Question 13
A metal wire of
cross-sectional area 0.20 mm2 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
Reference: Past Exam Paper – June 2016 Paper 13 Q21
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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