Question

The following four wires are made of the same material. Which of these will have the largest extension when the same tension is applied?
(A) Length = 50 cm, Diameter = 0.5 mm
(B) Length = 100 cm, Diameter = 1 mm
(C) Length = 200 cm, Diameter = 2 mm
(D) Length = 300 cm, Diameter = 3 mm

Answer 1

Hint We will first use the young’s law to calculate the elongation of wire using \[\gamma = \dfrac{{stress}}{{strain}}\] , we will put the equations of stress and strain and simplify to find the elongation equation.
After that we will see that the elongation is directly proportional to \[\dfrac{L}{{{D^2}}}\] , \[{\text{L}}\]is the actual length and D is the diameter.
Then we will calculate the \[\dfrac{L}{{{D^2}}}\] value for each wire and the wire which will have maximum value of \[\dfrac{L}{{{D^2}}}\] has the largest extension.

Complete step by step solution
As given that same tension is applied on the wires, for extension we use young’s law equation i.e. \[\gamma = \dfrac{{stress}}{{strain}}\] , where \[\gamma \] is the young’s modulus.
Now as we know that \[stress = \dfrac{F}{A}\] , where F is the force and A is the area. Then we know \[A = \pi {r^2}\] , where r is the radius , also \[r = \dfrac{D}{2}\] , where D is the diameter .
So \[A = \dfrac{{\pi {D^2}}}{4}\] . Now we know the formula of strain i.e. \[strain = \dfrac{{\Delta L}}{L}\] , where \[\Delta L\] is the elongation and L is the actual length.
Now substituting the equations of stress and strain in young’s modulus and solving the equation for \[\Delta L\] we get:
 \[\gamma = \dfrac{{4F}}{{\pi {D^2}}}\dfrac{L}{{\Delta L}}\]
Now we find \[\Delta L\] i.e. \[\Delta L = \dfrac{{4F}}{{\pi {D^2}}}\dfrac{L}{\gamma }\] .
As we can see that \[\Delta L\alpha \dfrac{L}{{{D^2}}}\] i.e. elongation is directly proportional to \[\dfrac{L}{{{D^2}}}\] .
So, the wire which has the largest extension has the largest \[\dfrac{L}{{{D^2}}}\] value.
Now we check each and every option one by one and find which has the largest \[\dfrac{L}{{{D^2}}}\] value
For wire A , \[\dfrac{L}{{{D^2}}} = \dfrac{{50}}{{{{0.5}^2}}} = 20000\]
For wire B, \[\dfrac{L}{{{D^2}}} = \dfrac{{100}}{{{{0.1}^2}}} = 10000\]
For wire C, \[\dfrac{L}{{{D^2}}} = \dfrac{{200}}{{{{0.2}^2}}} = 5000\]
For wire D, \[\dfrac{L}{{{D^2}}} = \dfrac{{300}}{{{{0.3}^2}}} = 3333.33\].
Thus, wire A has the largest extension.

So, the correct option is A.


Note Always remember the stress is Force/Area. One tends to consider that tension is a stress. But note that tension is a force, it has to be divided by the cross-section area to give the tensile stress.
Note that young’s modulus is a specific form of Hooke’s law of elasticity, that states that force needed to extend a spring by some distance (x) scales linearly with respect to that distance.
Also remember that young’s modulus is valid only in the range in which stress is proportional to the strain, and the material returns to its original dimensions when the external force is removed.