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