Question

An iron tyre is to be fitted onto a wooden wheel $1\text{m}$ in diameter. The diameter of tyre is $6\text{mm}$ smaller than that of the wheel. The tyre should be heated so that its temperature (in${}^{\circ }\text{C})$ increases by a minimum of (Given : coefficient of volume expansion of iron is $3.6\times {10}^{-5}{/}^{\circ }\text{C})$

Answer 1

Initial diameter of tyre, $D=1\text{m}-6\text{mm}=1000\text{mm}-6\text{mm}=994\text{mm}$
Hence initial readius of tyre,
$R=\frac{994}{2}=497\text{mm}$

Required change in diamger to get fitted on to wheel is,
$\Delta D=6\text{mm}$
or, $\Delta R=\frac{6}{2}=3\text{mm}$
Let assume that after increasing temperature by $\Delta T$, tyre will fit onto wheel.
Increment in the length (circumference) of iron tyre is,
$\Delta L=L\alpha \Delta T$
$\Rightarrow 2\pi \Delta R=\left(2\pi R\right)\alpha \Delta T$
$(\because \alpha =\frac{\gamma }{3})$
$\Rightarrow 2\pi \Delta R=\left(2\pi R\right)\left(\frac{\gamma }{3}\right)\Delta T$
$\Rightarrow \Delta T=\frac{3\Delta R}{R}.\frac{1}{\gamma }$
Given, $\gamma =3.6\times {10}^{-5}{/}^{\circ }\text{C}$

$\Rightarrow \Delta T=\frac{3\times 3}{3.6\times {10}^{-5}\times 497}=\frac{250000}{497}$
$\therefore \Delta T={503.01}^{\circ }\text{C}$

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: When temperature is inereased the length of all the}\\ \text{dimensions of body will increase. In case of tyre, it can}\\ \text{be visualized as a body of linear length equal to its}\\ \text{circumference}2\pi R.\end{array}$