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.

Question 9

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

Reference: Past Exam Paper – March 2016 Paper 12 Q19

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.

E_steel = 2 × E_copper

F × 50 / (d_steel² ×e_steel) = 2 × F × 50 / (d_copper² ×e_copper)

1 / (d_steel² ×e_steel) = 2 / (d_copper² ×e_copper)

e_steel / e_copper = d_copper² / 2×d_steel² = 2.0² / (2×1.0²) 

e_steel / e_copper = 4 / 2 = 2