Respuesta :
Complete Question:
Two solid rods have the same length and are made of the same material with circular cross sections. Rod 1 has a radius r, and rod 2 has a radius r / 2. If a compressive force F is applied to both rods, their lengths are reduced by ΔL1 and ΔL2, respectively. The ratio ΔL1 / ΔL2 is equal to.
Answer:
[tex]\frac{\triangle L_1}{\triangle L_2} =\frac{1}{4}[/tex]
Explanation:
Since the two rods have the same length, L₁ = L₂ = L
Radius of rod 1, r₁ = r
Radius of rod 2, r₂ = r/2
Cross sectional area of rod 1, [tex]A_1 = \pi r^2[/tex]
Cross sectional area of rod 2, [tex]A_2 = \frac{\pi r^2}{4}[/tex]
The same compressive force is applied to the two rods, F₁ = F₂ = F
The rods are said to be made of the same material, this means that they have the same young's modulus.
Young's modulus of rod 1, [tex]Y_1 = \frac{FL}{A_1 \triangle L_1}[/tex]
Young's modulus of rod 2, [tex]Y_2 = \frac{FL}{A_2 \triangle L_2}[/tex]
Since Y₁ = Y₂
[tex]\frac{FL}{A_1 \triangle L_1} = \frac{FL}{A_2 \triangle L_2}\\\\ \frac{1}{A_1 \triangle L_1}= \frac{1}{A_2 \triangle L_2}\\\\[/tex]
[tex]\frac{\triangle L_1}{\triangle L_2} = \frac{A_2}{A_1}\\[/tex]..............(*)
Put A₁ and A₂ into (*)
[tex]\frac{\triangle L_1}{\triangle L_2} = \frac{\pi r^2/4 }{\pi r^2}\\\\\frac{\triangle L_1}{\triangle L_2} =\frac{1}{4}[/tex]
