The Young Modulus (AQA A Level Physics): Revision Note

A metal wire that is supported vertically from a fixed point has a load of 92 N applied to the lower end.

The wire has a cross-sectional area of 0.04 mm² and obeys Hooke’s law.

The length of the wire increases by 0.50%. What is the Young modulus of the metal wire?

A.    4.6 × 10⁷ Pa              B.    4.6 × 10¹² Pa              C.    4.6 × 10⁹ Pa               D.    4.6 × 10¹¹ Pa

[1]

Answer:

$0.04{\mathrm{mm}}^2=0.04\times {10}^{-6}{\mathrm{m}}^2$

$\mathrm{Tensile}\mathrm{stress}=\frac{F}{A}=\frac{92}{0.04\times {10}^{-6}}=2.3\times {10}^9\mathrm{Pa}$

$\mathrm{Tensile}\mathrm{strain}=0.5\%=0.005$

$\mathrm{Young}\mathrm{modulus}=\frac{\mathrm{Tensile}\mathrm{stress}}{\mathrm{Tensile}\mathrm{strain}}=\frac{2.3\times {10}^9}{0.005}=4.6\times {10}^{11}\mathrm{Pa}$

• Therefore, the correct answer is D [1 mark]

To remember whether stress or strain comes first in the Young modulus equation, try thinking of the phrase ‘When you’re stressed, you show the strain’ i.e. Stress ÷ Strain.

The graph below shows the stress-strain graph for a copper wire.

Use the graph to calculate the Young modulus of copper.

[2]

Answer:

Step 1: Determine the stress and strain where the linear region ends

• The Young modulus is the gradient of the linear region of a stress-strain graph

[1 mark]

Step 2: Calculate the gradient of the graph in this region

$\mathrm{Young}\mathrm{modulus}=\frac{112.5\times {10}^6}{0.9\times {10}^{-3}}=1.25\times {10}^{11}\mathrm{Pa}$ [1 mark]