Answer:
a)diameter of the thread = 0.546mm
b) stress in the thread = 36.3MPa
Step-by-step explanation:
Hooke's law

strain = 1.1% = 0.011
Stress = strain* Young’s modulus
= 3.3 × 10⁹ × 0.011
= 36.3 × 10⁶Pa
stress = 36.3MPa
stress in the thread = 36.3MPa
b) Now,
stress = Force / area
Force = 8.5N
Stress = 36.3 × 10⁶
Area = Force / stress
= 8.5 / 36.3 × 10⁶
= 0.234 × 10⁻⁶m²
Area =

diameter of the thread = 0.546mm