Question

One end of a horizontal copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When arrangement is stretched by applying force at two ends, the ratio of elongation in the thin wire to that in the thick wire is:
(A) 0.25
(B) 0.50
(C) 2
(D) 4

Answer 1

Hint:-we know that the ratio of elongation is directly proportional to length, force and indirectly proportional to Young’s modulus as well as square of radius. So in order to solve this question, first we will find the value of elongation by applying the hooke's law individually then we will find the value of their ratio. At last we will substitute the values according to the question.

Complete step-by-step solution:-
As the forces are applied on the wire and they have been deformed, the longitudinal stress and strain are produced here in the wire. Firstly, let us understand the terms that have been mentioned here.
Stress: The net elastic force acting per unit area of the surface subject to deformation is defined as stress.
$Stress = \dfrac{{Force}}{{Area}}$
This stress is an internal property of material which opposes the deformation.
Strain: The change in dimension per unit original dimension of a body subject to deforming forces is called Strain.
$Strain = \dfrac{{Change{\text{ }}in{\text{ }}length}}{{Original{\text{ }}length}}$
This is a scalar quantity and in wire longitudinal strain is produced.
Now, according Hooke’s law, the stress is directly proportional to strain, which is understood as under deforming forces, the more is the stress, the more is strain i.e. the more stress produced in material will change the dimension in more quantity,
$Stress \propto strain$
$\dfrac{{Stress}}{{Strain}} = Y$
This constant is known as the modulus of elasticity. Now this constant is Young’s modulus of elasticity for wire as there is change in the length of the wire.
$
  \Upsilon = \dfrac{{Longitudinal{\text{ }}Stress}}{{Longitudinal{\text{ }}strain}} \\
  \Upsilon = \dfrac{{F.L}}{{A.l}} \\
 $
Now the elongation of wire $l$ is given by
\[
  l = \dfrac{{F.L}}{{A.\Upsilon }} \\
  l = \dfrac{{F.L}}{{\pi {r^2}.\Upsilon }} \\
 \]
For first wire, we have
\[{l_1} = \dfrac{{{F_1}.{L_1}}}{{\pi {r_1}^2.{\Upsilon _1}}}\]
Similarly for second wire we have
\[{l_2} = \dfrac{{{F_2}.{L_2}}}{{\pi {r_2}^2.{\Upsilon _2}}}\]
So, the ratio of elongation will be,
$\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{F_1}}}{{{F_2}}}.\dfrac{{{L_1}}}{{{L_2}}}.\dfrac{{{\Upsilon _2}}}{{{\Upsilon _1}}}.\left[ {\dfrac{{r_2^2}}{{r_1^2}}} \right]$
As both the wires are of the same material, the Young Modulus and the force applied is the same for both. Apart from that,
$
  \dfrac{{{L_1}}}{{{L_2}}} = 2 \\
  \dfrac{{{r_1}}}{{{r_2}}} = 2 \\
 $
On substituting the values we get
$\dfrac{{{l_1}}}{{{l_2}}} = 1.2.1.\dfrac{1}{{{2^2}}}$\dfrac
Finally we get
$\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{1}{2} \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = 0.5$
, the ratio of elongation in the thin wire to that in the thick wire is 0.50
So, the correct option is B.

Note:- Always remember that young’s modulus is the property of material which depends on the material only. Higher the Young‘s modulus of the higher the elasticity and elasticity is the property of materials by which it regains its original position after the deforming force has been removed.