An iron tyre is to be fitted onto a wooden wheel 1.0 m in diameter. The diameter of the tyre is 6 mm smaller than that of wheel. The tyre should be heated so that its temperature increases by a minimum of (coefficient of volume expansion of iron is $3.6\times {10}^{-5}{/}^{\circ }\mathrm{C}$)

Initial diameter of tyre = (1000 - 6) mm;
                                   = 994 mm, so initial radius of tyre

$\mathrm{R}=\frac{994}{2}=497\mathrm{mm}$

and change in diameter $∆$D = 6 mm; so

$\mathrm{\Delta R}=\frac{6}{2}=3\mathrm{mm}$

Given that after increasing temperature by $∆$T tyre will fit onto wheel. Increment in the length (circumference) of the iron tyre

$\mathrm{\Delta L}=\mathrm{L}\times \mathrm{\alpha }\times \mathrm{\Delta T}=\mathrm{L}\times \frac{\mathrm{\gamma }}{3}\times \mathrm{\Delta T}\left(\text{ As }\mathrm{\alpha }=\frac{\mathrm{\gamma }}{3}\right)$

$\begin{array}{l}\Rightarrow 2\mathrm{\pi \Delta R}=2\mathrm{\pi R}\left(\frac{\mathrm{\gamma }}{3}\right)\mathrm{\Delta T}\\ \Rightarrow \mathrm{\Delta T}=\frac{3}{\mathrm{\gamma }}\frac{\mathrm{\Delta R}}{\mathrm{R}}=\frac{3\times 3}{3.6\times {10}^{-5}\times 497}\\ [\text{ As }\mathrm{\Delta R}=3\mathrm{mm}\text{ and }\mathrm{R}=497\mathrm{mm}]\end{array}$

$\Rightarrow \mathrm{\Delta }T={500}^{\circ }\mathrm{C}$