Question

When a wire is stretched, an amount of work is done. What is the amount of work done in stretching a wire through 0.1mm, if its length is 2m and area of cross-section, ${10^{ - 6}}{m^2}$ (Y=2$ \times {10^{11}}\dfrac{N}{{{m^2}}}$ )
A) $5 \times {10^{ - 1}}J$
B) $5 \times {10^{ - 2}}J$
C) $5 \times {10^{ - 3}}J$
D) $5 \times {10^{ - 4}}J$

Answer 1

Hint: Hooke’s law is the only thing which is sufficient to get the answer of the question. Hooke’s law states that force per unit area is directly proportional to change in length per unit length. Y is called young’s modulus which is a proportionality constant. And work done is given by $\dfrac{1}{2} \times$ change in length $\times$ force

Complete step-by-step answer:
Step 1:
It is important to know about the Hooke’s law before going into the details, as the whole question revolves around this law.
Hooke’s law: Extension happens when an object increases in length, and compression happens when it decreases in length. The extension of an elastic object, such as a spring, is described by Hooke's law: force = spring constant × extension. F = k e.
Step 2:
Now coming on the question, we are asked about the work done when a wire is stretched,
Formula for the work done is W =$\dfrac{1}{2}$ ×F×$\Delta l$…..(1) ,
where F is the force and $\Delta l$ is the change in length
We are given, $\Delta l$=0.1mm
Length(L) = 2m
Y=2$ \times {10^{11}}\dfrac{N}{{{m^2}}}$, Y is the young’s modulus (proportionality constant)
and Area of cross-section, ${10^{ - 6}}{m^2}$
Step 3:
To find the force applied on the string there is another formula which is stated above:
Force per unit area is directly proportional to change in length per unit length
Which means,$\dfrac{F}{A}$ =Y$\dfrac{{\Delta l}}{L}$
Putting the values of the following: $F=AY$$\dfrac{{\Delta l}}{L}$
Force is equal to
$F = \dfrac{{{{10}^{ - 6}} \times 2 \times {{10}^{11}} \times 0.1 \times {{10}^{ - 3}}}}{2} = 100N$
On solving it we will get force equal to 100N or 100 newtons.
Now putting the value of force in eq (1)
$W = \dfrac{1}{2} \times F \times \Delta l$
$W = \dfrac{1}{2} \times 100 \times 0 \cdot 1 = 5 \times {10^{ - 3}}J$

Hence, the correct option is Option C.

Note: The Young’s Modulus Y is not just an ordinary ratio of the stress and the strain. It represents a physical property in a material known as stiffness. The stiffness represents the resistance of a body to change its dimensions easily when an external force is applied. Hence, the higher the stiffness in the body, the more difficult it is to make the change in its physical dimensions.