Question

A 1 m long steel wire of cross-sectional $1m{m^2}$ is extended by 1 mm. If Y = $2 \times {10^{11}}N{m^{ - 2}}$, then, the work done is
A. 0.1 J
B. 0.2 J
C. 0.3 J
D. 0.4 J

Answer 1

Hint:To solve this question, the concept of Hooke’s law should be employed. The Hooke’s law gives us the stress, which can be used to calculate the force, whose value should be used to find the work done.
The formula of Hooke’s law is:
$\sigma = E\varepsilon $
where $\sigma $ is the stress, $\varepsilon $ is the strain and E is the Young’s Modulus.

Complete step-by-step answer:
Before we get into the definition of Hooke’s law, the definitions of the terms of the Hooke’s law must be made clear.
Stress, represented by $\sigma $, is the internal resistance force developed inside the body per unit area, when an external deforming force is applied.
$\sigma = \dfrac{F}{A}$
where F is force and A is area.
Strain, represented by $\varepsilon $, is the ratio of the change in the length of a body to the original length when subjected to external deforming force.
$\varepsilon = \dfrac{{\Delta L}}{L}$
Hooke’s law states that: The stress induced in a body is directly proportional to the strain in the body.
$\sigma \propto \varepsilon $
Removing the proportionality,
$\sigma = E\varepsilon $
where E is called the Young’s modulus which is a representation of the quality of stiffness in a material.
Given,
Young’s modulus, $E = Y = 2 \times {10^{11}}N{m^{ - 2}}$
Original length , $L = 1m$
Change in length, $\Delta L = 1mm = {10^{ - 3}}m$
Area of cross-section, $A = 1m{m^2} = {10^{ - 6}}{m^2}$
Now,
Strain induced in the material, $\varepsilon = \dfrac{{\Delta L}}{L} = \dfrac{{{{10}^{ - 3}}}}{1} = {10^{ - 3}}$
By applying the Hooke’s law, we can obtain the stress induced,
Stress, $\sigma = E\varepsilon $
$\sigma = 2 \times {10^{11}} \times {10^{ - 3}} = 2 \times {10^7}N{m^{ - 2}}$
Given that, the stress is equal to force per unit area,
$\sigma = \dfrac{F}{A}$
$2 \times {10^8} = \dfrac{F}{{{{10}^{ - 6}}}}$
$ \Rightarrow F = 2 \times {10^8} \times {10^{ - 6}} = 2 \times {10^2} = 200N$
Now, that we have the force and the displacement (i.e. change of length or deformation), we can obtain the work.
Work done, $W = Fs = F\Delta L$
Substituting,
$W = 200 \times {10^{ - 3}} = 0.2J$

Hence, the correct option is Option B.

Note: If you can observe the problem, there is the word stress induced. This has been introduced deliberately to clarify the difference between pressure and stress.
Pressure and stress have the same formula and units but their meanings are completely different.

Pressure is equal to the external force applied per unit area.

Now, when this external force is applied on the material, there is a reaction force induced in the body. The stress is equal to the reaction force induced per unit area.
So, we can say that pressure is externally applied, while stress is internally developed.