Answer:
[tex]49.4^{\circ}[/tex]
Explanation:
We are given that
Stress,[tex]\sigma_1=60 MPa=60\times 10^6 Pa[/tex]
Temperature,[tex]T_0=20^{\circ}[/tex]
Stress,[tex]\sigma_2=5MPa=5\times 10^6 Pa[/tex]
[tex]\alpha=17\times 10^{-6}/^{\circ}C[/tex]
Modulus of elasticity=[tex]E=110GPa=110\times 10^9 Pa[/tex]
We have to find the temperature must the wire be heated.
[tex]T=T_0-\frac{\Delta\sigma}{E\alpha}[/tex]
Using the formula
[tex]T=20-\frac{(5-60)\times 10^6}{110\times 10^9\times 17\times 10^{-6}}[/tex]
[tex]T=49.4^{\circ}[/tex]
