JEE Mains Physics · Mechanical Properties of Solids

Stress, Strain and Young's Modulus

A wire under a tension F stretches by ΔL = FL/(AY): stress F/A divided by strain ΔL/L is Young's modulus Y, a number fixed by the material and not by the size of the wire.

All Mechanical Properties of Solids notes JEE Mains Physics strategy

Why this matters

Twenty-seven PYQs, eight of them asking for a number, and four from 2026. Nine compute a stress, a strain or an extension, nine compare two wires by ratio, and nine read Young's modulus off a graph or ask what it depends on. Most of the arithmetic is unit work: a diameter turned into an area, and mm² or cm² turned into m².

Concept 1 of 3: Stress, strain and the extension of a wire

Stress is the force on each square metre of the cross-section, and strain is the stretch on each metre of length. Within the elastic limit the two are proportional, and the constant is Young's modulus. So a wire stretches more when it is longer, thinner or pulled harder, and less when its material is stiffer.

Definition

Stress $= F/A$ , longitudinal strain $= \Delta L/L$ , and $Y = \dfrac{F/A}{\Delta L/L}$ , so $\Delta L = \dfrac{FL}{AY}$ .
Area from the radius or the diameter: $A = \pi r^{2} = \pi d^{2}/4$ . Units: $1\ \text{mm}^{2} = 10^{-6}\ \text{m}^{2}$ , $1\ \text{cm}^{2} = 10^{-4}\ \text{m}^{2}$ .
A hollow column of radii r and R has $A = \pi(R^{2} - r^{2})$ . A load shared equally by n columns puts $Mg/n$ on each.
A wire pulled at both ends by F carries a tension F, not 2F.
A load m on a planet with gravity g′ pulls with mg′, so the same wire and load stretch in proportion to g′.
Error in Y from $Y = \dfrac{mgL}{\pi r^{2}\,\Delta L}$ : $\dfrac{\delta Y}{Y} = \dfrac{\delta m}{m} + \dfrac{\delta L}{L} + 2\dfrac{\delta r}{r} + \dfrac{\delta(\Delta L)}{\Delta L}$ . The radius counts twice.

Young's modulus

Y = \frac{F/A}{\Delta L/L} \qquad \Delta L = \frac{FL}{AY}

Worked example

A steel wire 2.5 m long with a cross-section of

1.5\ \text{mm}^{2}

carries a 12 kg load.

Y = 2 \times 10^{11}\ \text{N/m}^{2}

g = 10\ \text{m/s}^{2}

. Find the stress, the strain and the extension.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q27Moderate

Example 1 · Mechanical Properties of Solids · Stress, Strain and Young's Modulus

Two persons pull a wire towards themselves. Each person exerts a force of

200\text{ }N

on the wire. Young's modulus of the material of wire is

1 \times10^{11}\text{ }N{\text{ }m}^{- 2}

. Original length of the wire is

2\text{ }m

and the area of cross section is

2{\text{ }cm}^{2}

. The wire will extend in length by ___

\mu m

Pulled from both ends is not twice the tension

Two people pulling the ends with F each give a tension F, the same as a wall at one end and one person pulling with F. Using 2F doubles the answer.

A diameter put in as a radius

A = πd²/4. Putting the diameter into πr² makes the area four times too large and the extension four times too small.

Squared units

1 mm² is 10⁻⁶ m² and 1 cm² is 10⁻⁴ m². Converting only the length (10⁻³, 10⁻²) is the commonest slip.

Drill 8 more on stress, strain and the extension of a wire

Concept 2 of 3: Comparing two wires by ratio

When a question compares two wires, write ΔL = FL/(AY) as a ratio. Whatever the wires share cancels, and only the factors that differ remain. The area depends on the square of the diameter, so a diameter ratio enters squared.

Definition

$\Delta L \propto \dfrac{FL}{d^{2}Y}$ : take each factor's ratio, square the diameter (or radius) ratio, and multiply.
Same load and the same extension: $Y \propto L/A$ .
Same material and the same volume: $L \propto 1/A$ , so $\Delta L \propto F/A^{2}$ .
Same material, same length and the same stress: the same strain, so the same extension.

Two wires

\frac{\Delta L_1}{\Delta L_2} = \frac{F_1}{F_2}\cdot\frac{L_1}{L_2}\cdot\left(\frac{d_2}{d_1}\right)^{2}\cdot\frac{Y_2}{Y_1}

Worked example

Wire A is twice as long as wire B and has half its diameter. They are of the same material and carry the same load. Find the ratio of their extensions.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 7 Apr 2025 · Q19Moderate

Example 2 · Mechanical Properties of Solids · Stress, Strain and Young's Modulus

Two wires

A

and

B

are made of same material having ratio of lengths

\frac{L_{A}}{L_{B}}=\frac{1}{3}

and their diameters ratio

\frac{d_{A}}{d_{B}}= 2

. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

Forgetting to square the diameter

A diameter ratio of 2 is an area ratio of 4. Using 2 leaves the answer off by a factor of 2.

Same volume moves the length too

If a wire of fixed volume is made four times thicker in area, it becomes four times shorter. Both changes go into ΔL ∝ FL/A.

Drill 8 more on comparing two wires by ratio

Concept 3 of 3: What Young's modulus depends on, and reading it off a graph

Young's modulus belongs to the material, not to the wire. A longer or thinner wire stretches more under a load, but its Y is the same. To read Y off a graph, first write the slope from ΔL = FL/(AY), then decide whether Y sits on top of that slope or below it.

Definition

Y depends only on the material and its temperature. Changing L or A changes the extension under a load, not Y.
Heating loosens the bonds between atoms, so Y falls as the temperature rises.
In physics, more elastic means a larger Y: steel is more elastic than rubber. Steel's large Y and high elastic limit are why it is used for buildings and bridges.
A graph's slope is a ratio of its axes. Write that ratio with ΔL = FL/(AY), then put in the numbers.

Graph (y against x)	Slope	What it gives
Stress against strain	$Y$	The steepest line has the largest Y
Strain against stress	$1/Y$	The shallowest line has the largest Y The axes are swapped from the usual plot, so the order reverses.
Load against extension	$AY/L$	$Y = \text{slope} \times L/A$
Extension against load	$L/(AY)$	$Y = L/(A \times \text{slope})$ A line at 45° has slope 1 only in the units printed on the axes.
Extension/load against length	$1/(AY)$	$Y = 1/(A \times \text{slope})$
Y against the length or radius of the wire	Zero, a flat line	Y does not depend on the wire's size

Write the slope from

\Delta L = FL/(AY)

before reading any number off the axes.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 1 · Q6Moderate

Example 3 · Mechanical Properties of Solids · Stress, Strain and Young's Modulus

The Young's modulus of steel wire of radius

r

and length L is Y . If the radius r and length L of the wire are doubled then the value of Y

Reading a strain–stress slope as Y

With strain on the y-axis the slope is 1/Y. The steepest line there is the softest material, not the stiffest.

Thinking a thicker wire has a larger Y

A thicker wire stretches less under the same load because its stiffness AY/L is larger. Its Y, a property of the material, is unchanged.

Drill 8 more on what young's modulus depends on, and reading it off a graph

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Stress, strain and the extension of a wire
Young's modulus
$Y = \frac{F/A}{\Delta L/L} \qquad \Delta L = \frac{FL}{AY}$
Comparing two wires by ratio
Two wires
$\frac{\Delta L_1}{\Delta L_2} = \frac{F_1}{F_2}\cdot\frac{L_1}{L_2}\cdot\left(\frac{d_2}{d_1}\right)^{2}\cdot\frac{Y_2}{Y_1}$

Reference tables (1)

What Young's modulus depends on, and reading it off a graph6 rows

Graph (y against x)	Slope	What it gives
Stress against strain	$Y$	The steepest line has the largest Y
Strain against stress	$1/Y$	The shallowest line has the largest Y The axes are swapped from the usual plot, so the order reverses.
Load against extension	$AY/L$	$Y = \text{slope} \times L/A$
Extension against load	$L/(AY)$	$Y = L/(A \times \text{slope})$ A line at 45° has slope 1 only in the units printed on the axes.
Extension/load against length	$1/(AY)$	$Y = 1/(A \times \text{slope})$
Y against the length or radius of the wire	Zero, a flat line	Y does not depend on the wire's size

Write the slope from

\Delta L = FL/(AY)

before reading any number off the axes.

Watch out for (7)

Pulled from both ends is not twice the tension
→ Stress, strain and the extension of a wire
A diameter put in as a radius
→ Stress, strain and the extension of a wire
Squared units
→ Stress, strain and the extension of a wire
Forgetting to square the diameter
→ Comparing two wires by ratio
Same volume moves the length too
→ Comparing two wires by ratio
Reading a strain–stress slope as Y
→ What Young's modulus depends on, and reading it off a graph
Thinking a thicker wire has a larger Y
→ What Young's modulus depends on, and reading it off a graph

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