Mechanical Properties of Solids Notes for NEET — Class 11 NCERT

Introduction

Solids resist deformation. Stretch a steel wire or compress a rubber block, and it pushes back. The relationship between the force you apply and the deformation it produces — within reason — is what this chapter is about.

For NEET 2027, expect 1 to 2 questions. Young's modulus, the stress-strain curve, bulk modulus and elastic potential energy in a wire are the heavy hitters. Most questions are formula-driven — get the units and signs right and you will lock in marks.

Elastic and plastic behaviour

Apply a small force to a body and it deforms. Remove the force:

If the body returns to its original shape, the deformation is elastic.
If permanent deformation is left behind, the deformation is plastic.

The elastic limit is the maximum stress beyond which permanent deformation begins. Below it, behaviour is elastic; above it, plastic.

Stress and strain

Stress

Restoring force per unit area inside a deformed body:

σ = \frac{F}{A} .

SI unit: pascal (Pa) = N/m². Dimensions $[M L^{- 1} T^{- 2}]$ .

Three types of stress

Longitudinal (tensile or compressive): force perpendicular to the cross-section, along the length.
Shearing (tangential): force parallel to one face — produces angular distortion.
Volumetric (hydrostatic): equal pressure on all sides.

Strain

Fractional change in dimension. Dimensionless.

Longitudinal strain = \frac{Δ L}{L} .

Shearing strain = tan θ \approx θ (small angle) .

Volumetric strain = \frac{Δ V}{V} .

Hooke's law

Within the proportional limit, stress is directly proportional to strain:

σ = E ε,

where $E$ is the appropriate elastic modulus (Young's, shear or bulk) for the type of deformation. $E$ has the same units as stress.

The stress-strain curve

For a typical metal in tension:

O → A (proportional region): straight line; Hooke's law holds.
A → B (elastic region): still elastic but slightly nonlinear. $B$ is the elastic limit.
B → C (yield region): permanent (plastic) deformation begins. The yield point is at $B$ .
C → D (plastic / strain hardening): material continues to take more stress before failing.
D (ultimate strength): highest stress the material can withstand.
E (fracture point): material breaks.

Brittle materials (cast iron, glass) have very small plastic regions; they fracture soon after the elastic limit. Ductile materials (copper, steel, gold) have large plastic regions before fracture.

Strain marker: 15% of fracture strain

Current region

Proportional (Hooke's law)

Young's modulus

For longitudinal stress and strain:

Y = \frac{longitudinal stress}{longitudinal strain} = \frac{F L}{A Δ L} .

Steel: $Y \approx 2 \times 1 0^{11} Pa$ , copper: $\approx 1.1 \times 1 0^{11}$ , aluminium: $\approx 7 \times 1 0^{10}$ , rubber: $\approx 1 0^{7}$ . Higher Y → stiffer material.

F (N)

L (m)

A (m²)

ΔL (m)

Y (Pa)

Answer

Y = 2.0000e+11 Pa

Y = \frac{F L}{A Δ L} = \frac{100 \times 2}{0.000001 \times 0.001} = 2.000 e + 11

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Shear modulus (modulus of rigidity)

For shearing stress and strain:

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

For most materials, $G \approx Y /3$ . Liquids and gases have $G = 0$ (they cannot sustain shear).

Bulk modulus

For volumetric stress (pressure) and volumetric strain:

B = - \frac{Δ P}{Δ V / V} .

The minus sign keeps $B$ positive (volume decreases when pressure increases). Reciprocal of bulk modulus is the compressibility $K = 1/ B$ . Liquids and solids have large $B$ ; gases have small $B$ and are highly compressible.

Bulk modulus B: 2.20e+9 Pa

Pressure increase ΔP: 10 MPa

Quick presets

Fractional volume change

ΔV/V = 4.545e-1%

\frac{Δ V}{V} = \frac{Δ P}{B} = \frac{1.00 e + 7}{2.20 e + 9}

Compressibility (1/B)

4.545e-10 Pa⁻¹

Poisson's ratio

Stretch a wire and it gets a little thinner. The ratio:

σ = - \frac{lateral strain}{longitudinal strain} .

Theoretical limits: $- 1 \leq σ \leq 0.5$ . For most materials, $σ$ is between $0.2$ and $0.5$ . Steel: $\sim 0.3$ , rubber: $\sim 0.5$ (nearly incompressible).

Longitudinal strain (stretch): 2.0%

Poisson's ratio σ: 0.30

Steel ≈ 0.3 · Concrete ≈ 0.2 · Aluminium ≈ 0.33 · Rubber ≈ 0.5

Lateral strain

-0.600%

Volume change ΔV/V

0.800%

σ = - \frac{ε _{lat}}{ε _{long}} \Rightarrow ε_{lat} = - σ ε_{long}

Elastic potential energy in a stretched wire

Work done in stretching a wire (force varies linearly from 0 to $F$ ):

U = \frac{1}{2} F Δ L = \frac{1}{2} (stress) (strain) (volume) .

Per unit volume (energy density):

u = \frac{U}{V} = \frac{1}{2} σ ε = \frac{1}{2} Y ε^{2} = \frac{σ ^{2}}{2 Y} .

Young's modulus Y: 200 GPa

Rubber ≈ 0.01 GPa · Bone ≈ 14 GPa · Aluminium ≈ 70 GPa · Steel ≈ 200 GPa

Strain ε: 0.100%

Volume: 10 cm³

Energy density u

1.000e+5 J/m³

Total energy U

1.0000 J

Stress σ

2.00e+8 Pa

u = \frac{1}{2} Y ε^{2} = \frac{1}{2} (200 \times 1 0^{9}) (1.00 e - 3)^{2}

U = u V = 1.00 e + 5 \times 10 \times 1 0^{- 6} = 1.0000 J

Wires in series and in parallel

Series — same force, different elongations add

Two wires of the same area $A$ joined end to end, with lengths $L_{1}, L_{2}$ and moduli $Y_{1}, Y_{2}$ :

Δ L = Δ L_{1} + Δ L_{2} = \frac{F L _{1}}{A Y _{1}} + \frac{F L _{2}}{A Y _{2}} .

Parallel — same elongation, forces add

Two wires sharing a load with both ends fixed to the same supports:

F_{total} = F_{1} + F_{2} = \frac{Y _{1} A _{1}}{L _{1}} Δ L + \frac{Y _{2} A _{2}}{L _{2}} Δ L .

The stiffer wire (larger $Y A / L$ ) carries more of the load.

Total force F: 100 N

Wire 1 (steel-ish)

L₁: 1.0 m

r₁: 1.0 mm

Y₁: 200 GPa

Wire 2 (copper-ish)

L₂: 1.0 m

r₂: 1.0 mm

Y₂: 110 GPa

Total elongation

ΔL = 0.449 mm

ΔL₁

0.159 mm

ΔL₂

0.289 mm

Δ L = \frac{F L _{1}}{A _{1} Y _{1}} + \frac{F L _{2}}{A _{2} Y _{2}}

Quick reference of elastic moduli

Approximate values at room temperature. All in GPa (10⁹ Pa). Use these to sanity-check NEET problems.

Material

Y (GPa)

G (GPa)

B (GPa)

σ (Poisson)

Steel

200

160

0.3

Copper

110

140

0.34

Brass

100

110

0.34

Aluminium

0.33

Bone

—

0.3

Granite

—

0.25

Concrete

—

0.2

Wood (oak)

—

Rubber

0.05

—

0.5

Worked NEET problems

NEET-style problem · Young's modulus

Question

A steel wire of length

2 m

and cross-sectional area

1 0^{- 6} m^{2}

is stretched by

1 mm

under a load of

100 N

. Find Young's modulus.

Solution

Y = \frac{F L}{A Δ L} = \frac{( 100 ) ( 2 )}{( 1 0 ^{- 6} ) ( 1 0 ^{- 3} )} = \frac{200}{1 0 ^{- 9}} = 2 \times 1 0^{11} Pa .

NEET-style problem · Elastic PE

Question

A wire of Young's modulus

Y

and cross-section

A

is stretched by

Δ L

from its natural length

L

. Find the energy stored.

Solution

Force in the stretched wire: $F = (Y A / L) Δ L$ .

U = \frac{1}{2} F Δ L = \frac{Y A ( Δ L ) ^{2}}{2 L} .

NEET-style problem · Bulk modulus

Question

A liquid of bulk modulus

2 \times 1 0^{9} Pa

is subjected to an additional pressure of

2 \times 1 0^{6} Pa

. The fractional decrease in volume is:

Solution

\frac{Δ V}{V} = \frac{Δ P}{B} = \frac{2 \times 1 0 ^{6}}{2 \times 1 0 ^{9}} = 1 0^{- 3} = 0.001 = 0.1%.

NEET-style problem · Stress comparison

Question

Two wires of the same material have radii in ratio

1 : 2

and lengths in ratio

1 : 1

. They are stretched by the same force. The ratio of their elongations is:

Solution

Elongation: $Δ L = F L / (A Y)$ — same $F$ , $L$ , $Y$ .

\frac{Δ L _{1}}{Δ L _{2}} = \frac{A _{2}}{A _{1}} = (\frac{r _{2}}{r _{1}})^{2} = 4.

The thinner wire stretches 4× more — its area is one-quarter.

NEET-style problem · Energy density

Question

The energy density (elastic PE per unit volume) in a wire stretched to a strain of

0.01

with Young's modulus

2 \times 1 0^{11} Pa

is:

Solution

u = \frac{1}{2} Y ε^{2} = \frac{1}{2} (2 \times 1 0^{11}) (0.01)^{2} = \frac{1}{2} (2 \times 1 0^{11}) (1 0^{- 4}) = 1 0^{7} J/m^{3} .

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Summary cheat sheet

Stress: $σ = F / A$ , units Pa. Strain: $ε = Δ L / L$ , dimensionless.
Hooke's law: $σ = E ε$ within the proportional limit.
Young's modulus: $Y = (F L) / (A Δ L)$ . Steel ≈ 2×10¹¹ Pa.
Shear modulus: $G = (F / A) / tan θ$ , fluids G = 0.
Bulk modulus: $B = - Δ P / (Δ V / V)$ . Compressibility = 1/B.
Poisson's ratio: $σ = - ε_{lat} / ε_{long}$ . Range: −1 to +0.5.
Elastic PE: $U = \frac{1}{2} F Δ L = \frac{Y A ( Δ L ) ^{2}}{2 L}$ .
Energy density: $u = \frac{1}{2} Y ε^{2} = σ^{2} / (2 Y)$ .
Wires in series: elongations add, force same. In parallel: forces add, elongation same.

Next: try the interactive widgets for stress-strain curves, Young's modulus and elastic PE, or work through the 30+ NEET PYQs with full solutions. To time yourself, take the free 10-question mock test.

Frequently asked questions

How many questions come from Mechanical Properties of Solids in NEET 2027?

You can expect 1 to 2 questions from this chapter in NEET 2027. The chapter has medium PYQ frequency. Young's modulus calculations, the stress-strain curve, bulk modulus and elastic PE in a wire are the most heavily tested concepts.

What is the difference between stress and strain?

Stress is the restoring force per unit area inside the body, with units of pascal (N per m squared). Strain is the fractional deformation, dimensionless. Stress is the cause; strain is the response. Hooke's law says they are proportional within the elastic region: stress equals modulus times strain.

What is Young's modulus and what does its value mean?

Young's modulus Y equals longitudinal stress divided by longitudinal strain. It measures resistance to length change under axial load. Steel has Y about 2 times 10 to the 11 pascals; rubber has Y about 1 times 10 to the 7 pascals. Higher Y means stiffer material — more force needed for the same fractional stretch.

How is bulk modulus different from Young's modulus?

Bulk modulus B equals volumetric stress (pressure) divided by volumetric strain (fractional volume change). It measures resistance to compression from all sides — relevant for fluids and solids under hydrostatic pressure. Young's modulus describes change in length under axial load. They are different quantities of the same material.

What is Poisson's ratio?

When you stretch a wire, it gets longer along the pull and slightly thinner across. Poisson's ratio sigma equals minus the lateral strain divided by the longitudinal strain. For most materials, sigma is between 0 and 0.5. Steel about 0.3, rubber close to 0.5 (nearly incompressible).

What is the elastic potential energy stored in a stretched wire?

A wire stretched within its elastic limit stores energy U equals one half times stress times strain times volume, equivalently one half times Y times strain squared times volume. Per unit volume, the energy density u equals one half times Y times strain squared. NEET asks this in two forms.

How do you find the equivalent Young's modulus for two wires in series and in parallel?

In series (same force, total elongation = sum), 1 over Y_eq equals one half of (1 over Y_1 plus 1 over Y_2) for equal cross-sections and lengths. In parallel (same elongation, total force = sum), Y_eq equals (Y_1 plus Y_2) over 2 for equal cross-sections. The detailed formula depends on lengths and areas — be careful in NEET problems.

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Mechanical Properties of SolidsNEET Physics · Class 11 · NCERT Chapter 8

Introduction

Elastic and plastic behaviour

Stress and strain

Stress

Three types of stress

Strain

Hooke's law

The stress-strain curve

Young's modulus

Shear modulus (modulus of rigidity)

Bulk modulus

Poisson's ratio

Elastic potential energy in a stretched wire

Wires in series and in parallel

Series — same force, different elongations add

Parallel — same elongation, forces add

Quick reference of elastic moduli

Worked NEET problems

Summary cheat sheet

Frequently asked questions

How many questions come from Mechanical Properties of Solids in NEET 2027?

What is the difference between stress and strain?

What is Young's modulus and what does its value mean?

How is bulk modulus different from Young's modulus?

What is Poisson's ratio?

What is the elastic potential energy stored in a stretched wire?

How do you find the equivalent Young's modulus for two wires in series and in parallel?

Continue with the next chapter notes

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