Q#10 (Past Exam Paper – November 2016 Paper 12 Q22)
Four solid steel rods, each of length 2.0 m and cross-sectional area 250 mm2, equally support an object weighing 10 kN. The weight of the object causes the rods to contract by 0.10 mm.
What is the Young modulus of steel?
A 2.0 × 108 N m-2
B 2.0 × 1011 N m-2
C 8.0 × 108 N m-2
D 8.0 × 1011 N m-2
Solution:
Answer: B.
Young modulus E = stress / strain
Young modulus E = FL / Ae
All quantities should be in SI units.
Force F = weight = 10 kN = 10 000 N
Length L = 2.0 m
Four rods are involved.
Cross-section area A = 4 × area of 1 rod = 4 × 250 mm2 = 4 × 250×10-6 m2
Compression e = 0.10 mm = 0.10×10-3 m
Young modulus of steel: E = FL / Ae
E = (10 000 × 2.0) / ((4 × 250×10-6) × (0.10×10-3))
E = 2.0×1011 N m-2
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