Q#13 (Past Exam Paper – June 2016 Paper 13 Q21)
A metal wire of cross-sectional area 0.20 mm$^2$ hangs vertically from a fixed point. A load of 84 N is then attached to the lower end of the wire. The wire obeys Hooke’s law and increases in length by 0.30%.
What is the Young modulus of the metal of the wire?
A 1.4 × 105 Pa
B 1.4 × 108 Pa
C 1.4 × 109 Pa
D 1.4 × 1011 Pa
Solution:
Answer: D.
Young modulus = stress / strain
Stress = force / area
Stress = 84 / (0.2×10-6)
Recall that 1 mm2 = 1 mm × 1 mm = 10-3 m × 10-3 m = 10-6 m2
The wire increases in length by 0.30 %.
This represents the strain.
Strain = extension / original length = 0.30 % = 0.30 / 100 = 0.0030
(the percentage needs to be expressed as a decimal)
Young modulus = stress / strain
Young modulus = 84 / [(0.2×10-6) × 0.0030]
Young modulus = 1.4×1011 Pa
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