Chapter 8 of 14

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Odisha (BSE)Class 11 Physics★ Unit VII · part of the 20-mark block covering Units VII to IX in CBSE Class 11 Physics (70 marks total); a reliable source of JEE Main numericals

Mechanical Properties of Solids

Every solid deforms under load — a steel girder, a concrete pillar, the bones in your leg. Usually you cannot see it happen, but it happens. This chapter puts numbers on how much, through stress, strain, and the three moduli that tell you how stubbornly a material resists being stretched, twisted or squeezed. CBSE Class 11 Physics Chapter 8.

⏱ 20 min readMedium difficulty✓ Free · NCERT-aligned

Worksheet

Podcast

Listen as a conversation

A teacher & a student talk through the chapter — in your language.

Mind Map

The chapter at a glance

Every concept and exam-favourite branch — revisable in 5 minutes.

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Study Guide

Night-before notes

Key points, glossary and the questions most likely to show up.

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VocabularyBuild your vocabularyLexiGenome mines the key words and schedules them with spaced repetition.Mine words →

By the end of this chapter you'll be able to…

1Define stress and strain, and say why strain has no units
2Tell tensile, shearing and hydraulic deformation apart from how the forces act
3State Hooke's law and the region of the curve where it actually holds
4Read a stress-strain curve for the elastic limit, yield point and ultimate strength
5Distinguish stiffness from strength, and ductile from brittle behaviour
6Apply Y = FL/(A dL) to wires, rods and columns
7Say which modulus applies to a given situation, and why liquids have no Young's modulus
8Use the bulk modulus for hydraulic compression without carrying the minus sign into the answer
9Define Poisson's ratio and compute the elastic energy stored in a stretched wire

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Why this chapter matters

Every solid you have met so far was rigid — it kept its shape whatever you did to it. That was a convenient lie, and this chapter drops it. A steel column holding up a building really is shorter than the unloaded column, and putting a number on that is the difference between a bridge that stands and one that does not.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Class 11 Physics: Laws of Motion

Free-body analysis and tension — needed to find the load on each wire

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Class 11 Physics: Work, Energy and Power

Work done against a varying force — where the half in the elastic energy comes from

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Class 11 Physics: Units and Measurements

Dimensional analysis — the fastest check that a modulus has come out in pascals

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Class 9 Science: Force and Laws of Motion

Force, weight and mass, which every stress calculation starts from

Mechanical Properties of Solids

1. What this chapter covers

Until now every body has been rigid — it kept its shape no matter what you did to it. That was always a convenient lie, and this chapter drops it.

Textbook section	Topic
8.1	Introduction — why rigid bodies are an idealisation
8.2	Stress and strain, and the three kinds of each
8.3	Hooke's law
8.4	The stress-strain curve
8.5	Elastic moduli — Young's, shear, bulk, Poisson's ratio, elastic potential energy
8.6	Applications of elastic behaviour of materials

Two carve-outs CBSE states explicitly

Topic	What the syllabus says
Shear modulus of rigidity	Qualitative idea only
Applications of elastic behaviour	Qualitative idea only

You still need to know what shear modulus is and when it applies — the exercises use it numerically — but the syllabus does not ask you to derive it.

2. Why a solid deforms at all

Apply forces to a body while it stays in static equilibrium, and it deforms. The textbook is blunt that the deformation may not be visible, but it is there — a steel column carrying a building really is shorter than the unloaded column, by about one part in a million.

The key idea is the restoring force. Deform a body and forces appear inside it that oppose the deformation, equal in magnitude and opposite in direction to what you applied. Those internal forces are what the whole chapter measures.

3. Stress and strain

Stress is the restoring force per unit area:

$stress = \frac{F}{A}$

Property	Value
SI unit	N m⁻², the pascal (Pa)
Dimensional formula	$[ML^{- 1} T^{- 2}]$

Strain is the fractional deformation — a change divided by an original. It is a ratio, so it is dimensionless and has no units. Half the errors in this chapter come from forgetting that.

The three ways a solid can be deformed

Type	How the forces act	Strain produced	Formula
Tensile / compressive (longitudinal)	Equal and opposite, perpendicular to opposite faces	Change in length	$Δ L / L$
Shearing	Equal and opposite, parallel to opposite surfaces	Change in shape, angle $θ$	$Δ x / L = θ$
Hydraulic	Perpendicular everywhere at once, same pressure all over	Change in volume	$Δ V / V$

This table decides which modulus a problem needs, and getting it wrong is the single most expensive mistake in the chapter. Ask one question: are the forces along the surface, across it, or all around it?

Across it, on two opposite faces → tensile or compressive → Young's modulus
Along the surface → shearing → shear modulus
All around, from a fluid → hydraulic → bulk modulus

A subtlety worth carrying

The chapter's Points to Ponder makes a point students routinely get wrong. Hang a weight $F$ from a wire fixed to the ceiling. The ceiling pulls up with $F$ and the weight pulls down with $F$ — so is the tension $2 F$ ?

No. The tension at any cross-section is $F$ , not $2 F$ , so the tensile stress is $F / A$ . Cut the wire anywhere and each half pulls the other with $F$ ; the two forces are what hold it in equilibrium, not something to be added together.

And stress is not a vector. Unlike a force, a stress cannot be assigned a direction — the same stress acts across a surface from both sides at once.

4. Hooke's law

For small deformations, stress is directly proportional to strain:

$stress \propto strain ⟹ stress = k \times strain$

The words "for small deformations" are load-bearing. Hooke's law is not a law of nature that always holds — it is valid only in the linear part of the stress-strain curve. Past that point it simply stops being true, and any calculation built on it stops being true with it.

The constant of proportionality is the modulus of elasticity, and it is a characteristic of the material rather than of the particular object.

5. Reading the stress-strain curve

This graph is worth more than any formula in the chapter, because almost every conceptual question comes from it.

Region or point	What it means
Proportional limit	End of the straight line. Hooke's law holds up to here and no further
Elastic limit (yield point)	Last point from which the material returns to its original shape on unloading
Yield strength $σ_{y}$	The stress at that yield point
Plastic region	Beyond it the deformation is permanent — a permanent set remains after unloading
Ultimate tensile strength $σ_{u}$	The maximum stress the material can bear — the peak of the curve
Fracture point	Where it breaks

Stiffness and strength are different properties, and this is where they separate.

Stiffness is the slope of the straight part. A stiff material resists deforming at all.
Strength is the height of the peak. A strong material resists breaking.

A material can be stiff and weak, or flexible and strong. Reading strength off the slope, or stiffness off the peak, is the classic error.

Ductile against brittle

Type	Behaviour	Example
Ductile	Large plastic region between yielding and fracture — deforms a lot before breaking, giving warning	Copper, mild steel
Brittle	Fracture occurs soon after the elastic limit, with almost no plastic region	Glass, cast iron

Elastomers are the odd case. Rubber and the elastic tissue of the aorta can be stretched to several times their length and still return — but they do not obey Hooke's law over most of that range, and they have no well-defined plastic region. The chapter shows the aorta's curve for exactly this reason.

6. The three elastic moduli

Young's modulus — resistance to stretching

$Y = \frac{tensile stress}{longitudinal strain} = \frac{F / A}{Δ L / L} = \frac{F L}{A Δ L}$

Strain is dimensionless, so Y carries the same units as stress — pascals.

Substance	$Y$ (10⁹ N m⁻²)
Steel	200
Iron (wrought)	190
Copper	110
Aluminium	70
Glass	65
Concrete	30
Wood	13
Bone	9.4

Metals have large Young's moduli, which is why a large force produces only a small change in length. The chapter's own illustration: increasing the length of a thin steel wire of 0.1 cm² cross-section by just 0.1% takes a force of 2000 N.

A warning about the word "elastic." In daily speech we call the thing that stretches more the more elastic one. The textbook calls this a misnomer. In physics, the material that stretches less under a given load is the more elastic — so steel is far more elastic than rubber, despite every intuition to the contrary.

Shear modulus — resistance to a change of shape

$G = \frac{shearing stress}{shearing strain} = \frac{F / A}{θ}$

CBSE marks this qualitative only, so you need what it means rather than a derivation. Shear moduli run roughly a third of the corresponding Young's modulus: steel 84 GPa, copper 42, aluminium 25, lead 5.6.

Where it catches people out: stretching a coiled spring. The spring gets longer, so it looks like a Young's modulus problem — but the wire itself is not getting longer, it is being twisted. That is shear, and the shear modulus governs it.

Bulk modulus — resistance to being squeezed

$B = - \frac{p}{Δ V / V}$

The minus sign is deliberate. Increasing the pressure decreases the volume, so $p$ and $Δ V$ always carry opposite signs — the minus makes $B$ come out positive. Never carry that minus into your final answer.

The reciprocal is the compressibility, $k = 1/ B$ .

Material	$B$ (10⁹ N m⁻²)
Nickel	260
Steel	160
Copper	140
Iron	100
Aluminium	72
Glass	37
Water	2.2
Air (at STP)	1.0 × 10⁻⁴

Read the bottom of that table. Solids are far harder to compress than liquids, and liquids are about twenty thousand times harder to compress than air — because a gas is mostly empty space, while in a liquid the molecules already sit almost touching.

Which moduli apply to what

Modulus	Solids	Liquids	Gases
Young's	Yes	No	No
Shear	Yes	No	No
Bulk	Yes	Yes	Yes

Young's and shear moduli need a definite length and a definite shape, and only solids have those. A liquid has neither, which is why only the bulk modulus survives for fluids.

Poisson's ratio

Stretch a wire and it gets thinner. The strain perpendicular to the force is the lateral strain, and within the elastic limit it is proportional to the longitudinal strain.

$σ = \frac{lateral strain}{longitudinal strain} = \frac{Δ d / d}{Δ L / L}$

It is a pure number with no units, depending only on the material — about 0.28 to 0.30 for steels, and around 0.33 for aluminium alloys.

Elastic potential energy in a stretched wire

Stretching a wire means doing work against the interatomic forces, and that work is stored as elastic potential energy.

$U = \frac{1}{2} \times F \times l u = \frac{1}{2} \times stress \times strain$

where $u$ is the energy stored per unit volume. The factor of one half appears for the same reason as in a spring — the force grows from zero to its final value as the stretch develops, so the average force doing the work is half the final one.

7. Why any of this matters in practice

CBSE marks this section qualitative only, so know the reasoning rather than the numbers.

Why bridges and buildings use girders shaped like the letter I. A beam under load sags, and the bending puts the top face in compression and the bottom face in tension, while the middle does almost nothing. So the material is concentrated in the top and bottom flanges where it is working, and the web between them is made thin. The result is a beam nearly as strong as a solid one for a fraction of the weight and cost.

Why a rope is made of many thin strands rather than one thick one. Twisted thin fibres share the load, and a flaw in one strand does not run through the whole rope.

Why cranes use thick steel cables. The maximum load is set by the stress the material can take, and stress is force over area — so carrying more load safely means more area.

Why mountains cannot grow past a certain height. At the base of a very tall mountain the compressive stress from the weight above would exceed what rock can bear, and the rock would flow. That sets a natural ceiling on mountain height on any given planet.

Summary

Rigid bodies were always an idealisation. Every real solid deforms under load, usually invisibly.
Deform a body and internal restoring forces appear, equal and opposite to what you applied.
Stress is restoring force per unit area, in Pa, with dimensions $[ML^{- 1} T^{- 2}]$ . Strain is a ratio and is dimensionless.
Three deformations: tensile or compressive (forces across opposite faces), shearing (forces along the surface), hydraulic (pressure all around).
Identify which one a problem is before choosing a modulus — that single decision is where most marks are lost.
Hang a weight from a wire and the tension at any cross-section is $F$ , not $2 F$ .
Stress is not a vector; it cannot be given a direction the way a force can.
Hooke's law holds only in the linear region of the stress-strain curve.
On that curve, the slope is stiffness and the height of the peak is strength. They are different properties.
Ductile materials deform a lot before fracture; brittle ones break soon after the elastic limit; elastomers stretch hugely without obeying Hooke's law.
$Y = F L / (A Δ L)$ — steel 200 GPa, copper 110, aluminium 70.
In physics the more elastic material is the one that stretches less. Calling rubber more elastic than steel is a misnomer.
$G = F / (A θ)$ governs shape change — and stretching a coiled spring is a shear problem, not a Young's modulus one.
$B = - p / (Δ V / V)$ ; the minus sign only keeps $B$ positive and never belongs in the answer. Compressibility is $1/ B$ .
Young's and shear moduli exist only for solids; the bulk modulus applies to solids, liquids and gases.
Poisson's ratio is dimensionless — about 0.28 to 0.30 for steel.
Energy stored per unit volume is $\frac{1}{2} \times$ stress $\times$ strain.
I-section girders put material where the bending stress is, which is why they are shaped that way.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Stress

stress = F/A,  in N/m^2 or Pa

Restoring force per unit area. Dimensional formula [M L^-1 T^-2]. Not a vector — a stress cannot be given a direction the way a force can.

The three strains

longitudinal dL/L; shearing dx/L = theta; volume dV/V

Every strain is a ratio, so all three are dimensionless and carry no units. Marks are lost by attaching units to a strain.

Hooke's law

stress = k x strain

Valid ONLY in the linear part of the stress-strain curve. Past the proportional limit it simply stops being true.

Young's modulus

Y = (F/A)/(dL/L) = F L /(A dL)

Resistance to stretching. Steel 200, iron 190, copper 110, aluminium 70, glass 65, concrete 30, wood 13, bone 9.4 (all in 10^9 Pa).

Shear modulus (modulus of rigidity)

G = (F/A)/theta

Resistance to a change of shape. CBSE marks this qualitative only. Roughly a third of Y: steel 84, copper 42, aluminium 25, lead 5.6 GPa.

Bulk modulus

B = -p/(dV/V)

Resistance to hydraulic compression. The minus sign exists only to make B positive, since a pressure rise shrinks the volume — never carry it into the answer.

Compressibility

k = 1/B

Fractional change in volume per unit increase in pressure — the reciprocal of the bulk modulus.

Bulk moduli worth knowing

steel 160, copper 140, glass 37, water 2.2, air 1.0e-4 (10^9 Pa)

Solids beat liquids, and liquids beat air by about 20,000 times, because a gas is mostly empty space while a liquid's molecules already almost touch.

Which modulus applies where

Young's and shear: solids only. Bulk: solids, liquids and gases.

Young's and shear moduli need a definite length and shape, and only a solid has those. This is a standard one-mark question.

Poisson's ratio

sigma = (dd/d)/(dL/L)

Lateral strain over longitudinal strain. A pure number with no units, depending only on the material — 0.28 to 0.30 for steels, about 0.33 for aluminium alloys.

Elastic potential energy in a stretched wire

U = (1/2) F l

The half appears because the force grows from zero to its final value as the stretch develops, so the average force doing the work is half the final one.

Elastic energy per unit volume

u = (1/2) x stress x strain = (1/2) Y (strain)^2

Energy density, in J/m^3. Since it goes as the square of the strain, doubling the extension stores four times the energy.

Stress in a wire hanging under its own weight

stress at the top = rho g L

Independent of the cross-sectional area, which is why a thicker wire is no help — the maximum hanging length is set by breaking stress divided by rho g.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Choosing the modulus from the answer rather than from the forces

✓ Decide by how the forces act. Perpendicular to two opposite faces means Young's modulus; parallel to the surface means shear modulus; pressure from all sides means bulk modulus. A shear problem can still ask for a distance, so the shape of the answer proves nothing.

WATCH OUT

✗ Giving strain a unit

✓ Every strain is a length over a length, or a volume over a volume, so all strains are pure numbers. Writing a strain in metres or as a percentage without converting is a guaranteed mark lost.

WATCH OUT

✗ Saying rubber is more elastic than steel

✓ In physics the more elastic material is the one that deforms LESS under a given load. Rubber stretches far more for the same stress, so its Young's modulus is smaller by a factor of thousands. The textbook explicitly calls the everyday usage a misnomer.

WATCH OUT

✗ Taking the tension in a hanging wire as 2F

✓ A wire fixed to the ceiling with weight F hanging from it has tension F at every cross-section, not 2F. The ceiling's pull and the weight are what keep it in equilibrium; they are not added together.

WATCH OUT

✗ Using Hooke's law beyond the linear region

✓ Hooke's law and every modulus derived from it hold only in the straight portion of the stress-strain curve. Beyond the proportional limit stress is no longer proportional to strain, so the whole calculation is invalid.

WATCH OUT

✗ Reading strength off the slope of the stress-strain curve

✓ The slope gives stiffness, which is Young's modulus. Strength is the height of the peak — the maximum stress before fracture. They are independent, and a material can be stiff but weak.

WATCH OUT

✗ Thinking the material that stretches further along the curve is stronger

✓ Extending a long way before breaking describes ductility, not strength. Strength is how high the curve reaches, not how far right it goes.

WATCH OUT

✗ Carrying the minus sign of the bulk modulus into the answer

✓ B = -p/(dV/V) is written with a minus precisely so that B comes out positive, since an increase in pressure produces a decrease in volume. A negative bulk modulus is always an error.

WATCH OUT

✗ Quoting a Young's modulus for water

✓ Liquids have no definite length or shape, so neither Young's modulus nor shear modulus is defined for them. Only the bulk modulus applies to a liquid or a gas.

WATCH OUT

✗ Treating a hollow column as a solid one

✓ The load-bearing area of a tube is the annulus, pi(R^2 - r^2), not pi R^2. Ignoring the hole overstates the area and understates the stress and strain.

WATCH OUT

✗ Forgetting to square the millimetre conversion in an area

✓ An area given in mm x mm must be converted with a factor of 10^-6 to reach m^2, not 10^-3. This single slip moves an answer by a factor of a million.

WATCH OUT

✗ Assuming a thicker wire can hang to a greater length

✓ The stress at the top of a wire hanging under its own weight is rho g L, with no area in it. Thickening the wire adds weight in exactly the same proportion as it adds area, so the maximum length is unchanged.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s

Exercises

Young's modulus ratios, reading stress-strain graphs, elongation of loaded wires, shear deflection, hollow columns, and bulk-modulus compression of water, glass and copper (Q8.1-8.16)

Questions

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Mechanical Properties of Solids?

10 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

10 questions~7 min worth ~20 marks in Odisha (BSE) exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Rigid bodies are an idealisation — every real solid deforms under load, usually invisibly.
•Deforming a body sets up internal restoring forces equal and opposite to the applied ones.
•Stress is restoring force per unit area, in N/m^2 or Pa, with dimensions [M L^-1 T^-2].
•Stress is not a vector — unlike a force it cannot be assigned a direction.
•Strain is a fractional change and is always dimensionless.
•Tensile or compressive stress: forces perpendicular to opposite faces, changing length.
•Shearing stress: forces parallel to opposite surfaces, changing shape.
•Hydraulic stress: pressure perpendicular everywhere at once, changing volume.
•A wire hung from the ceiling with weight F has tension F at every cross-section, not 2F.
•Hooke's law says stress is proportional to strain, and holds only in the linear region.
•Proportional limit ends Hooke's law; the elastic limit is the last point of full recovery.
•Beyond the elastic limit a permanent set remains after unloading.
•Ultimate tensile strength is the peak of the curve; fracture follows it.
•The slope of the curve is stiffness; the height of the peak is strength. They are independent.
•Ductile materials have a long plastic region (copper, mild steel); brittle ones break soon after yielding (glass, cast iron).
•Elastomers such as rubber and aortic tissue stretch enormously but do not obey Hooke's law over most of the range.
•Y = FL/(A dL). Steel 200, iron 190, copper 110, aluminium 70 GPa.
•The more elastic material is the one that stretches less — calling rubber more elastic than steel is a misnomer.
•G = F/(A theta) governs change of shape; CBSE marks it qualitative only.
•Stretching a coiled spring twists its wire, so the shear modulus governs it, not Young's modulus.
•B = -p/(dV/V); the minus sign only keeps B positive and never belongs in an answer.
•Compressibility k = 1/B.
•Bulk moduli: steel 160, copper 140, glass 37, water 2.2, air 1.0e-4 GPa.
•Solids resist compression far more than liquids, and liquids about 20,000 times more than air, because a gas is mostly empty space.
•Young's and shear moduli apply only to solids; the bulk modulus applies to solids, liquids and gases.
•Poisson's ratio is lateral strain over longitudinal strain — a pure number, 0.28 to 0.30 for steels.
•Elastic energy U = (1/2) F l, and per unit volume u = (1/2) x stress x strain.
•Energy goes as the square of the extension, so doubling the stretch stores four times the energy.
•Stress at the top of a wire hanging under its own weight is rho g L, independent of thickness.
•I-section girders put material in the top and bottom flanges where the bending stress is greatest.

Odisha (BSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: Unit VII sits inside the 20-mark block covering Units VII to IX (CBSE Class 11 Physics, 70 marks)

Question type	Marks each	Typical count	What it tests
Elastic moduli	3-5	1	Young's modulus numericals on wires, rods and columns
Stress and strain	2-3	1	Definitions, units, dimensions, and identifying the type of deformation
Stress-strain curve	2-3	1	Elastic limit, yield point, ultimate strength, ductile against brittle
Bulk modulus	2-3	1	Hydraulic compression and fractional volume change
Elastic energy	2-3	1	Energy stored in a stretched wire and energy per unit volume

Prep strategy

Before touching a formula, decide whether the deformation is tensile, shearing or hydraulic — that choice picks the modulus
Learn the stress-strain curve as a labelled diagram; most conceptual marks in this chapter come from it
Keep stiffness and strength separate: slope against peak height
Convert mm to m before squaring for an area, and remember strain has no units
Note that shear modulus and the applications section are marked qualitative only by CBSE

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Why bridge girders are I-shaped

Bending compresses a beam's top and stretches its bottom while the middle does almost nothing, so the material is moved into flanges where the stress actually is.

The height limit on mountains

At the base of a very tall mountain the compressive stress from the weight above would exceed what rock can bear, so it would flow. That sets a ceiling on how high mountains can grow.

Ropes made of many thin strands

Twisted fibres share the load between them, and a flaw in one strand does not propagate through the whole rope the way a crack would through a single thick one.

Why water is treated as incompressible

Its bulk modulus of 2.2 GPa means 80 atmospheres of pressure changes the density by only 0.4 per cent — the quantitative justification for an assumption used throughout fluid mechanics.

Bone as a structural material

Bone has a Young's modulus of about 9.4 GPa but an ultimate strength of 170 MPa — flexible compared with steel yet strong for its weight, which is why skeletons work.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Read the geometry of the forces before writing anything: perpendicular to faces means Young's, along the surface means shear, all around means bulk. That one decision determines the whole answer.

Convert millimetres to metres before squaring for an area. A cross-section in mm squared needs a factor of 10 to the minus six, not minus three.

Never attach a unit to a strain, a fractional volume change or Poisson's ratio — all three are pure numbers.

For a hollow column, use the annulus pi(R squared minus r squared). Forgetting the hole is the most common numerical slip in this chapter.

Check whether a question asks for a fraction or an actual amount. Fractional volume change needs no dimensions at all; a volume contraction must be multiplied by the body's volume.

In a stress-strain graph question, state which feature you are reading — slope for modulus, peak for strength — before quoting a number.

If a bulk modulus answer comes out negative, you have carried the minus sign of the definition into the result.

For a loaded arrangement of several wires, work out what hangs below each one before computing any stress. The upper wire always carries more.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

STRETCH

Derive the elongation of a uniform rod hanging under its own weight and show it equals half the value obtained by concentrating the whole weight at the free end.

STRETCH

Find the depression at the centre of a horizontal wire fixed at both ends when a mass is hung from its midpoint, for small depressions.

STRETCH

Show that the elastic energy density is (1/2) x stress x strain by integrating the work done during stretching.

STRETCH

Relate Young's modulus, bulk modulus, shear modulus and Poisson's ratio for an isotropic solid, and check the values for steel are mutually consistent.

STRETCH

Analyse a compound bar of two different materials rigidly joined side by side and loaded axially, using equal strain to divide the load.

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JEE Main & Advanced practice

Competitive-level problems on this chapter, above the board pattern. Try each one on paper before opening the solution.

JEE MainYoung's modulusSingle correct

Two wires A and B are made of the same material. Wire B has twice the length and twice the diameter of wire A. If the same load is hung from each, the ratio of the elongation of A to that of B is:

(a) $1 : 1$ (b) $2 : 1$ (c) $1 : 2$ (d) $4 : 1$

Stuck? Show the approach

Write $Δ L = F L / (A Y)$ and track how each symbol scales. Remember that doubling the diameter multiplies the area by four, not by two — that is the whole question.

Show the full solution

For a wire under load,

$Δ L = \frac{F L}{A Y}$

The material is the same, so $Y$ is common, and the load $F$ is the same.

For wire A: $Δ L_{A} = \frac{F L}{A Y}$

For wire B: the length is $2 L$ , and since the diameter doubles the area becomes $4 A$ :

$Δ L_{B} = \frac{F ( 2 L )}{( 4 A ) Y} = \frac{F L}{2 A Y} = \frac{Δ L _{A}}{2}$

Therefore

$\frac{Δ L _{A}}{Δ L _{B}} = 2$

Answer: (b) 2 : 1

The trap

Scaling the area with the diameter instead of its square. Doubling the diameter gives four times the area, so the length and area effects do not cancel — they leave a factor of two.

JEE MainElastic energySingle correct

A wire is stretched by 1 mm within its elastic limit and stores elastic potential energy $U$ . If the same wire is stretched by 3 mm instead, the energy stored becomes:

(a) $3 U$ (b) $6 U$ (c) $9 U$ (d) $U /3$

Stuck? Show the approach

Elastic energy is not proportional to the extension — it is proportional to its square, for the same reason a spring's energy goes as $x^{2}$ .

Show the full solution

The energy stored in a stretched wire is

$U = \frac{1}{2} F l$

while the force itself is proportional to the extension, $F = (\frac{Y A}{L}) l$ . Substituting,

$U = \frac{Y A}{2 L} l^{2}$

so $U \propto l^{2}$ for a given wire. Tripling the extension therefore multiplies the energy by $3^{2} = 9$ .

$\frac{U _{2}}{U _{1}} = (\frac{3 mm}{1 mm})^{2} = 9$

Answer: (c) 9U

The trap

Reading $U = \frac{1}{2} F l$ as linear in $l$ and answering $3 U$ . The force is not constant — it grows as the wire stretches, so both factors scale and the energy goes as the square.

JEE MainBreaking stressNumerical

A steel wire has a breaking stress of $7.8 \times 1 0^{8}$ N m⁻² and density $7800$ kg m⁻³. Find the maximum length of the wire that can hang vertically from a support without breaking under its own weight. Take $g = 9.8$ m s⁻².

Stuck? Show the approach

The most heavily stressed point is the very top, which carries everything below it. Write the stress there and watch what happens to the cross-sectional area.

Show the full solution

Consider a wire of cross-sectional area $A$ , length $L$ and density $ρ$ . The top cross-section supports the entire weight of the wire below it:

$W = (ρ A L) g$

So the stress at the top is

$σ = \frac{W}{A} = \frac{ρ A Lg}{A} = ρ g L$

The area cancels completely. Setting this equal to the breaking stress:

$L_{ma x} = \frac{σ _{b r e ak}}{ρ g} = \frac{7.8 \times 1 0 ^{8}}{7800 \times 9.8} = \frac{7.8 \times 1 0 ^{8}}{7.644 \times 1 0 ^{4}}$

$L_{ma x} \approx 1.02 \times 1 0^{4} m$

Answer: About 1.02 × 10⁴ m (roughly 10 km)

The trap

Assuming a thicker wire could hang further. Doubling the area doubles the weight it must carry, so the stress at the top is unchanged — the maximum length depends only on the material, never on the thickness.

JEE AdvancedElongation under self-weightNumerical

A uniform steel rod of length $L = 10$ m, density $ρ = 7860$ kg m⁻³ and Young's modulus $Y = 2.0 \times 1 0^{11}$ Pa hangs vertically from a support. Find the elongation produced by its own weight. Take $g = 9.8$ m s⁻².

Stuck? Show the approach

The tension is not the same everywhere — it is largest at the top and zero at the bottom. So you cannot use $Δ L = F L / A Y$ directly. Take a small element, find the load it carries, and integrate.

Show the full solution

Step 1 — Set up the element. Take an element of thickness $d x$ at a distance $x$ measured from the lower end. The weight it must support is the weight of everything below it:

$F (x) = ρ A g x$

Step 2 — Extension of that element. Applying the definition of Young's modulus to the element of original length $d x$ :

$d (Δ L) = \frac{F ( x ) d x}{A Y} = \frac{ρ A g x d x}{A Y} = \frac{ρ g x}{Y} d x$

The area cancels, which is why the answer will not contain it.

Step 3 — Integrate over the rod.

$Δ L = \int_{0}^{L} \frac{ρ g x}{Y} d x = \frac{ρ g}{Y} [\frac{x ^{2}}{2}]_{0}^{L} = \frac{ρ g L ^{2}}{2 Y}$

Step 4 — Substitute.

$Δ L = \frac{7860 \times 9.8 \times 1 0 ^{2}}{2 \times 2.0 \times 1 0 ^{11}} = \frac{7.70 \times 1 0 ^{6}}{4.0 \times 1 0 ^{11}}$

$Δ L \approx 1.93 \times 1 0^{- 5} m$

Note the structure of the result. It equals what you would get by hanging the rod's entire weight $ρ A Lg$ from a weightless rod and halving it — because the average tension along the rod is half the maximum.

Answer: ≈ 1.93 × 10⁻⁵ m

The trap

Using $Δ L = F L / (A Y)$ with $F$ equal to the rod's full weight, which doubles the answer. The load varies linearly from zero at the bottom to the full weight at the top, and only integration (or using the average) handles that.

JEE AdvancedComposite wiresMultiple correct

Two wires of the same material and the same length, with radii $r$ and $2 r$ , are joined end to end. A force $F$ is applied at the free ends. Which of the following are correct?

(a) The stress in the thinner wire is 4 times that in the thicker (b) The strain in the thinner wire is 4 times that in the thicker (c) The elongation of the thinner wire is 4 times that of the thicker (d) The elastic energy stored in the thinner wire is 4 times that in the thicker

Stuck? Show the approach

End to end means the wires are in series, so the same force passes through both — this is the fact everything else follows from. Then work through stress, strain, elongation and energy in that order.

Show the full solution

The key fact: joined end to end, both wires carry the same force $F$ .

(a) Stress. $σ = F / A$ and $A = π r^{2}$ , so

$\frac{σ _{t hin}}{σ _{t hi c k}} = \frac{π ( 2 r ) ^{2}}{π r ^{2}} = 4 ✓$

(b) Strain. $ε = σ / Y$ , and the material is the same so $Y$ is common:

$\frac{ε _{t hin}}{ε _{t hi c k}} = 4 ✓$

(c) Elongation. $Δ L = ε L$ , and the lengths are equal:

$\frac{Δ L _{t hin}}{Δ L _{t hi c k}} = 4 ✓$

(d) Energy. Here the answer changes. $U = \frac{1}{2} F Δ L$ , and $F$ is the same in both wires, so the energy ratio follows the elongation ratio alone:

$\frac{U _{t hin}}{U _{t hi c k}} = \frac{Δ L _{t hin}}{Δ L _{t hi c k}} = 4$

So (d) is also correct — but for a different reason than a student expects. Had the energy density been asked instead, the ratio would be $16$ , since $u = \frac{1}{2} σ ε$ and both $σ$ and $ε$ scale by 4.

All four statements are correct.

Answer: (a), (b), (c) and (d)

The trap

Assuming series wires share the stress equally, or reaching for $u = \frac{1}{2} σ ε$ when the question asks for total energy rather than energy per unit volume. Total energy uses $\frac{1}{2} F Δ L$ with $F$ common, giving 4; energy density would give 16.

JEE AdvancedBulk modulusNumerical

The density of water at the surface is $1.03 \times 1 0^{3}$ kg m⁻³ and the bulk modulus of water is $2.2 \times 1 0^{9}$ Pa. At what depth in a lake would the density of water be greater than the surface value by $0.10%$ ? Take $g = 9.8$ m s⁻² and neglect the variation of $g$ with depth.

Stuck? Show the approach

Work backwards. A fractional rise in density corresponds to a fractional fall in volume, which the bulk modulus converts to a pressure — and pressure at depth $h$ is $ρ g h$ .

Show the full solution

Step 1 — Convert the density change to a volume change. Mass is conserved when water is compressed, so $ρ V$ is constant. Differentiating,

$\frac{Δ ρ}{ρ} = - \frac{Δ V}{V}$

A density increase of $0.10%$ therefore means a volume decrease of the same fraction:

$\frac{Δ V}{V} = 1.0 \times 1 0^{- 3}$

Step 2 — Find the pressure that produces it.

$Δ p = B \frac{Δ V}{V} = (2.2 \times 1 0^{9}) (1.0 \times 1 0^{- 3}) = 2.2 \times 1 0^{6} Pa$

Step 3 — Convert pressure to depth. The gauge pressure at depth $h$ is $Δ p = ρ g h$ , so

$h = \frac{Δ p}{ρ g} = \frac{2.2 \times 1 0 ^{6}}{( 1.03 \times 1 0 ^{3} ) ( 9.8 )} = \frac{2.2 \times 1 0 ^{6}}{1.009 \times 1 0 ^{4}}$

$h \approx 218 m$

Sense check. A couple of hundred metres of water changes its density by only a tenth of a per cent, which is the quantitative reason water is treated as incompressible in almost every problem you will meet.

Answer: ≈ 2.2 × 10² m (about 218 m)

The trap

Setting the fractional density change equal to $+ Δ V / V$ and losing the sign, or forgetting that mass conservation is what links the two fractions at all. Density rises exactly as much as volume falls, in fractional terms.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh

JEE Main and Advanced (Elasticity)High

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Because every strain is one quantity divided by another of the same kind. Longitudinal strain is a change in length divided by an original length, so the metres cancel. Volume strain is a volume divided by a volume, and shearing strain is a sideways displacement divided by a height. In each case the units disappear, leaving a pure number. This is also why Young's modulus, the bulk modulus and the shear modulus all carry the same units as stress — dividing by a dimensionless strain does not change the units.

Yes, and the textbook says so directly, calling the everyday usage a misnomer. In physics, elasticity measures how strongly a material resists being deformed and how completely it returns to its original shape. Young's modulus is stress divided by strain, so a material that stretches very little under a large stress has a large modulus. Steel's is about 200 GPa; rubber's is of the order of 10 MPa, smaller by a factor of thousands. Rubber stretching further makes it less stiff, not more elastic.

Look at how the forces act on the body, not at what the answer will be. If two equal and opposite forces act perpendicular to opposite faces and change the length, it is Young's modulus. If they act parallel to the surface and change the shape without changing the volume, it is the shear modulus. If pressure acts perpendicular to every surface at once and changes the volume, it is the bulk modulus. A common trap is a shear problem that asks for a deflection in metres, which tempts students toward Young's modulus because the answer is a length.

Because pressure and volume change in opposite directions. Increasing the pressure on a body always decreases its volume, so if p is positive then the volume change is negative. Without the minus sign the ratio would come out negative, and a modulus that measures resistance to compression ought to be positive. The minus sign in B = -p/(dV/V) is there purely to fix that. It has done its job inside the definition, so a negative value in your final answer is always a mistake.

Young's modulus and the shear modulus both describe resistance to a change in length or shape, and both need the body to have a definite length and shape to begin with. A liquid has neither — it takes the shape of whatever holds it, and it cannot sustain a shearing stress at all, since it simply flows instead. What a liquid does have is a definite volume, and it resists having that volume reduced. So the bulk modulus is the only one of the three that is defined for liquids and gases.

No, and this surprises most students. The stress at the top of a hanging wire is the weight below divided by the cross-sectional area. That weight is density times area times length times g, so the area cancels and the stress is simply rho g L. Making the wire thicker adds exactly as much weight as it adds area, leaving the stress unchanged. The greatest length depends only on the material's breaking stress and its density — for steel it works out at about 10 km.

When a beam bends under load, its top surface is compressed and its bottom surface is stretched, while the material near the middle experiences very little stress and contributes almost nothing to the strength. An I-section concentrates the material in the top and bottom flanges, where the stress is greatest, and thins it down to a narrow web in between. The result is a beam almost as strong as a solid rectangular one but far lighter and cheaper. CBSE marks this section qualitative only, so the reasoning matters rather than any calculation.

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