QuestionA steel wire of original length 2.00 m and diameter 0.50 mm hangs vertically. When a load of 40 N is added, the wire extends by 2.0 mm. The wire stays within its limit of proportionality.(a) Calculate the cross-sectional area of the wire. (2 marks)
(b) Calculate the stress in the wire. (1 mark)
(c) Calculate the strain in the wire. (1 mark)
(d) Calculate the Young modulus of the steel. (1 mark)
(e) Explain why the diameter should be measured at several points along the wire. (1 mark)Worked answer(a) The radius is half the diameter: r = 0.25 mm = 0.25 × 10-3 mA = πr2 = π × (0.25 × 10-3)2 = 1.96 × 10-7 m2 (b) Stress:σ = F ÷ A = 40 ÷ 1.96 × 10-7 = 2.04 × 108 Pa (c) Strain:ε = Δx ÷ x = 2.0 × 10-3 ÷ 2.00 = 1.0 × 10-3 (d) Young modulus:E = σ ÷ ε = 2.04 × 108 ÷ 1.0 × 10-3 = 2.0 × 1011 Pa (e) The wire may not have a uniform diameter, so taking the mean of several readings allows for this variation and reduces the effect of random errors.Common mistakes: using the diameter instead of the radius in A = πr2, and forgetting to convert millimetres to metres. Both give answers wrong by a large factor, so check that E for a metal comes out at around 1011 Pa.
(b) Calculate the stress in the wire. (1 mark)
(c) Calculate the strain in the wire. (1 mark)
(d) Calculate the Young modulus of the steel. (1 mark)
(e) Explain why the diameter should be measured at several points along the wire. (1 mark)Worked answer(a) The radius is half the diameter: r = 0.25 mm = 0.25 × 10-3 m