# An iron Tyre is to be filled onto a wooden wheel of 100 cm in diameter. The diameter of the Tyre...

## Question:

An iron tyre is to be filled onto a wooden wheel of 100 cm in diameter. The diameter of the tyre is 5mm smaller than that of the wheel. How much should the temperature of the tyre be increased for this purpose?

Coefficient of linear expansion of iron is {eq}12 \times 10 ^{-6}{/eq}

## Linear Expansion:

In physics, the linear expansion can be described as the variation in the length of the specimen that directly varies with the change in the temperature. The coefficient of linear expansion has different values for the different materials

Given data:

• The diameter of the wheel is {eq}{d_w} = 100\,{\rm{cm}} = 1000\,{\rm{mm}} {/eq}
• The diameter of the tire is {eq}{d_t} = \left( {1000 - 5} \right)\,{\rm{mm}} = 995\,{\rm{mm}} {/eq}
• The coefficient of linear expansion of iron is {eq}\alpha = 12 \times {10^{ - 6}}\,{\rm{/}}^\circ {\rm{C}} {/eq}

The radius of the wheel is given as

{eq}{r_w} = \dfrac{{{d_w}}}{2} {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} {r_w} &= \dfrac{{{d_w}}}{2}\\ {r_w} &= \dfrac{{1000}}{2}\\ {r_w} &= 500\,{\rm{mm}} \end{align*} {/eq}

Similarly the radius of the tire is given by

{eq}{r_t} = \dfrac{{{d_t}}}{2} {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} {r_t} &= \dfrac{{{d_t}}}{2}\\ {r_t} &= \dfrac{{995}}{2}\\ {r_t} &= 497.5\,{\rm{mm}} \end{align*} {/eq}

The expression for the coefficient of linear expansion is given by

{eq}2\pi{r_w} = 2\pi{r_t} + 2\pi{r_t}\alpha \Delta T {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} {r_w} &= {r_t} + {r_t}\alpha \Delta T\\ 500 &= 497.5 + 497.5 \times 12 \times {10^{ - 6}}\Delta T\\ \Delta T &= 418.76\,^\circ {\rm{C}} \end{align*} {/eq}

Thus the temperature increased for this purpose is {eq}\Delta T = 418.76\,^\circ {\rm{C}} {/eq} 