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Worked Examples · Example 10.2

Q.A blacksmith fixes iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are 5.243 m5.243\ \text{m} and 5.231 m5.231\ \text{m}, respectively at 27 ∘C27\ ^\circ\text{C}. To what temperature should the ring be heated so as to fit the rim of the wheel?

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The ring's diameter must expand by exactly 0.0120.012 m to match the rim. Using linear expansion with αiron=1.2×10−5\alpha_{\text{iron}}=1.2\times10^{-5}/°C, the required temperature rise is about 191.2∘191.2^\circC, so the ring must be heated to about 218.2∘218.2^\circC.

A ring fits onto a wheel rim only when its inner diameter equals (or very slightly exceeds) the rim's diameter. Heating the ring makes it expand, increasing its diameter until it slips over the rim.

Step 1: Required increase in diameter

ΔD=Drim−Dring=5.243−5.231=0.012 m.\Delta D = D_{\text{rim}} - D_{\text{ring}} = 5.243 - 5.231 = 0.012\ \text{m}.

Step 2: Apply linear expansion

Diameter behaves like any linear dimension under thermal expansion:

ΔD=D0 α ΔT,\Delta D = D_0\,\alpha\,\Delta T,

where D0=5.231D_0 = 5.231 m is the ring's initial diameter and α=1.2×10−5 ∘C−1\alpha = 1.2\times10^{-5}\ ^\circ\text{C}^{-1} is the standard linear expansion coefficient of iron.

Step 3: Solve for the temperature rise …

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