Physics 9702 · Paper 2

Measuring the Young Modulus

To measure $E$ for a metal we stretch a long, thin wire and record how much it extends for each load.
Long and thin is deliberate: a long, thin wire gives a bigger, easier to measure extension.
Plot stress against strain (or force against extension) and the gradient of the straight part gives $E$ .

The wire experiment, step by step

Clamp a long, thin wire at one end so it runs horizontally over a pulley at the bench edge, then hang masses to load it.
Measure the original length $L$ with a metre rule.
Measure the diameter $d$ with a micrometer at several places along the wire, then average, so $A=\pi(d/2)^2$ .
Measure the extension $x$ from a marker or pointer on the wire against a fixed millimetre scale as you add each load.
Plot the graph and take the gradient of the straight part to get $E$ .
Unload the wire to check it returns to its original length, so it is still in the elastic region.

Apparatus to measure the Young modulus of a wire: a long thin wire clamped at one end, run horizontally over a pulley at the bench edge and loaded with a mass, with a marker on the wire against an upright scale. — The set-up: a long thin wire clamped at one end, run over a pulley and loaded, with a pointer against a scale to read the tiny extension.

Your turn— tap to reveal the worked answer (9702/13/M/J/24 Q3)

An experiment measures $E$ for a wire using $E = 4FL/(\pi d^2 x)$ . The four measurements have similar percentage uncertainties. Which one puts the largest uncertainty into $E$ ? (9702/13/M/J/24 Q3)

$E=\dfrac{FL}{Ax}$ with $A=\pi(d/2)^2$ .
A micrometer reads $d$ precisely, but $A$ depends on $d^2$ , so the percentage error in $d$ is doubled in $A$ . That gives it the biggest effect on $E$ .

Why the same material gives k = EA/L

k=\dfrac{EA}{L}

Symbols

$k$ = spring constant (N m⁻¹)
$E$ = Young modulus (Pa)
$A$ = cross-sectional area (m²)
$L$ = original length (m)

Combine the two definitions $k=F/x$ and $E=FL/(Ax)$ : rearranging the second gives $F/x=EA/L$ , and the left side is exactly $k$ , so $k=EA/L$ (the same relation as $E=kL/A$ , rearranged). (9702/23/O/N/25 Q3(b))
So for the same material, a thicker sample (larger $A$ ) or a shorter one (smaller $L$ ) has a bigger $k$ , even though the Young modulus $E$ stays the same.

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