A force acts on a wire to produce extension e. The same force then acts on a second wire of the same material, but of half the diameter and three times the length of the first wire.

Question 20

A force acts on a wire to produce extension e. The same force then acts on a second wire of the same material, but of half the diameter and three times the length of the first wire. Both wires obey Hooke’s law.

 

What is the extension of the second wire?

A 3e                 B 4e                 C 6e                D 12e

 

 

 

Reference: Past Exam Paper – November 2015 Paper 12 Q21

 

 

 

Solution:

Answer: D.

Since the wires are of the same material, they have the same Young modulus.

Young modulus E = stress / strain

E = FL / Ae

Area = πd² / 4

E = 4FL / πd²e

 

Wire 1: Force = F        Length = L                  Extension = e              Diameter = d

Wire 2: Force = F        Length = 3L                Extension = e_new          Diameter = d/2

 

Young modulus of first wire = Young modulus of second wire

FL / πd²e = F (3L) / π(d/2)²e_new

1 / e = 3×4 / e_new

E_new = 12 e

A wire has a final length of 6.0 m after undergoing a strain of 200%. What is the original length of the wire?

 Question 19

A wire has a final length of 6.0 m after undergoing a strain of 200%.

 

What is the original length of the wire?

A 1.5 m                       B 2.0 m                       C 3.0 m                       D 4.0 m

 

 

 

Reference: Past Exam Paper – November 2015 Paper 11 Q23

 

 

 

Solution:

Answer: B.

Let the original length = L

Extension = 6 – L

Percentage strain = (extension / Original length) × 100%

200 % = ((6 – L) / L) × 100%

(6 – L) / L = 200/100

(6 – L) / L = 2

6 – L = 2L

6 = 2L + L

L = 6/3 = 2 m