A student determines the Young modulus of steel using a long wire. The wire has an original length of 2.00 m and a diameter of 0.50 mm. A load of 60 N produces an extension of 2.9 mm.
(a) Calculate the Young modulus of steel. [4]
(b) Explain why a long, thin wire is used in this experiment. [1]
- (a) $A = \pi \left(0.25\times10^{-3}\right)^2 = 1.96\times10^{-7}$ m$^2$(1)
- (a) stress = $\dfrac{60}{1.96\times10^{-7}} = 3.1\times10^{8}$ Pa(1)
- (a) strain = $\dfrac{2.9\times10^{-3}}{2.00} = 1.45\times10^{-3}$ or correct substitution into $E = \dfrac{FL}{A\Delta L}$ (covers both stress and strain marks)(1)
- (a) $E = \dfrac{\text{stress}}{\text{strain}} = 2.1\times10^{11}$ Pa ecf from their area(1)
- (b) to give a larger extension (for a given load), so the percentage uncertainty in the extension is smaller allow thin wire gives a larger stress / long wire gives a larger extension for the same strain(1)
- Answer: (a) $2.1\times10^{11}$ Pa (b) larger extension → smaller percentage uncertainty
- Examiner tip: Use the radius, not the diameter, in πr² — using 0.50 mm makes the answer four times too small.