Physics 9702 · Paper 2

The Young Modulus

The Young modulus $E$ measures how stiff the material itself is, whatever shape it is in.
It is stress over strain. We build it in two steps: first the ratio, then a form you can put real measurements into.

Stress over strain

E=\dfrac{\sigma}{\varepsilon}

Symbols

$E$ = Young modulus, the material's stiffness (Pa, often GPa)
$\sigma$ = stress (Pa)
$\varepsilon$ = strain (no unit)

$E$ is a property of the material only. A thick rod and a thin wire of the same metal share the same $E$ .
A bigger $E$ means a stiffer material: more stress needed for the same strain.
On a stress-strain graph, $E$ is the gradient of the straight part, so it only holds in the linear Hooke region.

Worked example

Young modulus from stress and strain

A material under test reaches stress $\sigma=2.0\times10^{7}$ Pa at strain $\varepsilon=1.0\times10^{-4}$ .

Use $E=\sigma/\varepsilon$ .
$E=\dfrac{2.0\times10^{7}}{1.0\times10^{-4}}=2.0\times10^{11}\text{ Pa}$ .

Answer

E=2.0\times10^{11}\text{ Pa}

Using E = FL/Ax

In a real test you measure the force, the lengths and the area, not stress and strain directly. So substitute $\sigma=F/A$ and $\varepsilon=x/L$ into $E=\sigma/\varepsilon$ :

E=\dfrac{FL}{Ax}

Symbols

$F$ = force applied (N)
$L$ = original length (m)
$A$ = cross-sectional area (m²)
$x$ = extension (m)

Keep $L$ (original length) and $x$ (extension) separate: it is easy to swap them.
If you already have $k$ , you can also use $E=\dfrac{kL}{A}$ , which comes from combining $k=F/x$ with $E=FL/Ax$ .

Your turn— tap to reveal the worked answer (9702/21/M/J/24 Q4)

Define strain, then find the Young modulus of a copper wire of length 4.0 m and cross-sectional area 4.5 × 10^-7 m² that extends 1.4 mm under a force of 18 N. (9702/21/M/J/24 Q4)

Strain is extension divided by original length, a ratio with no unit.
$E=\dfrac{FL}{Ax}=\dfrac{18\times4.0}{4.5\times10^{-7}\times1.4\times10^{-3}}$ .
$E=1.1\times10^{11}\text{ Pa}$ .

Answer

E=1.1\times10^{11}\text{ Pa}

Worked example

Young modulus from k, L and A

A spring-like wire has $k=90$ N m^-1, length 3.2 m and cross-sectional area 4.0 × 10^-7 m².

Use $E=kL/A$ .
$E=\dfrac{90\times3.2}{4.0\times10^{-7}}=7.2\times10^{8}\text{ Pa}$ .

Answer

E=7.2\times10^{8}\text{ Pa}

(9702/23/M/J/24 Q3(c)(ii))

Values to recognise, and the define question

A bigger $E$ means a stiffer material: it takes more stress to produce the same strain.
A define the Young modulus question asks for stress over strain in words. (9702/23/O/N/25 Q3(a))

Three samples of the same metal — a thick rod, a thin wire and a bar — grouped together to share one Young modulus; a sample of a different material sits apart with a different value. — The Young modulus belongs to the material, not the shape: the same metal — thick rod, thin wire or bar — shares one E. A different material has a different E.

Typical Young modulus values (approximate):

Material	Young modulus / GPa
steel	210
copper	130
aluminium	70
glass	70
rubber	0.01

Common mistake

Define

E

as stress over strain. Writing “force over extension” scores zero. Remember strain has no unit, so

E

is in pascals.

PreviousStress & Strain NextReading E off a Graph

The Young Modulus

The Young modulus $E$ measures how stiff the material itself is, whatever shape it is in.
It is stress over strain. We build it in two steps: first the ratio, then a form you can put real measurements into.

Stress over strain

E=\dfrac{\sigma}{\varepsilon}

Symbols

$E$ = Young modulus, the material's stiffness (Pa, often GPa)
$\sigma$ = stress (Pa)
$\varepsilon$ = strain (no unit)

$E$ is a property of the material only. A thick rod and a thin wire of the same metal share the same $E$ .
A bigger $E$ means a stiffer material: more stress needed for the same strain.
On a stress-strain graph, $E$ is the gradient of the straight part, so it only holds in the linear Hooke region.

Worked example

Young modulus from stress and strain

A material under test reaches stress $\sigma=2.0\times10^{7}$ Pa at strain $\varepsilon=1.0\times10^{-4}$ .

Use $E=\sigma/\varepsilon$ .
$E=\dfrac{2.0\times10^{7}}{1.0\times10^{-4}}=2.0\times10^{11}\text{ Pa}$ .

Answer

E=2.0\times10^{11}\text{ Pa}

Using E = FL/Ax

In a real test you measure the force, the lengths and the area, not stress and strain directly. So substitute $\sigma=F/A$ and $\varepsilon=x/L$ into $E=\sigma/\varepsilon$ :

E=\dfrac{FL}{Ax}

Symbols

$F$ = force applied (N)
$L$ = original length (m)
$A$ = cross-sectional area (m²)
$x$ = extension (m)

Keep $L$ (original length) and $x$ (extension) separate: it is easy to swap them.
If you already have $k$ , you can also use $E=\dfrac{kL}{A}$ , which comes from combining $k=F/x$ with $E=FL/Ax$ .

Your turn— tap to reveal the worked answer (9702/21/M/J/24 Q4)

Define strain, then find the Young modulus of a copper wire of length 4.0 m and cross-sectional area 4.5 × 10^-7 m² that extends 1.4 mm under a force of 18 N. (9702/21/M/J/24 Q4)

Strain is extension divided by original length, a ratio with no unit.
$E=\dfrac{FL}{Ax}=\dfrac{18\times4.0}{4.5\times10^{-7}\times1.4\times10^{-3}}$ .
$E=1.1\times10^{11}\text{ Pa}$ .

Answer

E=1.1\times10^{11}\text{ Pa}

Worked example

Young modulus from k, L and A

A spring-like wire has $k=90$ N m^-1, length 3.2 m and cross-sectional area 4.0 × 10^-7 m².

Use $E=kL/A$ .
$E=\dfrac{90\times3.2}{4.0\times10^{-7}}=7.2\times10^{8}\text{ Pa}$ .

Answer

E=7.2\times10^{8}\text{ Pa}

(9702/23/M/J/24 Q3(c)(ii))

Values to recognise, and the define question

A bigger $E$ means a stiffer material: it takes more stress to produce the same strain.
A define the Young modulus question asks for stress over strain in words. (9702/23/O/N/25 Q3(a))

Typical Young modulus values (approximate):

Material	Young modulus / GPa
steel	210
copper	130
aluminium	70
glass	70
rubber	0.01

Common mistake

Define

E

as stress over strain. Writing “force over extension” scores zero. Remember strain has no unit, so

E

is in pascals.