Suppose that a rigid aluminum wire were to be strung out in a loop that just fits snugly around the equator (assuming a perfectly spherical Earth with a radius of $\left. 6.37 \times 10 ^ { 6 } \mathrm {~m} \right)$
If the temperature of the wire is increased by $0.50 ^ { \circ } \mathrm { C }$ and the increase in length is distributed equally over the entire length, how far off the ground will the wire loop be if it remained centered on the earth? The coefficient of linear expansion of aluminum is $24 \times 10 ^ { - 6 } K ^ { - 1 }$
A) 76 mm
B) 76 m
C) 7.6 mm
D) 76 cm
E) 760 m

Question

Suppose that a rigid aluminum wire were to be strung out in a loop that just fits snugly around the equator (assuming a perfectly spherical Earth with a radius of $\left. 6.37 \times 10 ^ { 6 } \mathrm {~m} \right)$
If the temperature of the wire is increased by $0.50 ^ { \circ } \mathrm { C }$ and the increase in length is distributed equally over the entire length, how far off the ground will the wire loop be if it remained centered on the earth? The coefficient of linear expansion of aluminum is $24 \times 10 ^ { - 6 } K ^ { - 1 }$
A)  76 mm
B)  76 m
C)  7.6 mm
D)  76 cm
E)  760 m

Quizplus · Accepted Answer

The original circumference of the wire loop is equal to the circumference of the Earth, which is $2\pi r$, where $r = 6.37 \times 10^6 \, \text{m}$. The change in length ($\Delta L$) of the wire due to thermal expansion can be calculated using the formula $\Delta L = \alpha L_0 \Delta T$, where $\alpha = 24 \times 10^{-6} \, K^{-1}$, $L_0 = 2\pi r$, and $\Delta T = 0.50^\circ C$. Substituting the values, we get $\Delta L = 24 \times 10^{-6} \times 2\pi \times 6.37 \times 10^6 \times 0.5$. Simplifying this gives $\Delta L \approx 76 \, \text{m}$. This means the wire would be lifted uniformly off the ground by approximately 76 meters if it expanded equally in all directions while remaining centered on the Earth.

Suppose That a Rigid Aluminum Wire Were to Be Strung 6.37×106 m)\left. 6.37 \times 10 ^ { 6 } \mathrm {~m} \right)6.37×106 m)

Suppose That a Rigid Aluminum Wire Were to Be Strung $\left. 6.37 \times 10 ^ { 6 } \mathrm {~m} \right)$