Respuesta :
Answer:
the ratio of lengths of the two rods, Aluminum to Invar is 11.27
Step-by-step explanation:
coefficient of linear expansion of aluminum, [tex]\alpha _{Al} = 23 \times 10^{-6} /K[/tex]
Coefficient of linear expansion of Invar, [tex]\alpha _{Iv} = 1.2 \times 10^{-6}/K[/tex]
Linear thermal expansion is given as;
[tex]\Delta L = L_0 \times \alpha\times \Delta T\\\\where;\\\\L_0 \ is \ the \ original \ length \ of \ the \ metal\\\\\Delta L \ is \ the \ increase \ in \ length[/tex]
The increase in length of Invar is given as;
[tex]\Delta L_{Iv} = L_0_{Iv} \times \alpha _{Iv}\times \Delta T_{Iv}[/tex]
The increase in length of the Aluminum;
[tex]\Delta L_{ Al} = L_0_{Al} \times \alpha _{Al} \times \Delta T_{Al}\\\\from \ the\ given \ question, \ the \ relationship \ between \ the \ rods \ is \ given \ as\\\\ L_0_{Al} \times \alpha _{Al} \times \frac{1}{3} \Delta T_{Iv}= 2( L_0_{Iv} \times \alpha _{Iv} \times \Delta T_{Iv})\\\\ L_0_{Al} \times \alpha _{Al} \times \Delta T_{Iv}= 6( L_0_{Iv} \times \alpha _{Iv} \times \Delta T_{Iv})\\\\ L_0_{Al} \times \alpha _{Al} \times \Delta T_{Iv} = 6L_0_{Iv} \times 6\alpha _{Iv} \times 6 \Delta T_{Iv}\\\\[/tex]
[tex]\frac{L_0_{Al}}{6L_0_{Iv} } = \frac{6\alpha _{Iv} \ \times \ 6 \Delta T_{Iv}}{\alpha _{Al} \ \times \ \Delta T_{Iv}} \\\\\frac{L_0_{Al}}{6L_0_{Iv} } = \frac{6\alpha _{Iv} \ \times \ 6}{\alpha _{Al} \ } \\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{\alpha _{Iv} }{\alpha _{Al} } )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{1.2 \times 10^{-6} }{23\times 10^{-6} } )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = 6^3(\frac{1.2}{23} )\\\\\frac{L_0_{Al}}{L_0_{Iv} } = \frac{259.2}{23} \\\\\frac{L_0_{Al}}{L_0_{Iv} } = 11.27[/tex]
Therefore, the ratio of lengths of the two rods, Aluminum to Invar is 11.27
The ratio of the lengths of the two rods which is length of aluminum to length of Invar rod is; 11.27
Formula for linear thermal expansion is;
ΔL = L × α × ΔT
Where;
ΔL is change in original length
L is original length
α is coefficient of linear expansion
ΔT is change in temperature
We are told that increase in length of aluminum rod is twice the increase in length of an Invar rod with only a third of the temperature increase.
Thus;
ΔL = 2ΔL
ΔT for the aluminum rod = ⅓ΔT for the Invar rod.
Thus, we have;
L_al × α_al × ⅓ΔT = 2L_in × 2α_in × 2ΔT
ΔT will cancel out to give;
⅓(L_al × α_al) = 2L_in × 2α_in × 2
Multiply both sides by 3 to get;
(L_al × α_al) = 6L_in × 6α_in × 6
From online tables, the linear coefficient of expansion of aluminum is 23 × 10^(-6) C¯¹
While the coefficient of thermal expansion for Invar rod is 1.2 × 10^(-6) K¯¹
Thus;
L_al × 23 × 10^(-6) = 6L_in × (6 × 1.2 × 10^(-6)) × 6
L_al/L_in = (6 × 6 × 1.2 × 10^(-6) × 6)/(23 × 10^(-6))
L_al/L_in = 11.27
Read more on coefficient of linear expansion at; https://brainly.com/question/6985348
