Q#20 (Past Exam Paper – November 2015 Paper 12 Q21)
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
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 = enew
Diameter = d/2
Young modulus of first wire = Young modulus of second wire
FL / πd2e = F (3L) / π(d/2)2enew
1 / e = 3×4 / enew
Enew = 12 e
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