PhysicsCBSE Class 11 · Chapter 8

Mechanical Properties of Solids Class 11: Notes, Elastic Moduli and Numericals

Updated 14 October 2026 · Fact-checked

Mechanical Properties of Solids is Chapter 8 of CBSE Class 11 Physics. It covers elasticity, stress and strain, Hooke's law, the stress-strain curve, Young's modulus, bulk modulus, a qualitative idea of the shear modulus, Poisson's ratio, elastic potential energy and, qualitatively, how elastic behaviour is used in engineering.

What Mechanical Properties of Solids is about

Up to now solid bodies were perfectly rigid. Real solids stretch, squeeze and twist a little when forces act on them, and spring back when the forces are removed. This property is elasticity. Putty and clay, which stay deformed, are plastic.

To compare materials fairly we use stress (restoring force per unit area) and strain (fractional change in size or shape). For small deformations Hooke's law says stress is proportional to strain, and the constant of proportionality is a modulus of elasticity. Young's modulus describes stretching or compressing a wire or rod, the bulk modulus describes squeezing the volume by pressure, and the shear modulus (qualitatively) describes a change of shape. The stress-strain curve shows the proportional limit, yield point, ultimate strength and fracture, and separates ductile from brittle materials.

You also meet Poisson's ratio (sideways thinning when a wire is stretched) and the elastic potential energy stored in a stretched wire, $\frac{1}{2} \times stress \times strain \times volume$ . For 2026-27 the shear modulus and applications are qualitative only, and the experiment to determine Young's modulus has been removed. This chapter leads into fluids, and its ideas reappear in the speed of sound in solids ( $Y / ρ$ ) in the Waves chapter.

Weightage: This chapter is part of Unit VII, Properties of Bulk Matter. The syllabus groups Unit VII with Unit VIII, Thermodynamics and Unit IX, Behaviour of Perfect Gases and Kinetic Theory of Gases: together they carry 20 of the 70 theory marks in the school annual exam.Source: CBSE Physics (042) syllabus 2026-27, Class XI course structure, Unit VII Properties of Bulk Matter

Key concepts

Elasticity, plasticity and stress

NCERT section 8.1 and 8.2

When a deforming force acts on a solid, its particles are pushed slightly out of place and the interatomic forces pull back. The property of a body to regain its original size and shape when the force is removed is elasticity; bodies that do not regain their shape are plastic.

Stress is the restoring force per unit area: $stress = \frac{F}{A}$ . SI unit N m $^{- 2}$ = pascal (Pa); dimensions $[M L^{- 1} T^{- 2}]$ .

Longitudinal stress: force perpendicular to the cross-section. It is tensile if the body is stretched and compressive if it is squeezed.
Shearing (tangential) stress: force parallel to the surface.
Hydraulic stress: a uniform pressure from all sides (a body in a fluid); it equals the pressure.

Worked example

A load of 100 N hangs from a wire of cross-sectional area 1 mm $^{2}$ . Find the stress in the wire. [1 mark]

Try it first, then show the solutionSolution

A = 1 mm $^{2}$ = $1 0^{- 6}$ m $^{2}$
Stress $= \frac{100}{1 0 ^{- 6}} = 1 0^{8} P a$ (tensile)

Answer: $1 0^{8}$ N m $^{- 2}$

Strain and Hooke's law

NCERT section 8.2 and 8.3

Strain is the fractional change in a dimension; it has no unit or dimensions.

Longitudinal strain $= \frac{Δ L}{L}$ .
Shearing strain $= \frac{Δ x}{L} = tan θ \approx θ$ , where Δx is the sideways shift of one face relative to the opposite face a distance L away.
Volume strain $= \frac{Δ V}{V}$ .

Hooke's law: for small deformations, stress is proportional to strain:
$stress = k \times strain$
where k is the modulus of elasticity. Its unit is the same as stress (Pa). Hooke's law holds for most materials only up to the proportional limit.

Worked example

A 2.0 m wire stretches by 1.0 mm under a load. Find the longitudinal strain. [1 mark]

Try it first, then show the solutionSolution

Strain $= \frac{Δ L}{L} = \frac{1.0 \times 1 0 ^{- 3}}{2.0}$
Strain $= 5 \times 1 0^{- 4}$ (no unit)

Answer: $5 \times 1 0^{- 4}$

The stress-strain curve

NCERT section 8.4

Stretching a metal wire gradually and plotting stress against strain gives a characteristic curve:

1. O to A, straight line: Hooke's law holds; A is the proportional limit. The slope here is Young's modulus.
2. A to B: not quite straight but still elastic; on removing the load the wire returns to its original length. B is the yield point (elastic limit); the stress there is the yield strength.
3. Beyond B: strain grows rapidly for a small increase in stress. If the load is removed at C, the wire does not return to its length: it has a permanent set (plastic deformation).
4. D: the ultimate tensile strength, the largest stress the wire can bear. Beyond D the wire thins (necks) and breaks at E, the fracture point.

If D and E are far apart, the material is ductile (copper, aluminium: can be drawn into wires). If they are close, it is brittle (glass, cast iron). Elastomers such as rubber and the aorta stretch hugely but do not obey Hooke's law over most of the range.

Worked example

On the stress-strain curves of two materials, A has a much larger plastic region (between the yield point and fracture) than B. Which is ductile and which is brittle? [1 mark]

Try it first, then show the solutionSolution

A large gap between the elastic limit and fracture means the material can be permanently stretched a lot before breaking.

Answer: A is ductile; B is brittle

Young's modulus

NCERT section 8.5.1

Young's modulus is the ratio of tensile (or compressive) stress to longitudinal strain within the proportional limit:
$Y = \frac{F / A}{Δ L / L} = \frac{F L}{A Δ L}$

So the extension of a wire is $Δ L = \frac{F L}{A Y}$ : longer and thinner wires stretch more; stiffer materials (larger Y) stretch less.

Typical values: steel about $2 \times 1 0^{11}$ Pa, copper about $1.1 \times 1 0^{11}$ Pa, aluminium about $0.7 \times 1 0^{11}$ Pa. Metals have large Y; rubber has a very small Y. Steel is more elastic than rubber in the physics sense: for the same stress it shows much less strain.

Worked example

A 5 kg mass hangs from a steel wire 2 m long and 0.8 mm in diameter. Find the extension. ( $Y = 2 \times 1 0^{11}$ Pa, g = 9.8 m s $^{- 2}$ ) [2 marks]

Try it first, then show the solutionSolution

$A = π (0.4 \times 1 0^{- 3})^{2} \approx 5.03 \times 1 0^{- 7} m^{2}$ ; F = 49 N
$Δ L = \frac{F L}{A Y} = \frac{49 \times 2}{5.03 \times 1 0 ^{- 7} \times 2 \times 1 0 ^{11}} \approx 9.7 \times 1 0^{- 4} m$

Answer: About 0.97 mm

Shear modulus (qualitative) and bulk modulus

NCERT section 8.5.2 and 8.5.3

Shear modulus (modulus of rigidity) G is the ratio of shearing stress to shearing strain: $G = \frac{F / A}{θ}$ . It describes resistance to a change of shape at constant volume. For most materials G is roughly a third of Y. Only solids have a shear modulus; liquids and gases cannot resist a shearing force. (The syllabus asks only for this qualitative idea.)

Bulk modulus B describes resistance to a change of volume under a uniform pressure p:
$B = - \frac{p}{Δ V / V}$
The minus sign makes B positive, because an increase in pressure decreases the volume. The reciprocal, $k = \frac{1}{B}$ , is the compressibility.

Solids are the least compressible (largest B), liquids next, gases by far the most compressible. Water: $B \approx 2.2 \times 1 0^{9}$ Pa; steel about $1.6 \times 1 0^{11}$ Pa.

Worked example

Water is subjected to an extra pressure of $1.0 \times 1 0^{7}$ Pa (about 100 atmospheres). By what fraction does its volume decrease? ( $B = 2.2 \times 1 0^{9}$ Pa) [1 mark]

Try it first, then show the solutionSolution

$\frac{Δ V}{V} = \frac{p}{B} = \frac{1.0 \times 1 0 ^{7}}{2.2 \times 1 0 ^{9}}$
$\approx 4.5 \times 1 0^{- 3}$ , i.e. about 0.45%

Answer: About 0.45% decrease

Poisson's ratio

NCERT section 8.5.4

When a wire is stretched, it gets longer and also thinner. The strain perpendicular to the force is the lateral strain $\frac{Δ d}{d}$ .

Poisson's ratio is the ratio of lateral strain to longitudinal strain:
$σ = \frac{Δ d / d}{Δ L / L}$

It is a ratio of two strains, so it has no unit and no dimensions. It depends only on the material; for steel it is about 0.28 to 0.30, and for most materials it lies between 0 and 0.5.

Worked example

A wire 1 m long and 2 mm in diameter is stretched by 0.5 mm, and its diameter decreases by $3 \times 1 0^{- 4}$ mm. Find Poisson's ratio. [2 marks]

Try it first, then show the solutionSolution

Longitudinal strain $= \frac{0.5}{1000} = 5 \times 1 0^{- 4}$ ; lateral strain $= \frac{3 \times 1 0 ^{- 4}}{2} = 1.5 \times 1 0^{- 4}$
$σ = \frac{1.5 \times 1 0 ^{- 4}}{5 \times 1 0 ^{- 4}} = 0.3$

Answer: 0.3

Elastic potential energy in a stretched wire

NCERT section 8.5.5

Work done in stretching a wire is stored as elastic potential energy. As the extension grows from 0 to l, the force grows from 0 to $F = \frac{Y A l}{L}$ , so
$W = \int_{0}^{l} \frac{Y A}{L} x d x = \frac{1}{2} \frac{Y A l ^{2}}{L} = \frac{1}{2} F l$

Rewriting with stress $= \frac{Y l}{L}$ and strain $= \frac{l}{L}$ :
$U = \frac{1}{2} \times stress \times strain \times volume$
and the energy per unit volume is $u = \frac{1}{2} \times stress \times strain = \frac{1}{2} Y (strain)^{2}$ .

Worked example

A wire stretches by 2 mm when a 250 N load is applied. How much elastic energy is stored? [1 mark]

Try it first, then show the solutionSolution

$U = \frac{1}{2} F l = \frac{1}{2} \times 250 \times 2 \times 1 0^{- 3}$
U = 0.25 J

Answer: 0.25 J

Applications of elastic behaviour (qualitative)

NCERT section 8.6

Engineers design so that working stresses stay well below the elastic limit, with a factor of safety.

Crane ropes: the rope must carry the load without permanent stretch: area ≥ load ÷ yield strength. A single thick wire would be rigid, so ropes are made of many thin wires twisted together: strong and flexible.
Beams and bridges: a loaded beam sags (bends). The sag decreases greatly with the depth of the beam (it goes as $\frac{1}{d ^{3}}$ ) and with a higher Y, but only as $\frac{1}{b}$ with breadth. So beams are deep rather than wide, and I-shaped girders put material where it is needed, saving weight while resisting bending and buckling.
Height of mountains: the rock at the base must not exceed its elastic limit under the weight above it. This limits mountains on Earth to roughly 10 km.
Columns: columns with distributed (flared) ends support more than columns with rounded ends.

Worked example

Why are the beams in buildings and the rails of a railway track made I-shaped? [2 marks]

Try it first, then show the solutionSolution

Sag (bending) decreases sharply with the depth of the beam, so material placed far from the middle resists bending best.
An I-section puts most metal in the top and bottom flanges and saves weight in the middle, giving high strength for less material without buckling.

Answer: Depth resists bending; the I-shape gives strength with less weight.

Physics in real life: Mechanical Properties of Solids

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Limit on mountain height

Suppose the rock at the base of a mountain can bear a stress of about $3 \times 1 0^{8}$ Pa before flowing, and the rock's density is 3000 kg m $^{- 3}$ .

Estimate the greatest height a mountain could have on the Earth. (g = 9.8 m s $^{- 2}$ )

Try it first, then show the solutionSolution

Stress at the base $\approx h ρ g$ (weight of a column per unit area)
$h = \frac{3 \times 1 0 ^{8}}{3000 \times 9.8} \approx 1.0 \times 1 0^{4} m$

Answer: About 10 km, close to the height of Mount Everest

Memory hook: Rock has a stress limit, so mountains have a height limit.

Choosing a crane rope

Imagine a crane at a building site that must lift a 10 tonne ( $1 0^{4}$ kg) load with a steel rope whose yield strength is $3 \times 1 0^{8}$ Pa.

Suppose the rope must not exceed its yield strength. What is the smallest radius it can have? (g = 9.8 m s $^{- 2}$ )

Try it first, then show the solutionSolution

Need $\frac{m g}{π r ^{2}} \leq 3 \times 1 0^{8}$
$r \geq \frac{9.8 \times 1 0 ^{4}}{π \times 3 \times 1 0 ^{8}} \approx 1.0 \times 1 0^{- 2} m$
In practice a much thicker rope (with a factor of safety) made of many thin strands is used.

Answer: At least about 1 cm; real ropes are about 3 cm for safety

Memory hook: Strength from area, flexibility from strands.

Bulk modulus under the sea

Suppose a steel instrument case is lowered 1000 m into the ocean.

By what fraction does its volume shrink? (ρ of sea water ≈ 1000 kg m $^{- 3}$ , g = 9.8 m s $^{- 2}$ , B of steel = $1.6 \times 1 0^{11}$ Pa)

Try it first, then show the solutionSolution

Extra pressure $p = ρ g h = 1000 \times 9.8 \times 1000 = 9.8 \times 1 0^{6} P a$
$\frac{Δ V}{V} = \frac{p}{B} = \frac{9.8 \times 1 0 ^{6}}{1.6 \times 1 0 ^{11}} \approx 6.1 \times 1 0^{- 5}$

Answer: About 0.006%: solids are very hard to compress

Memory hook: Big B, tiny squeeze.

Topics in this chapter

Mechanical Properties of Solids formula sheet

Stress and strain: $stress = \frac{F}{A}, strain = \frac{Δ L}{L}, \frac{Δ x}{L}, \frac{Δ V}{V}$
Hooke's law: $stress = k \times strain$
Young's modulus: $Y = \frac{F L}{A Δ L}, Δ L = \frac{F L}{A Y}$
Shear modulus: $G = \frac{F / A}{θ}$; Qualitative idea only for 2026-27.
Bulk modulus and compressibility: $B = - \frac{p}{Δ V / V}, k = \frac{1}{B}$
Poisson's ratio: $σ = \frac{Δ d / d}{Δ L / L}$
Elastic potential energy: $U = \frac{1}{2} F Δ L = \frac{1}{2} \times stress \times strain \times volume$
Energy per unit volume: $u = \frac{1}{2} stress \times strain = \frac{1}{2} Y (strain)^{2}$

How the school exam asks Mechanical Properties of Solids

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Definitions and the stress-strain curve	Define stress, strain, elastic limit, yield point; label the curve; ductile vs brittle; elastomers.	Draw the curve with O, A, B, C, D, E labelled and one line on each region.
Young's modulus numericals	Extension of a wire; Y from data; comparing two wires of different lengths or diameters; wires in series.	$Δ L = F L / A Y$ ; take ratios to cancel what is common; convert mm to m and diameter to radius.
Bulk modulus	Fractional volume change under pressure; pressure needed for a given compression; density at depth.	$Δ V / V = p / B$ ; pressure at depth = ρgh.
Poisson's ratio and elastic energy	Poisson's ratio from data; energy stored in a stretched wire; energy density.	Ratio of strains; $U = \frac{1}{2} F Δ L$ .
Applications	Why steel is more elastic than rubber; why beams are I-shaped; why ropes are stranded; mountain heights.	Link each to Y, the sag dependence on depth, flexibility, or the elastic limit.

Tricks to remember Mechanical Properties of Solids

Short memory aids. Say each one aloud once, then use it on a question the same day.

Y, B, G: stretch, squeeze, shear
Young's modulus for length, Bulk modulus for volume, rigidity (G) for shape.
Use it when: Matching moduli to deformations.
Strain has no unit, modulus has stress's unit
Strain is a ratio; modulus = stress ÷ strain, so it is in pascals.
Use it when: Unit and dimension MCQs.
Half F times extension
Energy stored = ½ × final force × extension, like the area of a triangle under the F-x line.
Use it when: Elastic energy.
Curve order: P-Y-U-F
Proportional limit, Yield point, Ultimate strength, Fracture.
Use it when: Labelling the stress-strain curve.
More elastic = harder to deform
In physics, steel is more elastic than rubber because it has a larger Y.
Use it when: Assertion-reason items.

Mechanical Properties of Solids: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Unit of stress · 1 mark · MCQExam level

The SI unit of stress is:
(a) N m
(b) N m $^{- 2}$
(c) N m $^{- 1}$
(d) no unit

Try it first, then show the solutionSolution

Stress = force/area.

Answer: (b) N m $^{- 2}$

Example 2: Steel or rubber · 1 mark · Assertion-reasonExam level

Assertion (A): Steel is more elastic than rubber.
Reason (R): For the same stress, the strain in steel is much smaller than in rubber, so its Young's modulus is larger.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

Elasticity is judged by the modulus; R gives exactly this reason.

Answer: (a)

Example 3: Extension of a steel wire · 3 marks · Short answerHarder exam level

A 10 kg mass hangs from a steel wire 3.0 m long and 1.0 mm in diameter. Find the stress, strain and extension. ( $Y = 2.0 \times 1 0^{11}$ Pa, g = 9.8 m s $^{- 2}$ ) [3 marks]

Try it first, then show the solutionSolution

$A = π (5 \times 1 0^{- 4})^{2} = 7.85 \times 1 0^{- 7} m^{2}$ ; stress $= \frac{98}{7.85 \times 1 0 ^{- 7}} \approx 1.25 \times 1 0^{8} P a$
Strain $= \frac{stress}{Y} = \frac{1.25 \times 1 0 ^{8}}{2.0 \times 1 0 ^{11}} \approx 6.2 \times 1 0^{- 4}$
Extension $= 6.2 \times 1 0^{- 4} \times 3.0 \approx 1.9 \times 1 0^{- 3} m$

Answer: Stress ≈ $1.25 \times 1 0^{8}$ Pa; strain ≈ $6.2 \times 1 0^{- 4}$ ; extension ≈ 1.9 mm

Example 4: Comparing two wires · 2 marks · Very short answerHarder exam level

A wire of length L and radius r stretches by 1 mm under a load. A second wire of the same material has length 2L and radius 2r. How much does it stretch under the same load? [2 marks]

Try it first, then show the solutionSolution

$Δ L \propto \frac{L}{r ^{2}}$
$Δ L_{2} = 1 mm \times \frac{2}{4} = 0.5 mm$

Answer: 0.5 mm

Example 5: Water under pressure · 3 marks · Short answerHarder exam level

One litre of water is subjected to an extra pressure of 100 atm ( $1.013 \times 1 0^{7}$ Pa). Find the decrease in its volume and the percentage change in its density. ( $B = 2.2 \times 1 0^{9}$ Pa) [3 marks]

Try it first, then show the solutionSolution

$Δ V = \frac{p V}{B} = \frac{1.013 \times 1 0 ^{7} \times 1 0 ^{- 3}}{2.2 \times 1 0 ^{9}} \approx 4.6 \times 1 0^{- 6} m^{3}$ = 4.6 mL
Fractional change $\approx 0.46%$
Mass is unchanged, so density rises by about 0.46%

Answer: About 4.6 mL; density up by about 0.46%

Example 6: Stress-strain curve and breaking load · 5 marks · Long answerHarder exam level

(a) Sketch the stress-strain curve of a ductile metal and explain the regions. (b) What is meant by 'permanent set'? (c) A copper wire 2 m long and 1 mm $^{2}$ in cross-section stretches by 2 mm under a 110 N load. Find Y. (d) Would the wire stretch more or less under the same load if it were 4 m long? [5 marks]

Try it first, then show the solutionSolution

(a) Linear region (Hooke's law) to the proportional limit; elastic but non-linear to the yield point; plastic region to the ultimate strength; necking and fracture
(b) If the wire is loaded beyond the yield point and unloaded, it does not return to zero strain: the leftover strain is the permanent set
(c) $Y = \frac{F L}{A Δ L} = \frac{110 \times 2}{1 0 ^{- 6} \times 2 \times 1 0 ^{- 3}}$
$Y = 1.1 \times 1 0^{11} P a$
(d) $Δ L \propto L$ , so twice as much (4 mm)

Answer: Y = $1.1 \times 1 0^{11}$ Pa; it would stretch twice as much

Example 7: Energy density in a wire · 3 marks · Short answerHOTS / competency

A steel wire 2 m long with cross-section 2 mm $^{2}$ is stretched by 0.5 mm. Find the energy stored per unit volume and in the whole wire. ( $Y = 2 \times 1 0^{11}$ Pa) [3 marks]

Try it first, then show the solutionSolution

Strain $= \frac{0.5 \times 1 0 ^{- 3}}{2} = 2.5 \times 1 0^{- 4}$ ; stress $= Y \times strain = 5 \times 1 0^{7} P a$
$u = \frac{1}{2} \times 5 \times 1 0^{7} \times 2.5 \times 1 0^{- 4} = 6.25 \times 1 0^{3} J m^{- 3}$
Volume $= 2 \times 1 0^{- 6} \times 2 = 4 \times 1 0^{- 6} m^{3}$ ; $U = 6.25 \times 1 0^{3} \times 4 \times 1 0^{- 6} = 0.025 J$

Answer: $6.25 \times 1 0^{3}$ J m $^{- 3}$ ; 0.025 J

Example 8: Cutting a wire in half · 2 marks · Very short answerHOTS / competency

A wire breaks under a load of 400 N. (a) What load will break a wire of the same material and thickness but half the length? (b) What load will break one of the same length but twice the diameter? (c) Under a given load below breaking, how does the extension of the half-length wire compare? [2 marks]

Try it first, then show the solutionSolution

Breaking load = breaking stress × area, independent of length: (a) 400 N; (b) area × 4: 1600 N
(c) $Δ L \propto L$ , so it stretches half as much

Answer: (a) 400 N (b) 1600 N (c) half the extension

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

A lift cable

Suppose a lift of total mass 1000 kg hangs from a steel cable 50 m long with a radius of 1 cm. Take $Y = 2 \times 1 0^{11}$ Pa, g = 9.8 m s $^{- 2}$ and the elastic limit of steel as $2.5 \times 1 0^{8}$ Pa.

(i) · 1 mark
Find the stress in the cable when the lift is at rest.
Answer
Answer: About $3.1 \times 1 0^{7}$ Pa
$\frac{9800}{π \times 1 0 ^{- 4}}$
(ii) · 1 mark
Find the extension of the cable.
Answer
Answer: About 7.8 mm
$\frac{F L}{A Y} = \frac{9800 \times 50}{3.14 \times 1 0 ^{- 4} \times 2 \times 1 0 ^{11}}$
(iii) · 1 mark
Is the cable within its elastic limit, and by what factor?
Answer
Answer: Yes, about 8 times below it
$\frac{2.5 \times 1 0 ^{8}}{3.1 \times 1 0 ^{7}} \approx 8$
(iv) · 1 mark
Why is such a cable made of many thin strands rather than one solid rod?
Answer
Answer: For flexibility and safety
Strands give the same area (strength) but bend easily round pulleys, and one broken strand does not snap the cable.

Testing a new alloy

Suppose a lab tests a wire of a new alloy, 1 m long and 0.5 mm $^{2}$ in cross-section. Up to a load of 50 N the extension rises in proportion to the load, reaching 0.5 mm at 50 N. The wire breaks at 150 N after stretching a lot.

(i) · 1 mark
Find the Young's modulus of the alloy.
Answer
Answer: $2 \times 1 0^{11}$ Pa
$\frac{50 \times 1}{0.5 \times 1 0 ^{- 6} \times 0.5 \times 1 0 ^{- 3}}$
(ii) · 1 mark
What is the stress at the proportional limit?
Answer
Answer: $1 0^{8}$ Pa
$\frac{50}{0.5 \times 1 0 ^{- 6}}$
(iii) · 1 mark
What is the ultimate tensile strength?
Answer
Answer: $3 \times 1 0^{8}$ Pa
$\frac{150}{0.5 \times 1 0 ^{- 6}}$
(iv) · 1 mark
Is the alloy ductile or brittle?
Answer
Answer: Ductile
It stretched a lot before breaking (large plastic region).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Two wires of the same material have lengths in the ratio 2 : 1 and diameters in the ratio 1 : 2. They carry equal loads. Find the ratio of their extensions. [2 marks] (2 marks)

Step	Marks
$Δ L = \frac{F L}{A Y} \propto \frac{L}{d ^{2}}$ (same F and Y)	1
$\frac{Δ L _{1}}{Δ L _{2}} = \frac{2}{1} \times \frac{2 ^{2}}{1 ^{2}} = 8$ , so 8 : 1	1

Show that the elastic potential energy per unit volume of a stretched wire is ½ × stress × strain. [3 marks] (3 marks)

Step	Marks
At extension x the force is $F = \frac{Y A x}{L}$ ; work for a further dx is $F d x$	1
$W = \int_{0}^{l} \frac{Y A}{L} x d x = \frac{1}{2} \frac{Y A l ^{2}}{L}$	1
Divide by volume AL: $u = \frac{1}{2} Y (\frac{l}{L})^{2} = \frac{1}{2} \times (\frac{Y l}{L}) \times \frac{l}{L} = \frac{1}{2} stress \times strain$	1

(a) Draw a typical stress-strain curve for a metal wire and mark the proportional limit, yield point, ultimate tensile strength and fracture point. (b) Explain the difference between a ductile and a brittle material using the curve. (c) A steel wire of diameter 2 mm has an ultimate tensile strength of $4 \times 1 0^{8}$ Pa. Find the largest load it can carry, and the safe working load with a factor of safety of 5. [5 marks] (5 marks)

Step	Marks
(a) Curve: straight to A (proportional limit), curving to B (yield point), plastic region to D (ultimate strength), dropping to E (fracture); axes labelled	1
Regions explained: elastic up to B, permanent set beyond B	1
(b) Ductile: D and E far apart (large plastic region), e.g. copper; brittle: fracture soon after the elastic limit, e.g. glass	1
(c) $A = π (1 0^{- 3})^{2} = 3.14 \times 1 0^{- 6} m^{2}$ ; maximum load $= 4 \times 1 0^{8} \times 3.14 \times 1 0^{- 6} \approx 1257 N$	1
Safe working load $= \frac{1257}{5} \approx 251 N$	1

Where students lose marks in Mechanical Properties of Solids

The slips examiners see most often, and what to write instead.

Mistake: Using diameter as radius in the area
Rushing.
Wrong working
Diameter 1 mm: $A = π (1 \times 1 0^{- 3})^{2}$
Right working
$A = π (0.5 \times 1 0^{- 3})^{2}$
Do this: Halve the diameter first; the error makes the answer four times wrong.
Usually costs: 1-2 marks.
Mistake: Forgetting unit conversions (mm², cm)
Mixed units in data.
Do this: 1 mm $^{2}$ = $1 0^{- 6}$ m $^{2}$ ; 1 cm $^{2}$ = $1 0^{- 4}$ m $^{2}$ .
Usually costs: Whole numerical.
Mistake: Saying rubber is more elastic than steel
Everyday language.
Do this: Elasticity in physics is measured by the modulus; steel's Y is far larger.
Usually costs: 1 mark.
Mistake: Writing calculations with the shear modulus
Older syllabi included shear numericals.
Do this: For 2026-27 only a qualitative idea of the shear modulus is needed.
Usually costs: Time.
Mistake: Dropping the minus sign idea in the bulk modulus
Not thinking about the direction of volume change.
Do this: Pressure up, volume down; the minus sign keeps B positive.
Usually costs: Half a mark in definitions.

NCERT exercise guide: Mechanical Properties of Solids

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercises 8.1-8.4 (Young's modulus and curves)

What it tests: Comparing steel and copper wires, reading Y and yield strength from a graph, comparing materials from their stress-strain graphs, and true/false statements.

How to approach it: Y = slope of the linear part; a steeper line means a larger Y; the higher yield point means a stronger material.

Same kind as Exercise 8.1

A steel wire and a copper wire of the same length and diameter carry the same load. Which stretches more, and by what factor? ( $Y_{s t ee l} = 2.0 \times 1 0^{11}$ Pa, $Y_{co pp er} = 1.1 \times 1 0^{11}$ Pa)

Show the solutionSolution

$Δ L \propto \frac{1}{Y}$ when F, L and A are the same
$\frac{Δ L _{C u}}{Δ L _{s t ee l}} = \frac{2.0}{1.1} \approx 1.8$

Answer: The copper wire, about 1.8 times as much

Exercises 8.5-8.11 (loaded wires and structures)

What it tests: Elongation of joined wires, columns under load, a copper bar pulled in tension, a ski-lift cable, a bar on three wires and a mass whirled on a wire.

How to approach it: Find the force in each wire from equilibrium (or $m v^{2} / r$ ), then $Δ L = F L / A Y$ .

Same kind as Exercise 8.5

A steel wire and a brass wire of equal length and equal cross-section are joined end to end and hung with a load at the bottom. Find the ratio of their extensions. ( $Y_{s t ee l} = 2.0 \times 1 0^{11}$ Pa, $Y_{b r a ss} = 0.91 \times 1 0^{11}$ Pa)

Show the solutionSolution

Both carry the same tension, and have the same L and A
$\frac{Δ L _{s t ee l}}{Δ L _{b r a ss}} = \frac{Y _{b r a ss}}{Y _{s t ee l}} = \frac{0.91}{2.0} \approx 0.46$

Answer: Steel stretches about 0.46 times as much as brass

Exercises 8.12-8.16 (bulk modulus)

What it tests: Bulk modulus of water from data, density at depth, volume change of glass and copper under pressure.

How to approach it: $Δ V / V = p / B$ ; density rises as volume falls: $ρ^{'} = ρ (1 + p / B)$ for small changes.

Same kind as Exercise 8.16

By how much must the pressure on water be increased to compress it by 0.20%? ( $B = 2.2 \times 1 0^{9}$ Pa)

Show the solutionSolution

$p = B \frac{Δ V}{V} = 2.2 \times 1 0^{9} \times 2.0 \times 1 0^{- 3}$
$p = 4.4 \times 1 0^{6} P a$ (about 43 atm)

Answer: $4.4 \times 1 0^{6}$ Pa

Practice questions on Mechanical Properties of Solids

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
The Young's modulus of a perfectly rigid body is:
(a) zero
(b) 1
(c) infinite
(d) depends on the load
Hint
No strain for any stress.
Answer
(c) infinite
Q2 · 1 mark · MCQExam level
Poisson's ratio has:
(a) the unit of stress
(b) the unit of Y
(c) no unit
(d) the unit m $^{- 1}$
Hint
Ratio of two strains.
Answer
(c) no unit
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): The bulk modulus of a gas is much smaller than that of a solid.
Reason (R): Gases are much more compressible than solids.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
k = 1/B.
Answer
(a)
Q4 · 2 marks · Very short answerExam level
Distinguish between the elastic limit and the proportional limit.
Hint
Linear vs recoverable.
Answer
Up to the proportional limit, stress is proportional to strain (Hooke's law). Up to the elastic limit (yield point), the body returns to its original size on unloading even though the graph may no longer be straight.
Q5 · 2 marks · Very short answerHarder exam level
A wire 2 m long with cross-section 1 mm $^{2}$ stretches by 2 mm under a 200 N load. Find the Young's modulus.
Hint
Y = FL/AΔL.
Answer
$Y = \frac{200 \times 2}{1 0 ^{- 6} \times 2 \times 1 0 ^{- 3}} = 2 \times 1 0^{11}$ Pa.
Q6 · 2 marks · Very short answerHarder exam level
A rubber cord 10 cm long with a diameter of 4 mm is stretched by 1 cm, and its diameter decreases by 0.08 mm. Find Poisson's ratio for the rubber.
Hint
Lateral strain / longitudinal strain.
Answer
Longitudinal strain = 1/10 = 0.1; lateral strain = 0.08/4 = 0.02; σ = 0.02/0.1 = 0.2.
Q7 · 2 marks · Very short answerHarder exam level
A load of 100 N stretches a wire by 1 mm. How much energy is stored? How much is stored if the load is doubled (still within the elastic limit)?
Hint
Double load, double extension.
Answer
$U = \frac{1}{2} \times 100 \times 1 0^{- 3} = 0.05$ J. With 200 N the extension is 2 mm, so $U = \frac{1}{2} \times 200 \times 2 \times 1 0^{- 3} = 0.2$ J (four times).
Q8 · 2 marks · Very short answerHarder exam level
A solid sphere of volume 1 m $^{3}$ is lowered into the sea to a depth where the extra pressure is $2 \times 1 0^{7}$ Pa. Its bulk modulus is $1 0^{11}$ Pa. Find the change in its volume.
Hint
ΔV = pV/B.
Answer
$Δ V = \frac{2 \times 1 0 ^{7} \times 1}{1 0 ^{11}} = 2 \times 1 0^{- 4}$ m $^{3}$ (a decrease of 200 cm $^{3}$ ).
Q9 · 2 marks · Very short answerHarder exam level
Explain why a beam supporting a load is placed with its larger dimension vertical (deep rather than wide).
Hint
Sag depends on depth cubed.
Answer
The sag of a loaded beam falls as $1/ (b d^{3})$ : increasing the depth d reduces sag far more than increasing the breadth b, so a deep beam bends less for the same amount of material.
Q10 · 3 marks · Short answerHOTS / competency
Two wires A and B of the same length are made of steel ( $Y = 2 \times 1 0^{11}$ Pa) and copper ( $Y = 1.1 \times 1 0^{11}$ Pa). A has twice the diameter of B. Find the ratio of their extensions under equal loads.
Hint
$Δ L \propto 1/ (d^{2} Y)$ .
Answer
$\frac{Δ L _{A}}{Δ L _{B}} = \frac{d _{B}^{2} Y _{B}}{d _{A}^{2} Y _{A}} = \frac{1 \times 1.1}{4 \times 2} \approx 0.14$ : steel wire A stretches about 0.14 times as much as B.
Q11 · 3 marks · Short answerHOTS / competency
Why do we use a factor of safety when designing a bridge, and why is a bridge designed to stay well inside the elastic region rather than up to the yield point?
Hint
Fatigue, unexpected loads, permanent set.
Answer
Real loads can exceed estimates (traffic jams, wind, earthquakes), materials have flaws, and repeated loading can weaken them. Working well inside the elastic region ensures the bridge always springs back; going past the yield point would leave a permanent sag and could lead to failure.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define stress and give its SI unit.Show answer
Restoring force per unit area; N m $^{- 2}$ (pascal).
2.State Hooke's law.Show answer
Within the proportional limit, stress is proportional to strain.
3.What is the yield point?Show answer
The point beyond which the material is permanently deformed; the stress there is the yield strength.
4.Write the formula for Young's modulus.Show answer
$Y = F L / (A Δ L)$ .
5.What is compressibility?Show answer
The reciprocal of the bulk modulus.
6.Define Poisson's ratio.Show answer
Lateral strain divided by longitudinal strain.
7.Write the elastic energy per unit volume.Show answer
½ × stress × strain.
8.Which state of matter has no shear modulus?Show answer
Fluids (liquids and gases).

Best order to study it

Learn elasticity, plasticity and the three kinds of stress.
Learn the three kinds of strain and Hooke's law.
Draw and explain the stress-strain curve; ductile, brittle and elastomers.
Learn Young's modulus and practise extension problems.
Learn the bulk modulus and compressibility; the shear modulus qualitatively.
Learn Poisson's ratio and the elastic energy of a stretched wire.
Finish with qualitative applications: ropes, beams, I-girders, mountains.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you name and define the three types of stress?
Can you explain why strain has no unit?
Can you draw and label the stress-strain curve?
Can you tell a ductile from a brittle material using the curve?
Can you find the extension of a wire using Y?
Can you compare extensions of two wires of different sizes?
Can you find the fractional volume change using B?
Can you calculate Poisson's ratio from data?
Can you derive the energy stored in a stretched wire?
Can you explain why girders are I-shaped?

Mechanical Properties of Solids: frequently asked questions

What was removed from this chapter for 2026-27?

The section on elastic behaviour of solids (old 9.2) and the experiment to determine Young's modulus were removed. The shear modulus and the applications of elasticity are needed only as qualitative ideas.

Is Poisson's ratio in the syllabus?

Yes. The CBSE content names Poisson's ratio, and NCERT section 8.5.4 defines it as lateral strain divided by longitudinal strain.

Why is the bulk modulus defined with a minus sign?

An increase in pressure makes the volume decrease, so ΔV is negative. The minus sign makes B a positive number.

Why do metals have large Young's moduli?

Their atoms are held by strong interatomic forces, so a large stress is needed for even a small strain.

How is this chapter tested?

In the school annual exam through definitions, the stress-strain curve, short numericals on Young's and bulk modulus, and reasoning questions on applications. The idea of Y also gives the speed of sound in solids in the Waves chapter.

Based on: CBSE Physics (042) syllabus 2026-27 (Classes XI-XII); NCERT Physics Part II, Textbook for Class XI, Chapter 8 (Reprint 2026-27)