Question

For a wire of length ${\text{Z}}$, maximum change in length under stress condition is $2{\text{ mm}}$. What is the change in length under the same conditions when length of wire is halved?
A. $1{\text{ mm}}$
B. ${\text{2 mm}}$
C ${\text{4 mm}}$
D. ${\text{8 mm}}$

Answer 1

An object or medium under stress becomes deformed. The quantity that describes this deformation is called strain. Strain is given as a fractional change in either length (under tensile stress) or volume (under bulk stress). Therefore, strain is dimensionless.To solve this problem try to establish the Hooke’s law of elasticity that is the relation between young modulus of the wire and the force applied on wire, area of wire, length of wire and the change in length of wire.

Complete step by step answer:
Given, maximum change in length of wire under stress condition is ${\text{2mm}}$
Now, according to the Hooke's law,
Young’s modulus ${\text{Y = }}\dfrac{{{\text{stress}}}}{{{\text{strain}}}}{\text{ }}.......{\text{(i)}}$
As we know, ${\text{stress = }}\dfrac{F}{A}{\text{ }}.......{\text{(ii)}}$
and ${\text{strain = }}\dfrac{L}{l}{\text{ }}......{\text{(iii)}}$
Here in the above equations, $F$ is force, $A$ is area, $l$ is change in length and $L$ is original length

Therefore, from $({\text{i}}),({\text{ii}}){\text{ and (iii)}}$ , we get
${\text{Y = }}\dfrac{{{\text{F/A}}}}{{L/l}}{\text{ }}.......{\text{(iv)}}$
Note that it is given that under the same conditions the length of wire is halved and other properties like force on wire and the area of wire remain the same. So, the young’s modulus in both cases remains same (as the material of wire remains same).So, in both initial and final conditions Young’s modulus remains the same as ${\text{Y}}$ , force applied be $F$ and area of wire be $A$ .

Let the change in wire be ${l_1}$ and ${l_2}$ in both conditions respectively.According to question, ${l_1}{\text{ = 2 mm}}$
Initial condition, when the length of wire is ${\text{L}}$,
${\text{Y = }}\dfrac{{{\text{F/A}}}}{{L/{l_1}}}{\text{ }}......{\text{(v)}}$
Second condition when the length of wire is halved that is $\dfrac{{\text{L}}}{2}$
${\text{Y = }}\dfrac{{{\text{F/A}}}}{{L/2{l_2}}}{\text{ }}......{\text{(vi)}}$
From the equation $({\text{v}})$ and $({\text{vi}})$
${l_1}\;{\text{ = 2}}{l_2}$
As ${l_1}{\text{ = 2 mm}}$
$\therefore {l_2}{\text{ = 1 mm}}$

Therefore, option A is the correct answer.

Note: The greater the stress, the greater the strain, however, the relation between strain and stress does not need to be linear. Only when stress is sufficiently low is the deformation it causes in direct proportion to the stress value. The proportionality constant in the relation is called the elastic modulus. The elastic modulus for tensile stress is called Young’s modulus. Young's modulus depends upon the material and property of the wire. So under the same conditions, Young’s modulus of the same wire remains the same.