A student determined the Young modulus of a material in the form of a wire. For each mass added, the student determined the force on the wire. The student plotted the graph shown. The table gives the Young modulus for some materials. Deduce which material the wire was made from. original length of wire $$\(=1.80 \mathrm{~m}\)$$ diameter of wire $$\(=0.17 \mathrm{~mm}\)$$ (5) .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................. ..............................................................................................................................................................................................................................................
Exam No：wph11-01-que-20240112 Year：2024 Question No：17(c)
Answer:
EITHER
Use of \(A=\pi r^{2}\)
Use of \(\Delta \sigma=\frac{\Delta F}{A}\) with initial force greater than 0 N
Use of \(\Delta \varepsilon=\frac{\Delta x}{x}\)
Use of \(E=\frac{\Delta \sigma}{\Delta \varepsilon}\)
\(E=107(\mathrm{GPa}) \approx 106(\mathrm{GPa})\) so material is brass
OR
Use of \(A=\pi r^{2}\)
Determines gradient of graph
Calculates \(\frac{x}{A}\)
Use of \(E=\frac{\Delta \sigma}{\Delta \varepsilon}\)
\(E=107(\mathrm{GPa}) \approx 106(\mathrm{GPa})\) so material is brass
Example of calculation
\[
A=\pi \times\left(\frac{0.17 \times 10^{-3} \mathrm{~m}}{2}\right)^{2}=2.27 \times 10^{-8} \mathrm{~m}^{2}
\]
\[
\begin{array}{l}
\Delta \sigma=\frac{14 \mathrm{~N}}{2.27 \times 10^{-8} \mathrm{~m}^{2}}=6.17 \times 10^{8} \mathrm{~Pa} \\
\Delta \varepsilon=\frac{0.0104 \mathrm{~m}}{1.80 \mathrm{~m}}=5.78 \times 10^{-3} \\
E=\frac{6.17 \times 10^{8} \mathrm{~Pa}}{5.78 \times 10^{-3}}=1.07 \times 10^{11} \mathrm{~Pa}
\end{array}
\]
\(107 \mathrm{GPa} \approx 106 \mathrm{GPa}\) so material is brass.
Use of \(A=\pi r^{2}\)
Use of \(\Delta \sigma=\frac{\Delta F}{A}\) with initial force greater than 0 N
Use of \(\Delta \varepsilon=\frac{\Delta x}{x}\)
Use of \(E=\frac{\Delta \sigma}{\Delta \varepsilon}\)
\(E=107(\mathrm{GPa}) \approx 106(\mathrm{GPa})\) so material is brass
OR
Use of \(A=\pi r^{2}\)
Determines gradient of graph
Calculates \(\frac{x}{A}\)
Use of \(E=\frac{\Delta \sigma}{\Delta \varepsilon}\)
\(E=107(\mathrm{GPa}) \approx 106(\mathrm{GPa})\) so material is brass
Example of calculation
\[
A=\pi \times\left(\frac{0.17 \times 10^{-3} \mathrm{~m}}{2}\right)^{2}=2.27 \times 10^{-8} \mathrm{~m}^{2}
\]
\[
\begin{array}{l}
\Delta \sigma=\frac{14 \mathrm{~N}}{2.27 \times 10^{-8} \mathrm{~m}^{2}}=6.17 \times 10^{8} \mathrm{~Pa} \\
\Delta \varepsilon=\frac{0.0104 \mathrm{~m}}{1.80 \mathrm{~m}}=5.78 \times 10^{-3} \\
E=\frac{6.17 \times 10^{8} \mathrm{~Pa}}{5.78 \times 10^{-3}}=1.07 \times 10^{11} \mathrm{~Pa}
\end{array}
\]
\(107 \mathrm{GPa} \approx 106 \mathrm{GPa}\) so material is brass.
Knowledge points:
CH2 - Materials
Solution:
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