
An iron tyre is to be fitted on to a wooden wheel 1 m in diameter. The diameter of the tyre is 6mm smaller than that of the wheel. The tyre should be heated so that its temperature increases by a minimum of (the coefficient of cubical expansion of iron is \[3.6\times {{10}^{-5}}/{}^\circ C\].
(A) 167\[{}^\circ C\]
(B) 334\[{}^\circ C\]
(C) 500\[{}^\circ C\]
(D) 1000\[{}^\circ C\]
Answer
570.6k+ views
Hint: This is a problem in which upon heating there takes place expansion of the material. We are given with the cubical expansion of iron thus, here there is an expansion of volume. There can be expansion of length as well as area too.
Complete step by step answer:
Given diameter, d=1 mm
Radius, r=\[\frac{d}{2}\]=0.5 mm
r= \[0.5\times {{10}^{-3}}m\]
cubical expansion \[\gamma =3.6\times {{10}^{-5}}/{}^\circ C\]
The tyre is to be fitted on a wheel whose diameter is 1m. So, the initial diameter is (1000-6) =994 mm
converting this into m we get 0.994m. Using the formula,
\[\begin{align}
&\Rightarrow 3\alpha =\gamma \\
&\Rightarrow 3\alpha =3.6\times {{10}^{-5}}/{}^\circ C \\
&\Rightarrow \alpha =1.2\times {{10}^{-5}}/{}^\circ C \\
\end{align}\]
Where \[\alpha \]is the coefficient of linear expansion. Using, the formula
\[\begin{align}
&\Rightarrow l={{l}_{0}}(1+\alpha \Delta T) \\
&\Rightarrow 1000=994(1+1.2\times {{10}^{-5}}\Delta T) \\
&\Rightarrow 1.00603=1+1.2\times {{10}^{-5}}\Delta T \\
&\Rightarrow 0.00603=1.2\times {{10}^{-5}}\Delta T \\
&\Rightarrow \Delta T\approx 500{}^\circ C \\
\end{align}\]
Thus, the temperature change comes out to be approximately 500\[{}^\circ C\].
So, the correct option is (C).
Note: Here, \[\Delta T\]is the change in temperature that is the difference between final value and initial value of temperatures. We need to keep in mind although we have to take the values of temperature always in kelvin but when we talk about difference then it does not matter what units are.these coefficients are unique for every material. For copper and iron under the same conditions of temperature and pressure, their values will be different.
Complete step by step answer:
Given diameter, d=1 mm
Radius, r=\[\frac{d}{2}\]=0.5 mm
r= \[0.5\times {{10}^{-3}}m\]
cubical expansion \[\gamma =3.6\times {{10}^{-5}}/{}^\circ C\]
The tyre is to be fitted on a wheel whose diameter is 1m. So, the initial diameter is (1000-6) =994 mm
converting this into m we get 0.994m. Using the formula,
\[\begin{align}
&\Rightarrow 3\alpha =\gamma \\
&\Rightarrow 3\alpha =3.6\times {{10}^{-5}}/{}^\circ C \\
&\Rightarrow \alpha =1.2\times {{10}^{-5}}/{}^\circ C \\
\end{align}\]
Where \[\alpha \]is the coefficient of linear expansion. Using, the formula
\[\begin{align}
&\Rightarrow l={{l}_{0}}(1+\alpha \Delta T) \\
&\Rightarrow 1000=994(1+1.2\times {{10}^{-5}}\Delta T) \\
&\Rightarrow 1.00603=1+1.2\times {{10}^{-5}}\Delta T \\
&\Rightarrow 0.00603=1.2\times {{10}^{-5}}\Delta T \\
&\Rightarrow \Delta T\approx 500{}^\circ C \\
\end{align}\]
Thus, the temperature change comes out to be approximately 500\[{}^\circ C\].
So, the correct option is (C).
Note: Here, \[\Delta T\]is the change in temperature that is the difference between final value and initial value of temperatures. We need to keep in mind although we have to take the values of temperature always in kelvin but when we talk about difference then it does not matter what units are.these coefficients are unique for every material. For copper and iron under the same conditions of temperature and pressure, their values will be different.
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