Deformation of Solids: Question 7

Worked solution

Part (a): Experimental method

Clamp one end of a long, thin wire made of the test material to a rigid support (e.g. a bench clamp) so that the wire hangs vertically, with a small marker attached near its free end. A second, identical wire is hung alongside as a fixed reference wire, carrying no load; a vernier scale is attached between the two wires so that any effects common to both wires, such as a small temperature change or sagging of the support, are compensated for and do not affect the extension reading.

• Original length $L$: measured with a metre rule, from the fixed clamp down to the marker, before any load is applied.
• Cross-sectional area $A$: found from the diameter of the wire, measured with a micrometer screw gauge, using $A=\pi r^2$ (see part (b) for how to do this accurately).
• Load $F$: known weights are added to the free end of the test wire in equal increments.
• Extension $x$: for each load, the vernier scale reading (relative to the unloaded reference wire) gives the extension directly.

A graph of load against extension is plotted; provided it is a straight line through the origin (confirming the wire is within the limit of proportionality), the gradient gives the spring constant, and stress ($F/A$) and strain ($x/L$) can then be used to find the Young modulus.

Part (b): Precaution for measuring diameter

The diameter should be measured with the micrometer screw gauge at several different points along the length of the wire, and in two perpendicular directions at each point, and the readings averaged.

This is necessary because the wire’s diameter may not be perfectly uniform along its length, and because the diameter appears squared in the cross-sectional area formula $A=\pi r^2$. A small error in a single diameter measurement would therefore be doubled (as a percentage) in the area, and hence in the final calculated value of the Young modulus, so averaging several measurements reduces this error.

Part (c): Calculating the Young modulus

The wire has diameter $d=0.36\text{ mm}$, so its radius is: $r = \frac{d}{2} = \frac{0.36\times10^{-3}}{2} = 0.18\times10^{-3}\text{ m} = 1.8\times10^{-4}\text{ m}$

The cross-sectional area is: $A = \pi r^2 = \pi \times (1.8\times10^{-4})^2 = \pi \times 3.24\times10^{-8}$ $A = 1.018\times10^{-7}\text{ m}^2$

The stress produced by the $18\text{ N}$ load is: $\sigma = \frac{F}{A} = \frac{18}{1.018\times10^{-7}} = 1.768\times10^{8}\text{ Pa}$

Converting the extension to metres, $x = 0.90\text{ mm} = 9.0\times10^{-4}\text{ m}$, the strain is: $\varepsilon = \frac{x}{L} = \frac{9.0\times10^{-4}}{1.80} = 5.0\times10^{-4}$

So the Young modulus is: $E = \frac{\sigma}{\varepsilon} = \frac{1.768\times10^{8}}{5.0\times10^{-4}}$ $E = 3.54\times10^{11}\text{ Pa}$

As a check, using $E=\dfrac{FL}{Ax}$ directly: $E = \frac{18\times1.80}{1.018\times10^{-7}\times9.0\times10^{-4}} = \frac{32.4}{9.16\times10^{-11}} = 3.54\times10^{11}\text{ Pa}$ Both methods agree.

Final answers

• (a) Method: long wire, reference wire and vernier scale to compensate for common effects, metre rule for $L$, micrometer for $A$, added loads for $F$, vernier readings for $x$.
• (b) Measure diameter at several points/directions and average, because it is squared in the area formula.
• (c) Young modulus $= \boxed{3.54\times10^{11}}\text{ Pa}$