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Friday, January 10, 2020

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.


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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