Question

If Young’s modulus for steel is $2.0\times {10}^{11}N{m}^{-2}$ , then how much weight be suspended from a steel wire of length $2.0m$ and diameter $1.0mm$ so that the length of the wire be increased by $1.0mm$ ? ( $g=9.8m{s}^{-2}$ )

Answer 1

Hint: You can explain in brief what Hooke’s law states with its equation, i.e. Strain $$\propto$$ Stress. Then write down this equation in terms of linear extension $${\displaystyle \frac{Strain}{Stress}}=Y$$. Then calculate the stress using the equation Stress $$={\displaystyle \frac{F}{A}}$$ and the strain using the equation Strain $$={\displaystyle \frac{\mathrm{\Delta }l}{l}}$$. Then put these values in the equation $${\displaystyle \frac{Strain}{Stress}}=Y$$ and then use the equation $$F=mg$$ to reach the solution.

Complete step-by-step solution:
The Young modulus is based on Hooke’s law
According to Hooke’s law, the strain in a material is directly proportional to the stress applied.
Strain $$\propto$$ Stress
$${\displaystyle \frac{Strain}{Stress}}=Y$$ …………….....(Equation 1)
Here, $$Y=$$ Young’s modulus
The above equation is just for the linear extension of material, we have different equations for the areal and volumetric extension of a material.
Given in the problem
$$Y=2.0\times {10}^{11}N{m}^{-2}$$
$$l=2m$$
$$\mathrm{\Delta }l=1mm={\displaystyle \frac{1}{1000}}m$$
Radius $$={\displaystyle \frac{d}{2}}={\displaystyle \frac{1}{2}}mm={\displaystyle \frac{1}{2000}}m$$
Now we know that,
Stress $$={\displaystyle \frac{F}{A}}$$
Here, $$F=$$ Force
$$A=$$ Area of cross-section
We also know that
Strain $$={\displaystyle \frac{\mathrm{\Delta }l}{l}}$$
Here, $$\mathrm{\Delta }l=$$ Change in the length of the wire
$$l=$$ The original length of the wire
Substituting the values of stress and strain in equation 1, we get
$${\displaystyle \frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{\mathrm{\Delta }l}{l}}}}=Y$$
$$F={\displaystyle \frac{\mathrm{\Delta }lYA}{l}}$$
Let $$m$$ be the weight suspended from the wire
$$mg={\displaystyle \frac{\mathrm{\Delta }lY\pi r^2}{l}}$$
$$m={\displaystyle \frac{\mathrm{\Delta }lY\pi r^2}{\lg }}$$
$$m={\displaystyle \frac{{10}^{-3}\times 2\times {10}^{11}\times 3.14\times {\left(0.5\times {10}^{-3}\right)}^2}{2\times 9.8}}$$
$$m=8.01kg$$
Hence, a $$8.01kg$$ weight should be suspended from a steel wire of length $2.0m$ and diameter $1.0mm$ so that the length of the wire increases by $1.0mm$.

Note: While attempting this solution, it was assumed that the wire would follow Hooke’s law (the strain will increase at a constant rate with the strain). But in reality, any material follows Hooke’s law for a certain range of stress, and stress and outside of this range strain do not remain directly proportional to stress. Even if Hooke’s law is followed if we keep on stretching the wire we will reach a point where the strain would become so large that the wire would break in two.