

5.2 Shear Modulus (Modulus of Rigidity, G)

The ratio of shearing stress to the corresponding shearing strain is the shear modulus G, also called the modulus of rigidity. It relates to a change in shape at constant volume and exists only for solids.

Shear Modulus

G = (F/A) / (Δx/L) = F L / (A Δx) = F / (A θ)

Shearing stress σs = G × θ | SI unit: N m⁻² (Pa) | For most materials G ≈ Y / 3

Material	G (GPa)	Material	G (GPa)
Aluminium	25	Lead	5.6
Brass	36	Nickel	77
Copper	42	Steel	84
Glass	23	Tungsten	150
Iron	70	Wood	10

5.3 Bulk Modulus (B) and Compressibility

When a body is submerged in a fluid, it feels a hydraulic stress equal to the fluid pressure. The volume decreases, giving a volume strain. The ratio of hydraulic stress to volume strain is the bulk modulus B.

Bulk Modulus and Compressibility

B = − p / (ΔV / V) | k = 1 / B

The negative sign means volume decreases (ΔV < 0) when pressure increases (p > 0), so B is always positive | Unit: N m⁻² (Pa)

Figure 5: Solids are the least compressible because neighbouring atoms are tightly coupled; liquids are less tightly bound; gas molecules are very poorly coupled to each other.

Solids	B (GPa)	Liquids	B (GPa)
Aluminium	72	Water	2.2
Brass	61	Ethanol	0.9
Copper	140	Carbon disulphide	1.56
Glass	37	Glycerine	4.76
Iron	100	Mercury	25
Nickel	260	Gas: Air (STP)	1.0 × 10⁻⁴
Steel	160

5.4 Comparison of the Three Elastic Moduli

Type of stress	Stress	Strain	Change in shape	Change in volume	Modulus	State of matter
Tensile / compressive	Equal, opposite forces normal to opposite faces (σ = F/A)	ΔL/L (longitudinal)	Yes	No	Y = FL/(AΔL) Young's modulus	Solid
Shearing	Equal, opposite forces parallel to opposite surfaces (σs = F/A)	Pure shear, θ	Yes	No	G = F/(Aθ) Shear modulus	Solid
Hydraulic	Pressure acting normally and equally everywhere	ΔV/V (volume)	No	Yes	B = −p/(ΔV/V) Bulk modulus	Solid, liquid, gas

6. Poisson's Ratio

A stretched wire also becomes thinner. The strain perpendicular to the applied force is the lateral strain. Simon Poisson showed that, within the elastic limit, lateral strain is directly proportional to longitudinal strain. Their ratio is Poisson's ratio.

Poisson's Ratio

σ = (Δd / d) / (ΔL / L) = (Δd / ΔL) × (L / d)

d = original diameter, Δd = contraction in diameter | L = original length, ΔL = elongation | A pure number with no unit

Figure 6: Stretching a wire lengthens it by ΔL (longitudinal strain) and reduces its diameter by Δd (lateral strain). Poisson's ratio depends only on the material.

7. Elastic Potential Energy in a Stretched Wire

When a wire is stretched, work is done against the inter-atomic forces. This work is stored in the wire as elastic potential energy. For a wire of length L and area A stretched by l, F = YA(l/L), and the work for a small extra stretch dl is dW = F dl. Integrating from 0 to l:

Elastic Potential Energy

W = ½ × stress × strain × volume | u = ½ σ ε = ½ Y ε²

u = energy per unit volume (J m⁻³) | Total energy U = ½ F ΔL

Figure 7: Because stress rises linearly with strain, the shaded triangle (½ × base × height) gives the energy stored per unit volume.

8. Applications of Elastic Behaviour of Materials

Every engineering design needs precise knowledge of how materials behave elastically. Three classic applications appear in the NCERT text and in exams.

8.1 Thickness of a Crane Rope

To lift 10 tonnes without permanent deformation the rope must stay below the yield strength. For mild steel σy ≈ 300 × 10⁶ N m⁻², so the area must satisfy A ≥ W/σy = Mg/σy ≈ 3.3 × 10⁻⁴ m², a radius of about 1 cm. A safety factor of about ten gives a radius near 3 cm. Because a single wire of this size would be practically rigid, the rope is made of many thin wires braided together.

8.2 Bending of Beams and the I-Shaped Girder

A bar of length l, breadth b and depth d loaded at its centre by W sags by:

Sag of a Loaded Beam

δ = W l³ / (4 b d³ Y)

To reduce bending: choose a large Y, a small span l, and increase the depth d (δ ∝ d⁻³) rather than breadth b (δ ∝ b⁻¹)

Figure 8: (a) A beam supported near its ends sags by δ. (b) The I-section gives a wide load-bearing surface and large depth while using less material. (c) A very deep, thin bar may buckle sideways.

Increasing the depth reduces sag strongly, but a deep thin bar can buckle. The I-shaped cross-section is the common compromise: large load-bearing surface, enough depth to prevent bending, and lower weight and cost. Similarly, a pillar with distributed (wider) ends supports more load than one with rounded ends.

8.3 Why Mountains Are at Most About 10 km High

At the base of a mountain of height h the stress due to its weight is about hρg. This acts vertically with the sides free, so it is not a uniform bulk compression and contains a shear component of about hρg. Rocks flow when this exceeds their elastic limit (≈ 30 × 10⁷ N m⁻²). Putting ρ = 3 × 10³ kg m⁻³: h = 30 × 10⁷ / (3 × 10³ × 10) = 10 km, more than the height of Mt. Everest.

9. Formula Sheet, Points to Ponder and Common Mistakes

Quantity	Formula	SI unit
Stress	σ = F / A	N m⁻² (Pa)
Longitudinal strain	ΔL / L	none
Shearing strain	Δx / L = tan θ ≈ θ	none
Volume strain	ΔV / V	none
Hooke's law	stress = k × strain	—
Young's modulus	Y = FL / (AΔL)	N m⁻²
Shear modulus	G = F / (Aθ) = FL / (AΔx)	N m⁻²
Bulk modulus	B = −p / (ΔV/V)	N m⁻²
Compressibility	k = 1 / B	N⁻¹ m²
Poisson's ratio	σ = (Δd/d) / (ΔL/L)	none
Energy per unit volume	u = ½ σ ε = ½ Y ε²	J m⁻³
Energy stored in wire	U = ½ F ΔL	J
Minimum rope area	A ≥ Mg / σy	m²
Beam sag	δ = W l³ / (4 b d³ Y)	m

Points to Ponder

Hooke's law is valid only in the linear part of the stress–strain curve.
Young's modulus and shear modulus are relevant only for solids; bulk modulus applies to solids, liquids and gases.
A material that stretches less for a given load is more elastic. Steel is more elastic than rubber.
Metals generally have larger Young's moduli than alloys and elastomers.
A deforming force in one direction can produce strain in other directions too, which is described by Poisson's ratio.
Stress is not a vector; tension in a hanging wire is F, not 2F.

Common Exam Mistakes

Writing a unit for strain or Poisson's ratio. Both are pure numbers.
Using diameter instead of radius in A = πr². Check the question carefully, or use A = πd²/4.
Forgetting to convert GPa to Pa (× 10⁹) and cm² to m² (× 10⁻⁴) before substituting.
Calling rubber "more elastic than steel" because it stretches more.
Applying Hooke's law beyond the proportional limit.

10. Lab Experiments on Mechanical Properties of Solids

Practical work cements the ideas above. Try these experiments (demo links will be updated):

Young's Modulus

Determination of Young's modulus of a wire using Searle's apparatus

Load a wire in equal steps, measure the extension and find Y = MgL / (πr²l).

Hooke's Law

Verification of Hooke's law using a helical spring

Plot load against extension to confirm F ∝ x and find the spring constant.

Virtual Lab

Virtual simulation: stress–strain curve of a metal wire

Increase the load and watch the elastic, yield, plastic and fracture regions appear.

11. Solved Numericals (NCERT Examples and Exercises)

Work through each problem on paper first, then open the solution. Always convert to SI units before substituting.

1Stress, elongation and strain in a steel rod (NCERT Example 8.1)

A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate (a) stress, (b) elongation, (c) strain. Y of structural steel = 2.0 × 10¹¹ N m⁻².

Show step-by-step solution

(a) Stress = F/A = F/(πr²) = (100 × 10³)/(3.14 × (10 × 10⁻³)²) = 3.18 × 10⁸ N m⁻²

(b) ΔL = (F/A)L/Y = (3.18 × 10⁸ × 1)/(2 × 10¹¹) = 1.59 × 10⁻³ m = 1.59 mm

(c) Strain = ΔL/L = 1.59 × 10⁻³ ≈ 0.16 %

2Ratio of Young's moduli (NCERT Exercise 8.1)

A steel wire of length 4.7 m and cross-sectional area 3.0 × 10⁻⁵ m² stretches by the same amount as a copper wire of length 3.5 m and area 4.0 × 10⁻⁵ m² under a given load. Find Y(steel) : Y(copper).

Show step-by-step solution

Y = FL/(AΔL). F and ΔL are equal, so Y ∝ L/A.

Y_s/Y_c = (L_s/A_s) × (A_c/L_c) = (4.7/3.0 × 10⁻⁵) × (4.0 × 10⁻⁵/3.5) = 1.79 ≈ 1.8

3Two wires joined end to end (NCERT Example 8.2)

A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are joined end to end. When stretched by a load the net elongation is 0.70 mm. Find the load. (Y_c = 1.1 × 10¹¹, Y_s = 2.0 × 10¹¹ N m⁻²)

Show step-by-step solution

Both wires carry the same tension W and have the same A, so W/A = Y_c(ΔL_c/L_c) = Y_s(ΔL_s/L_s).

ΔL_c/ΔL_s = (Y_s/Y_c)(L_c/L_s) = (2.0/1.1)(2.2/1.6) = 2.5

With ΔL_c + ΔL_s = 7.0 × 10⁻⁴ m: ΔL_c = 5.0 × 10⁻⁴ m and ΔL_s = 2.0 × 10⁻⁴ m.

W = A·Y_c·ΔL_c/L_c = π(1.5 × 10⁻³)² × (5.0 × 10⁻⁴ × 1.1 × 10¹¹)/2.2 = 1.8 × 10² N

4Shear displacement of a lead slab (NCERT Example 8.4)

A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force of 9.0 × 10⁴ N on its narrow face. The lower edge is riveted to the floor. How much will the upper edge be displaced? (G for lead = 5.6 × 10⁹ N m⁻²)

Show step-by-step solution

Area of the face = 0.5 m × 0.1 m = 0.05 m². Shear stress = 9.0 × 10⁴/0.05 = 1.8 × 10⁶ N m⁻².

Shear strain Δx/L = stress/G, so Δx = (1.8 × 10⁶ × 0.5)/(5.6 × 10⁹) = 1.6 × 10⁻⁴ m = 0.16 mm

5Compression of ocean water (NCERT Example 8.5)

The average depth of the Indian Ocean is about 3000 m. Find the fractional compression ΔV/V of water at the bottom. B(water) = 2.2 × 10⁹ N m⁻², g = 10 m s⁻².

Show step-by-step solution

p = hρg = 3000 × 1000 × 10 = 3 × 10⁷ N m⁻²

ΔV/V = p/B = (3 × 10⁷)/(2.2 × 10⁹) = 1.36 × 10⁻² = 1.36 %

6Thighbone under a human pyramid (NCERT Example 8.3)

In a circus pyramid the performer at the bottom supports 220 kg. Each thighbone is 50 cm long with an effective radius of 2.0 cm. Find the compression of each thighbone. Y(bone) = 9.4 × 10⁹ N m⁻².

Show step-by-step solution

Weight = 220 × 9.8 = 2156 N; each bone carries 1078 N. A = π(2 × 10⁻²)² = 1.26 × 10⁻³ m².

ΔL = FL/(YA) = (1078 × 0.5)/(9.4 × 10⁹ × 1.26 × 10⁻³) = 4.55 × 10⁻⁵ m (fractional change 0.0091 %).

7Maximum load on a chairlift cable (NCERT Exercise 8.9)

A steel cable of radius 1.5 cm supports a chairlift. If the maximum stress must not exceed 10⁸ N m⁻², what is the maximum load?

Show step-by-step solution

A = πr² = 3.14 × (1.5 × 10⁻²)² = 7.07 × 10⁻⁴ m²

F_max = stress × A = 10⁸ × 7.07 × 10⁻⁴ = 7.07 × 10⁴ N ≈ 7.1 × 10⁴ N

8Wire whirled in a vertical circle (NCERT Exercise 8.11)

A 14.5 kg mass is fastened to a steel wire of unstretched length 1.0 m and whirled in a vertical circle at 2 rev/s. Cross-section of the wire = 0.065 cm². Find the elongation at the lowest point. Y(steel) = 2.0 × 10¹¹ N m⁻².

Show step-by-step solution

ω = 2 × 2π = 4π = 12.57 rad s⁻¹. At the lowest point, tension T = mg + mω²L = 14.5 × 9.8 + 14.5 × (12.57)² × 1 = 142.1 + 2290 ≈ 2432 N.

ΔL = TL/(YA) = (2432 × 1)/(2.0 × 10¹¹ × 6.5 × 10⁻⁶) = 1.87 × 10⁻³ m

9Bulk modulus of water (NCERT Exercise 8.12)

Initial volume of water = 100.0 litre, pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), final volume = 100.5 litre. Compute the bulk modulus and compare it with that of air.

Show step-by-step solution

Pressure change p = 100 × 1.013 × 10⁵ = 1.013 × 10⁷ Pa. Volume strain = 0.5/100 = 5 × 10⁻³ (in magnitude).

B = p/(ΔV/V) = (1.013 × 10⁷)/(5 × 10⁻³) = 2.03 × 10⁹ Pa

B(air) = 1.0 × 10⁵ Pa, so B(water)/B(air) ≈ 2 × 10⁴. Water molecules are tightly coupled to their neighbours, whereas gas molecules are almost free, so water is about twenty thousand times harder to compress.

10Energy stored in a stretched wire

A steel wire of length 2 m and cross-sectional area 1 mm² carries a load of 100 N. Find (a) the elongation and (b) the elastic potential energy stored. Y = 2.0 × 10¹¹ N m⁻².

Show step-by-step solution

(a) ΔL = FL/(AY) = (100 × 2)/(10⁻⁶ × 2 × 10¹¹) = 1.0 × 10⁻³ m = 1 mm

(b) U = ½ F ΔL = ½ × 100 × 10⁻³ = 0.05 J. Check: σ = 10⁸ N m⁻², ε = 5 × 10⁻⁴, u = ½σε = 2.5 × 10⁴ J m⁻³, volume = 2 × 10⁻⁶ m³, so U = 0.05 J ✓.

11Thickness of a crane rope

How thick must a steel rope be for a crane to lift 10 tonnes without permanent deformation? Yield strength of mild steel = 300 × 10⁶ N m⁻².

Show step-by-step solution

A ≥ Mg/σy = (10⁴ × 9.8)/(300 × 10⁶) = 3.3 × 10⁻⁴ m², which corresponds to a radius of about 1 cm.

With a safety margin of about ten times the load, a radius of about 3 cm is recommended, built from many braided thin wires.

12Maximum height of a mountain on Earth

The elastic limit of a typical rock is 30 × 10⁷ N m⁻² and its density is 3 × 10³ kg m⁻³. Estimate the maximum height of a mountain (g = 10 m s⁻²).

Show step-by-step solution

The shear stress at the base is about hρg. Setting hρg = 30 × 10⁷:

h = (30 × 10⁷)/(3 × 10³ × 10) = 10⁴ m = 10 km, which is more than the height of Mt. Everest.

12. Practice Numericals with Answers

Try these NCERT exercise problems (Exercises 8.6 – 8.16). Material constants are from the tables above.

#	Problem	Answer
1	A copper piece of cross-section 15.2 mm × 19.1 mm is pulled with 44,500 N producing only elastic deformation. Find the strain. (Y copper = 1.1 × 10¹¹ N m⁻²)	≈ 1.4 × 10⁻³
2	An aluminium cube of edge 10 cm has one face fixed to a wall. A mass of 100 kg hangs from the opposite face. Find the vertical deflection. (G = 25 GPa)	≈ 3.9 × 10⁻⁷ m
3	Four identical hollow mild-steel columns (inner radius 30 cm, outer radius 60 cm) support a 50,000 kg structure. Find the compressional strain of each. (Y = 2.0 × 10¹¹ N m⁻²)	≈ 7.2 × 10⁻⁷
4	Find the density of water at a depth where the pressure is 80 atm, given surface density 1.03 × 10³ kg m⁻³ and B = 2.2 × 10⁹ N m⁻².	≈ 1.034 × 10³ kg m⁻³
5	Find the fractional volume change of a glass slab under 10 atm hydraulic pressure. (B glass = 37 GPa)	≈ 2.7 × 10⁻⁵
6	Find the volume contraction of a solid copper cube of edge 10 cm under 7.0 × 10⁶ Pa. (B copper = 140 GPa)	≈ 5 × 10⁻⁸ m³ (0.05 cm³)
7	By how much must the pressure on 1 litre of water change to compress it by 0.10 %? (B = 2.2 × 10⁹ N m⁻²)	≈ 2.2 × 10⁶ Pa

13. Question and Answer: Short and Long

Short Answer Questions (2 – 3 Marks)

Q1. Define elasticity and plasticity with one example each.

Answer

Elasticity is the property by which a body regains its original size and shape when the deforming force is removed (a steel spring). Plasticity is the property by which a body stays permanently deformed after the force is removed (putty or mud).

Q2. What are the three types of stress? Give one example for each.

Answer

(1) Tensile/compressive (longitudinal) stress: a stretched wire or a compressed pillar. (2) Shearing stress: a book pushed horizontally on a table. (3) Hydraulic stress: a solid sphere compressed uniformly by a fluid deep in the ocean.

Q3. Why does a thin steel wire used in a crane not stretch much even under a heavy load?

Answer

Steel has a large Young's modulus (about 2 × 10¹¹ N m⁻²). A large force is needed for a small change in length, so the strain stays within the elastic limit.

Q4. Distinguish between yield strength and ultimate tensile strength.

Answer

Yield strength σy is the stress at the yield point B beyond which plastic (permanent) deformation begins. Ultimate tensile strength σu is the maximum stress the material can bear, at point D, after which fracture follows at E.

Q5. Why is Young's modulus not defined for liquids?

Answer

A liquid has no definite shape or length and cannot sustain a tensile or shear stress at rest, so only the bulk modulus applies to liquids and gases.

Q6. Why is the cross-section of bridge girders I-shaped?

Answer

It saves material and weight while keeping a wide load-bearing top surface and a large depth. Since sag ∝ 1/d³, the depth resists bending, and the flanges stop the thin web from buckling.

Q7. Why are crane ropes made of many thin wires braided together rather than a single thick rod?

Answer

A single wire of the required radius would be practically a rigid rod. Braided thin wires are easier to make, more flexible and equally strong.

Q8. What does a negative sign in B = −p/(ΔV/V) mean?

Answer

When pressure increases the volume decreases, so ΔV is negative for positive p. The minus sign makes B a positive quantity.

Long Answer Questions (5 Marks)

Q1. Explain the stress–strain curve for a metal. Define yield point, ultimate tensile strength and fracture point.

Answer

O to A: stress ∝ strain, Hooke's law is obeyed, and the body regains its original dimensions on unloading. Slope = Young's modulus.
A to B: stress and strain are no longer proportional, but the body still returns to its original size. B is the yield point (elastic limit) and the stress there is the yield strength σy.
B to D: strain increases rapidly for a small increase in stress. If the load is removed at C, the body does not return to its original size; a permanent set remains (plastic deformation).
D: the ultimate tensile strength σu is the maximum stress. Beyond D, strain increases even for a reduced force, and the wire fractures at E.
If D and E are close the material is brittle; if they are far apart it is ductile.

Q2. Derive the expression for elastic potential energy stored in a stretched wire.

Answer

Let a wire of length L, area A and Young's modulus Y be stretched by a force F so that its extension is l. Then F = YA(l/L).

For a further small extension dl, the work done is dW = F dl = (YAl/L) dl.

Total work for extension 0 to l: W = ∫₀ˡ (YAl/L) dl = ½ · (YA/L) · l² = ½ · Y · (l/L)² · AL.

Since AL is the volume of the wire: W = ½ × Y × strain² × volume = ½ × stress × strain × volume.

This work is stored as elastic potential energy, so the energy per unit volume is u = ½ σ ε.

Q3. Define Young's modulus, shear modulus and bulk modulus and compare them.

Answer

See the comparison table in Section 7. In brief: Y = (F/A)/(ΔL/L) relates tensile stress to longitudinal strain (change in length and shape, solids only); G = (F/A)/θ relates shear stress to shear strain (change in shape, no volume change, solids only); B = −p/(ΔV/V) relates hydraulic pressure to volume strain (change in volume, no change in shape, solids, liquids and gases).

14. MCQs on Mechanical Properties of Solids (20 Questions)

Tap an option to check your answer. The explanation appears after you answer.

Score: 0 / 20 answered

1What is the SI unit of stress?

Answer: B. Stress = force / area, so its SI unit is N m⁻² = pascal (Pa).

2The dimensional formula of strain is:

Answer: C. Strain is a ratio of two similar quantities (ΔL/L), so it has no unit and no dimensions.

3The dimensional formula of Young's modulus is:

Answer: A. Y = stress/strain and strain is dimensionless, so Y has the dimensions of stress: [M L⁻¹ T⁻²].

4Hooke's law is valid:

Answer: C. Stress ∝ strain only in the linear part OA of the stress–strain curve.

5The slope of the linear part of a stress–strain graph gives:

Answer: B. Slope = stress/strain = Young's modulus.

6Shearing strain is given by:

Answer: C. Shear strain is the relative displacement of faces divided by the distance between them = tan θ ≈ θ for small angles.

7Which modulus of elasticity is applicable to solids, liquids and gases?

Answer: C. Only bulk modulus relates to volume change, which every state of matter can have.

8Which is more elastic: steel or rubber?

Answer: B. The material that deforms less for the same load (larger Y) is more elastic.

9Compressibility is defined as:

Answer: B. k = 1/B = −(1/Δp)(ΔV/V).

10The Young's modulus of a wire depends on:

Answer: D. Y is a characteristic of the material; length, radius and load do not change it.

11Two wires of the same material and area, with lengths L and 2L, carry the same load. The ratio of their extensions is:

Answer: B. ΔL = FL/(AY) ∝ L, so the ratio is L : 2L = 1 : 2.

12If the radius of a wire is doubled (same load, same length), its extension becomes:

Answer: C. ΔL ∝ 1/A ∝ 1/r². Doubling r gives ΔL/4.

13Elastic potential energy stored per unit volume in a stretched wire is:

Answer: B. u = ½ × stress × strain = ½ Y ε².

14Poisson's ratio has:

Answer: C. It is the ratio of two strains, hence dimensionless.

15On the stress–strain curve of a metal, the point of maximum stress is called:

Answer: C. Point D marks the ultimate tensile strength σu; fracture occurs later at E.

16Materials like rubber and aortic tissue that can be stretched to very large strains are called:

Answer: B. Elastomers have a large elastic region but do not obey Hooke's law over most of it.

17For most materials the shear modulus G is approximately:

Answer: C. G ≈ Y/3 (e.g. steel: G = 84 GPa, Y = 200 GPa).

18The sag δ of a beam loaded at the centre is proportional to:

Answer: B. δ = W l³/(4 b d³ Y), so δ ∝ d⁻³. Doubling the depth reduces sag 8 times.

19The order of bulk modulus is:

Answer: C. Solids are least compressible, gases most compressible.

20A wire hangs from a ceiling carrying a weight F. The tensile stress in it is:

Answer: A. Tension at any section is F (not 2F), so stress is F/A.

15. True or False, Fill in the Blanks and Assertion–Reason

True or False

Decide whether each statement is true or false and tap your choice.

Score

1Strain has the unit of metre.

Answer: False. Strain is a ratio of two lengths, so it is dimensionless and has no unit.

2The Young's modulus of rubber is greater than that of steel.

Answer: False. Steel has a far larger Y. Rubber stretches more but needs far less force per unit strain.

3The stretching of a helical coil is determined mainly by its shear modulus.

Answer: True. Stretching a coil twists the wire it is made of, which is a shear deformation.

4Hooke's law holds right up to the fracture point of a metal.

Answer: False. It holds only in the linear region OA; beyond that the curve is non-linear and then plastic.

5Young's modulus and shear modulus are meaningful only for solids.

Answer: True. Only solids have a definite length and shape that can resist tension and shear.

6Bulk modulus is meaningful for solids, liquids and gases.

Answer: True. All states of matter show a volume change under pressure.

7A material that stretches more is more elastic.

Answer: False. The material that stretches less for a given load (larger Y) is more elastic.

8In a wire hanging from a ceiling with weight F at the bottom, the tension is 2F.

Answer: False. The tension at any cross-section is F, so stress = F/A.

9Poisson's ratio has no unit.

Answer: True. It is the ratio of lateral strain to longitudinal strain.

10Glass is a ductile material.

Answer: False. Glass is brittle: its ultimate strength and fracture points are almost the same.

11Compressibility is the reciprocal of the bulk modulus.

Answer: True. k = 1/B.

12Stress is a vector quantity.

Answer: False. Stress cannot be given a single direction like a force, so it is not a vector.

13Steel is more elastic than copper.

Answer: True. Y(steel) = 2.0 × 10¹¹ N m⁻² > Y(copper) = 1.1 × 10¹¹ N m⁻².

14Ductile and brittle behaviour is judged from the gap between points D and E of the stress–strain curve.

Answer: True. A small gap means brittle, a large gap means ductile.

Fill in the Blanks

Think of the missing word first, then open the answer.

1The restoring force per unit area is called ______.

Show answer

stress

2The ratio of tensile stress to longitudinal strain is the ______ modulus.

Show answer

Young's

3Shear modulus is also called the modulus of ______.

Show answer

rigidity

4The reciprocal of bulk modulus is called ______.

Show answer

compressibility

5The stress corresponding to the yield point is called the ______ strength.

Show answer

yield

6Substances that can be stretched to large strains, like rubber, are called ______.

Show answer

elastomers

7Permanent deformation that remains after the load is removed is called a permanent ______.

Show answer

set

8For a stretched wire, elastic potential energy per unit volume = ½ × stress × ______.

Show answer

strain

Assertion–Reason Questions

Options: (a) Both A and R are true and R explains A. (b) Both are true but R does not explain A. (c) A is true, R is false. (d) A is false, R is true.

1Assertion (A): Steel is more elastic than rubber.
Reason (R): For a given load, steel is deformed less than rubber.

Show answer

Both A and R are true and R is the correct explanation of A. Elasticity is judged by resistance to deformation, so the larger Y of steel makes it more elastic.

2Assertion (A): A hollow shaft is stronger than a solid shaft of the same mass.
Reason (R): A hollow shaft has greater resistance to bending and twisting for the same amount of material.

Show answer

Both A and R are true and R is the correct explanation of A. Material placed away from the axis resists bending and twisting better; the same logic is behind I-beams.

3Assertion (A): Young's modulus of a wire does not change when its length is doubled.
Reason (R): Young's modulus depends only on the material of the wire.

Show answer

Both A and R are true and R is the correct explanation of A. Y = (F/A)/(ΔL/L); changing L changes ΔL proportionally, but the ratio is a material property.

4Assertion (A): Stress is a vector quantity.
Reason (R): Stress is force per unit area and force is a vector.

Show answer

A is false but R is true. Stress has no single direction; force on a section does, but stress itself is not a vector.

16. Chapter Summary: Mechanical Properties of Solids

Key Takeaways for Revision

Stress is restoring force per unit area (σ = F/A, unit Pa); strain is fractional change in dimension (no unit). Three types: tensile/compressive, shearing and hydraulic.
Hooke's law: for small deformations stress ∝ strain, and the constant is the modulus of elasticity.
Stress–strain curve: OA proportional, B yield point (yield strength σy), D ultimate tensile strength σu, E fracture. Far-apart D and E mean ductile; close mean brittle.
Young's modulus Y = FL/(AΔL); shear modulus G = F/(Aθ) ≈ Y/3; bulk modulus B = −p/(ΔV/V), with compressibility 1/B.
Poisson's ratio = lateral strain / longitudinal strain, a pure number (≈ 0.3 for steel).
Elastic energy per unit volume u = ½ σε; total U = ½ F ΔL.
Applications: crane rope A ≥ Mg/σy, beam sag δ = Wl³/(4bd³Y) so use I-sections, mountain height ≈ 10 km.
A material that deforms less under a given load is more elastic; steel beats rubber.

17. Frequently Asked Questions (FAQ)

Mechanical properties of solids describe how a solid responds to applied forces. The main ones are elasticity (regaining shape after the force is removed), plasticity (permanent deformation), stiffness (measured by the elastic moduli), strength (yield strength and ultimate tensile strength), ductility and brittleness.

Stress is the restoring force developed per unit area inside a deformed body: stress = F/A. Its SI unit is the pascal (Pa), equal to N m⁻², and its dimensional formula is [M L⁻¹ T⁻²]. The three kinds of stress are tensile or compressive (longitudinal), shearing (tangential) and hydraulic (volume) stress.

Strain is the fractional change in a dimension produced by stress. Longitudinal strain = ΔL/L, shearing strain = Δx/L = tan θ ≈ θ, and volume strain = ΔV/V. Because strain is a ratio of two similar quantities it has no unit and no dimensions.

Hooke's law states that, for small deformations, stress is directly proportional to strain: stress = k × strain. The constant k is called the modulus of elasticity. The law holds only in the linear (proportional) part of the stress–strain curve and is an empirical law.

Young's modulus Y is the ratio of tensile (or compressive) stress to longitudinal strain within the elastic limit: Y = σ/ε = (F/A)/(ΔL/L) = FL/(AΔL). Its SI unit is N m⁻² (Pa). For steel Y ≈ 2.0 × 10¹¹ N m⁻².

Shear modulus G is the ratio of shearing stress to shearing strain: G = (F/A)/(Δx/L) = F/(Aθ). It measures resistance to change in shape at constant volume. For most materials G ≈ Y/3, and it exists only for solids.

Bulk modulus B = −p/(ΔV/V) is the ratio of hydraulic stress (pressure) to volume strain. Compressibility k = 1/B is the fractional decrease in volume per unit increase in pressure. Solids have a much larger bulk modulus than liquids, and liquids much larger than gases.

Poisson's ratio is the ratio of lateral strain to longitudinal strain in a stretched wire: σ = (Δd/d)/(ΔL/L). It is a pure number with no unit. For steels it is about 0.28 to 0.30, and for aluminium alloys about 0.33.

On a stress–strain curve, if the ultimate tensile strength point (D) and the fracture point (E) are far apart, the material is ductile (copper, mild steel). If they are close together, the material is brittle (glass, cast iron) and breaks with almost no plastic deformation.

In physics, the more elastic material is the one that stretches less for a given load, i.e. the one with the larger Young's modulus. Steel (Y ≈ 2 × 10¹¹ N m⁻²) is far more elastic than rubber, even though rubber can be stretched to much larger strains. Rubber is an elastomer, not a more elastic material.

The work done in stretching a wire is stored as elastic potential energy. Energy per unit volume u = ½ × stress × strain = ½ σε = ½ Y ε². Total energy U = ½ × stress × strain × volume = ½ F ΔL.

The sag of a loaded beam is δ = W l³/(4 b d³ Y), so it depends on d⁻³ and only b⁻¹. Increasing depth reduces bending far more than increasing breadth. An I-section gives a large load-bearing top surface and enough depth to resist bending and buckling while saving material, weight and cost.

Mechanical Properties of Solids – Class 11 Physics Notes, Formulas, MCQs & Numericals

1. Introduction: Elasticity and Plasticity

Elasticity

Plasticity

2. Stress and Strain (Types, Formulas, Units)

Stress

Three Types of Stress and Strain

Strain Has No Unit

Hindi Terms for Revision (हिंदी शब्दावली)

3. Hooke's Law

Hooke's Law

4. Stress–Strain Curve

Reading the Curve Region by Region

Ductile vs Brittle

5. Elastic Moduli: Young's, Shear and Bulk Modulus

5.1 Young's Modulus (Y)

Young's Modulus

Wire Extension and Young's Modulus Calculator

5.2 Shear Modulus (Modulus of Rigidity, G)

Shear Modulus

5.3 Bulk Modulus (B) and Compressibility

Bulk Modulus and Compressibility

5.4 Comparison of the Three Elastic Moduli

6. Poisson's Ratio

Poisson's Ratio

7. Elastic Potential Energy in a Stretched Wire

Elastic Potential Energy

8. Applications of Elastic Behaviour of Materials

8.1 Thickness of a Crane Rope

8.2 Bending of Beams and the I-Shaped Girder

Sag of a Loaded Beam

8.3 Why Mountains Are at Most About 10 km High

9. Formula Sheet, Points to Ponder and Common Mistakes

Points to Ponder

Common Exam Mistakes

10. Lab Experiments on Mechanical Properties of Solids

Determination of Young's modulus of a wire using Searle's apparatus

Verification of Hooke's law using a helical spring

Virtual simulation: stress–strain curve of a metal wire

11. Solved Numericals (NCERT Examples and Exercises)

12. Practice Numericals with Answers

13. Question and Answer: Short and Long

Short Answer Questions (2 – 3 Marks)

Long Answer Questions (5 Marks)

14. MCQs on Mechanical Properties of Solids (20 Questions)

15. True or False, Fill in the Blanks and Assertion–Reason

True or False

Fill in the Blanks

Assertion–Reason Questions

16. Chapter Summary: Mechanical Properties of Solids

Key Takeaways for Revision

17. Frequently Asked Questions (FAQ)

Related Physics Topics

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