A steel wire of cross-sectional area 0.5${\text{m}}{{\text{m}}^{\text{2}}}$ is held between two fixed supports. If tension in the wire is negligible and it is just taut at 20 ${}^{\text{0}}{\text{C}}$, determine tension when temperature falls 0 ${}^{\text{0}}{\text{C}}$. Young's modulus of elasticity is $21 \times {10^{11}}$ dyne/${\text{c}}{{\text{m}}^{{\text{ - 2}}}}$and coefficient of linear expansion is $12 \times {10^{ - 6}}$per degree centigrade. Assume that the distance between the supports remains the same.
Answer
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Hint- To solve this question, we use the basic theory of Thermal expansion. As we know Thermal expansion occurs when an object expands and becomes larger due to a change in the object's temperature. Similarly, in this case it will affect the operation of steel wire when temperature changes to ${\text{20}}{}^{\text{0}}{\text{C}}$. Some basic formulas are used to get our desired result in this problem.
Formula used- $ \Rightarrow Strain = \dfrac{{\Delta L}}{L} = \alpha \Delta T$
Stress = Y × Strain
Given: Y=$21 \times {10^{11}}dyn/c{m^2}$
$\alpha = 12 \times {10^{ - 6}}/{}^0C$
Change in temperature $\Delta T = 20{}^0C$
Change in temperature $A = 0.5m{m^2} = 0.005c{m^2}$
Complete answer:
Let the initial length of steel wire be L and let L be the steel wire's length when the temperature is reduced to$0{}^0C$.
Decrease in length due to compression,$\Delta L = {L^I} - L$
Using$\Delta L = L\alpha \Delta T$
$ \Rightarrow Strain = \dfrac{{\Delta L}}{L} = \alpha \Delta T = 12 \times {10^{ - 6}} \times 20 = 2.4 \times {10^{ - 4}}$
From Hooke's law, Stress=Y×Strain
⟹ Tension T=Y×Strain × A
$\therefore T = 21 \times {10^{11}} \times 2.4 \times {10^{ - 4}} \times 0.005 = 252 \times {10^4}dyn$
⟹ T=25.2N (1N=${10^4}dyn$)
Therefore, tension of the steel wire is 25.2 N.
Note- The expansion can occur in length of iron pendulum in which case it is called Linear Expansion. And If we take a square tile and then after heat it, the expansion will be on two fronts that is length and breadth, and it is called Area Expansion. Similarly, if we take a cube shape structure and heat it, all its sides expand and now the body experiences an increase in the overall volume of the structure and it is called Volume Expansion.
Formula used- $ \Rightarrow Strain = \dfrac{{\Delta L}}{L} = \alpha \Delta T$
Stress = Y × Strain
Given: Y=$21 \times {10^{11}}dyn/c{m^2}$
$\alpha = 12 \times {10^{ - 6}}/{}^0C$
Change in temperature $\Delta T = 20{}^0C$
Change in temperature $A = 0.5m{m^2} = 0.005c{m^2}$
Complete answer:
Let the initial length of steel wire be L and let L be the steel wire's length when the temperature is reduced to$0{}^0C$.
Decrease in length due to compression,$\Delta L = {L^I} - L$
Using$\Delta L = L\alpha \Delta T$
$ \Rightarrow Strain = \dfrac{{\Delta L}}{L} = \alpha \Delta T = 12 \times {10^{ - 6}} \times 20 = 2.4 \times {10^{ - 4}}$
From Hooke's law, Stress=Y×Strain
⟹ Tension T=Y×Strain × A
$\therefore T = 21 \times {10^{11}} \times 2.4 \times {10^{ - 4}} \times 0.005 = 252 \times {10^4}dyn$
⟹ T=25.2N (1N=${10^4}dyn$)
Therefore, tension of the steel wire is 25.2 N.
Note- The expansion can occur in length of iron pendulum in which case it is called Linear Expansion. And If we take a square tile and then after heat it, the expansion will be on two fronts that is length and breadth, and it is called Area Expansion. Similarly, if we take a cube shape structure and heat it, all its sides expand and now the body experiences an increase in the overall volume of the structure and it is called Volume Expansion.
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