A blacksmith fixes iron ring on the rim of a wooden wheel of a horse cart. The diameter of the rim and ring are 5.255 m and 5.25 m respectively, at 20 ⁰C. The minimum temperature to which the ring is to be heated so as to fit it on the rim is nearly $(α_{i r o n} = 1.2 \times 10^{- 5} K^{- 1})$

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$60^{ο} C$

$80^{ο} C$

$100^{ο} C$

$120^{ο} C$

answer is B.

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Detailed Solution

$d_{2} = d_{1} (1 + α Δ t)$
$\begin{array}{l} Δ t = (\frac{5.255}{5.25} - 1) \times \frac{1}{12 \times 10^{- 6}} \\ = \frac{5 \times 10^{- 2} \times 1^{6}}{525 \times 12} = {79.4}^{0} C \\ t_{2} = {99.4}^{0} C \end{array}$

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