Four solid steel rods, each of length 2.0 m and cross-sectional area 250 mm2, equally support an object weighing 10 kN.

Question 10

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

Reference: Past Exam Paper – November 2016 Paper 12 Q22

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