This problem involves thermal expansion and elasticity. When the steel wire is cooled, it contracts. To regain its original length, a mass is hung, which stretches the wire due to its weight. The key is to find the mass 'm' such that the thermal contraction equals the elastic elongation.

Step 1: Calculate Thermal Contraction
The change in length due to cooling is given by: $\mathrm{\Delta L}=L\alpha \mathrm{\Delta T}$ where:
$\alpha ={10}^{-5}\text{/°C}$ (coefficient of linear expansion)
$\mathrm{\Delta T}=30°C-40°C=-10°C$ (negative sign indicates cooling)
Therefore, the contraction is: $\mathrm{\Delta L}{\left(\text{thermal}\right)}_{\text{contraction}}=L\left({10}^{-5}\right)(-10)=-{10}^{-4}L$ The magnitude of this contraction is $\left|\mathrm{\Delta L}\right|={10}^{-4}L$.

Step 2: Calculate Elastic Elongation
The elongation due to the mass 'm' is given by Hooke's Law: $\mathrm{\Delta L}=\frac{(FL)}{(YA)}$ where:
$F=mg$ (force due to the mass)
$Y={10}^{11}\text{N/m}{}^2$ (Young's modulus)
$A=\pi r^2$ (cross-sectional area)
Given radius $r=1\text{mm}={10}^{-3}\text{m}$, so $A=\pi {\left({10}^{-3}\right)}^2=\pi {10}^{-6}{\text{m}}^2$. Therefore, the elongation is: $\mathrm{\Delta L}{\left(\text{elastic}\right)}_{\text{elongation}}=\frac{(mgL)}{(YA)}$

Step 3: Equate the Two Length Changes
For the wire to regain its original length, the elastic elongation must equal the thermal contraction: $\frac{mgL}{YA}={10}^{-4}L$ Notice that 'L' cancels out from both sides. $\frac{mg}{YA}={10}^{-4}$

Step 4: Solve for Mass 'm'
Rearranging the equation: $m=\frac{{10}^{-4}YA}{g}$ Now substitute the known values: Y = 10 11, $A=\pi {10}^{-6}$, $g\approx 10\text{m/s}{}^2$. $m=\frac{\left({10}^{-4}\right)\left({10}^{11}\right)(\pi {10}^{-6})}{10}=\frac{{10}^{(-4+11-6)}\pi }{10}=\frac{{10}^1\pi }{10}=\pi$ Since $\pi \approx 3.14$, the mass 'm' is approximately 3.14 kg.

Related Topics

Thermal Expansion: The tendency of matter to change its shape, area, and volume in response to a change in temperature. The linear expansion is described by $\mathrm{\Delta L}=L\alpha \mathrm{\Delta T}$.

Elasticity: The property of a material to return to its original shape after the deforming force is removed. Hooke's Law states that stress is proportional to strain, with Young's modulus (Y) being the constant of proportionality: $\text{Stress}=Y\text{Strain}$, where Stress = Force/Area and Strain = Change in length/Original length.

Key Formulae

Thermal Contraction/Elongation: $\mathrm{\Delta L}=L\alpha \mathrm{\Delta T}$

Elastic Elongation: $\mathrm{\Delta L}=\frac{FL}{YA}=\frac{\left(mg\right)L}{YA}$