If a steel band were to fit snugly around the Earth's equator at 2 5 ∘ C , but then was heated to 4 5 ∘ C , how high above the Earth would the band be (assume equal everywhere)? Take the coefficient of linear expansion of steel to be 12 × 1 0 − 6 ∘ C − 1 .

2 π ( R + h ) = 2 π R ( 1 + α Δ T ) h = R α Δ T R = 6400 km Δ T = 45 − 25 = 20 ∘ C h = 6400 × 12 × 1 0 − 6 × 20 h = 1.536 km

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If a steel band were to fit snugly around the Earth's equator at $2 5^{\circ} C$ , but then was heated to $4 5^{\circ} C$ , how high above the Earth would the band be (assume equal everywhere)? Take the coefficient of linear expansion of steel to be $12 \times 1 0^{- 6}^{\circ} C^{- 1}$ .

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Answer:

Solution:

$2 π (R + h) = 2 π R (1 + α Δ T)$
$h = R α Δ T$
$R = 6400 km$
$Δ T = 45 - 25 = 20^{\circ} C$
$h = 6400 \times 12 \times 1 0^{- 6} \times 20$
$h = 1.536 km$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

⏱ 2mℹ️ Source: QB

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