Final answer:
To calculate the length of the aluminum wire needed to reach the proportionality limit of stress, the density of aluminum is used along with the formula for stress due to a hanging wire's own weight. The correct length is found to be approximately 2.96 m.
Step-by-step explanation:
The question you've asked concerns the topic of stress and strain in materials, which is a fundamental concept in mechanical engineering and physics. To find out how long an aluminum wire must be before the stress at the upper end reaches the proportionality limit, we will be using the concept of stress due to the weight of the wire itself.
Considering the density (ρ) of aluminum is 2.7 g/cm³, we first convert it to kg/m³ (as 1 g/cm³ = 1000 kg/m³), so ρ = 2700 kg/m³. Stress (σ) is defined as the force (F) per unit area (A), and for a wire with a uniform cross-section, the force due to its own weight is F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.81 m/s²).
As the wire hangs, its mass can be expressed as m = ρ*V, where V is the volume of the wire, and since V = A*l (with l being the length of the wire), we have m = ρ*A*l.
Therefore, the stress at the upper end of the wire due to its own weight is σ = (mg)/A = ρ*g*l. To find the length l before the stress reaches the proportionality limit (σ = 8.0×10⁷ N/m²), we can solve for l: l = σ / (ρ*g). Plugging in the numbers gives us:
l = (8.0×10⁷ N/m²) / (2700 kg/m³ × 9.81 m/s²) ≈ 2963.44 m
So, the wire must be approximately 2.96 m long to reach the proportionality limit at its upper end. Hence, the answer is (a) 2.96 m.