Final answer:
Physics concepts of elasticity and Hooke's Law are used to relate the stretch of a copper rod to the stretch of a steel rod under the same stress. Since copper and steel have different Young's moduli, the stretch of the steel rod can be calculated based on this difference and the given stretch of the copper rod. So, the correct option is d) (0.15 {mm}).
Step-by-step explanation:
The question pertains to the concept of elasticity in materials, which in the field of physics, is used to describe how different materials deform under stress and then return to their original shape once the stress is removed. The given scenario involves two rods of copper and steel subjected to the same stress and the need to determine the amount of stretch experienced by the steel rod in comparison to the copper rod.
To solve this, we use Hooke's Law and the concept of Young's modulus which relates stress and strain in a linearly elastic material. Copper and steel have different Young's moduli, meaning they respond differently to stress. As the copper rod stretches 0.15 mm, to determine how much the steel rod will stretch under the same conditions, the relationship between the moduli of steel and copper and their corresponding strains must be established.
Assuming the dimensions and the applied stress are the same for both rods, and neglecting any other form of deformation such as twisting or bending, we get:
Strain (copper) = (Stretch (copper)) / (Original length)
Strain (steel) = (Stretch (steel)) / (Original length)
Given that the Stress = Young's Modulus × Strain, and stress is constant for both:
(Young's Modulus of copper) × (Strain (copper)) = (Young's Modulus of steel) × (Strain (steel))
Solving for the stretch of the steel rod, we rearrange the formula, and using the given stretch for copper, calculate the stretch for steel.
Without the exact values of Young's Modulus for both materials, we can only say the steel rod will stretch an amount directly related to the ratio of the Young's moduli of steel and copper.
So, the correct option is d) (0.15 {mm}).