Question

An iron tyre is to be fitted onto a wooden wheel $1.0\text{m}$ in diameter. The diameter of the tyre is $6\text{mm}$ smaller than that of the wheel. Coefficient of volumetric expansion of iron is $3.6\times {10}^{-5}{/}^{\circ }\text{C}$. Then, to fit the iron tyre on the wooden wheel,

• A. Coefficient of linear expansion of iron is $1.2\times {10}^{-5}{/}^{\circ }\text{C}$.
• B. Coefficient of linear expansion of iron is $2.4\times {10}^{-5}{/}^{\circ }\text{C}$.
• C. The tyre should be heated so that its temperature increases by a minimum of ${500}^{\circ }\text{C}$.
• D. The tyre should be heated so that its temperature increases by a minimum of ${300}^{\circ }\text{C}$.

Answer 1

Answer: (A), (C)

The correct options are
A Coefficient of linear expansion of iron is $1.2\times {10}^{-5}{/}^{\circ }\text{C}$.
C The tyre should be heated so that its temperature increases by a minimum of ${500}^{\circ }\text{C}$.

Given, Coefficient of volume expansion of iron is $\gamma =3.6\times {10}^{-5}{/}^{\circ }\text{C}$
Thus,
Coefficient of linear expansion is $\alpha =\frac{\gamma }{3}=1.2\times {10}^{-5}{/}^{\circ }\text{C}$
Change in length of diameter of the iron tyre so that it fits the wooden wheel should be $\Delta d=6\text{mm}=0.006\text{m}$
So, original diameter of the iron tyre $=1-0.006=0.994\text{m}$
We know,
$\Delta d=d\times \alpha \times \Delta T$
$\Rightarrow 0.006=0.994\times 1.2\times {10}^{-5}\times \Delta T$
$\therefore \Delta T\approx {500}^{\circ }\text{C}$
Thus, options (a) and (c) are correct answers.