Q#9 (Past Exam Paper – March 2016 Paper 12 Q19)
The Young modulus of steel is twice that of copper.
A 50 cm length of copper wire of diameter 2.0 mm is joined to a 50 cm length of steel wire of diameter 1.0 mm, making a combination wire of length 1.0 m, as shown.
The combination wire is stretched by a weight added to its end. Both the copper and the steel wires obey Hooke’s law.
What is the ratio
extension of steel wire ?
extension of copper wire
A 4 B 2 C 1 D 0.5
Solution:
Answer: B.
Young modulus E = Stress / Strain
E = FL / Ae
The length L of both the copper wire and the steel wire is 50 cm each.
A weight is attached at the end of the combination wire. This causes a force F on each wire.
Cross-sectional area A = πd2 / 4
The cross-sectional area is proportional to the square of the diameter (d2). Since Young modulus E = FL / Ae, we can say that the Young modulus is proportional to 1/d2 (since the π/4 in the area is a constant).
The Young modulus of steel is twice that of copper.
Esteel = 2 × Ecopper
F × 50 / (dsteel2 ×esteel) = 2 × F × 50 / (dcopper2 ×ecopper)
1 / (dsteel2 ×esteel) = 2 / (dcopper2 ×ecopper)
esteel / ecopper = dcopper2 / 2×dsteel2 = 2.02 / (2×1.02)
esteel / ecopper = 4 / 2 = 2
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