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postnew [5]
3 years ago
7

The increase in length of an aluminum rod is twice the increase in length of an Invar rod with only a third of the temperature i

ncrease. Find the ratio of the lengths of the two rods.
Mathematics
1 answer:
cluponka [151]3 years ago
5 0

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, \alpha _{Al} = 23 \times 10^{-6} /K

Coefficient of linear expansion of Invar, \alpha _{Iv} = 1.2 \times 10^{-6}/K

Linear thermal expansion is given as;

\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

The increase in length of Invar is given as;

\Delta L_{Iv}  = L_0_{Iv} \times \alpha _{Iv}\times  \Delta T_{Iv}

The increase in length of the Aluminum;

\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}\\\\

\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

Therefore, the ratio of lengths of the two rods, Aluminum to Invar is 11.27

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