Question

A steel cylinder of diameter ‘exactly’ 1 $ \text{cm} $ at $ 30{}^\circ \text{C} $ is to be fitted into a hole in a steel plate. The diameter of the hole is $ 0\cdot 99970\text{ cm} $ at $ 30{}^\circ \text{C} $ . To what temperature should the plate be heated? Given $ {{\alpha }_{\text{Steel}}}=1\cdot 1\times {{10}^{-50}}\text{ }{{\text{C}}^{-1}} $ .

Answer 1

Hint: Thermal expansion is the tendency of matter to change its shape, area, volume and density in response to a change in temperature, usually not including phase transitions.
Linear thermal expansion is
$ \vartriangle \text{L}=\alpha \text{ L}\vartriangle \text{T} $
$ \vartriangle \text{L} $ =change in length L
$ \vartriangle \text{T} $ =change in temperature
$ \alpha $ =co-efficient of linear expansion which varies slightly with temperature.

Complete step by step solution
A cylinder has diameter 1 $ \text{cm} $ at $ 30{}^\circ \text{C} $ is to fitted into a hole of diameter $ 0\cdot 99975\text{ cm} $ in a steel plate at the same temperature.
For exactly fitting a cylinder into hole the diameter of the cylinder and hole should be equal. But in this case, the diameter is different. We have to increase the diameter.
$ \vartriangle \text{d} $ is increase in diameter
$ \begin{align}
 & \vartriangle \text{d}=1-0\cdot 9997 \\
 & \text{ }=0\cdot 0003\text{ cm } \\
\end{align} $
According to thermal expansion
$ \begin{align}
  & \vartriangle \text{d}={{\text{d}}_{0}}\alpha \text{ }\vartriangle \text{t} \\
 & {{\text{d}}_{0}}=0\cdot 9997\text{ cm} \\
 & \alpha =1\cdot 1\times {{10}^{-50}}\text{ }{{\text{C}}^{-1}} \\
\end{align} $
$ \begin{align}
  & \vartriangle \text{t}=\dfrac{\vartriangle \text{d}}{{{\text{d}}_{0}}\alpha }=\dfrac{0\cdot 0003}{0\cdot 9997\times 1\cdot 1\times {{10}^{-5}}} \\
 & \vartriangle \text{t}=\dfrac{3}{9997\times 1\cdot 1}\times {{10}^{5}} \\
 & \text{ }=\dfrac{3}{9996\cdot 7}\times {{10}^{5}} \\
 & \text{ }=27\cdot {{2}^{0}}\text{ C} \\
\end{align} $
To increase the diameter we have to heat the plate at $ 27\cdot {{2}^{0}}\text{ C} $
Additional information:
-In two-dimensions: For small temperature changes, the change in area $ \vartriangle \text{A} $ is given by
 $ \vartriangle \text{A}=2\alpha \text{A}\vartriangle \text{T} $
 $ \vartriangle \text{A} $ =change in area A
 $ \vartriangle \text{T} $ =change in temperature
 $ \alpha $ = coefficient of linear expansion
-In three-dimensions: The change in volume $ \vartriangle \text{V} $ is
 $ \vartriangle \text{V}=3\alpha \text{V}\vartriangle \text{T} $
This equation is usually written as
$ \vartriangle \text{V}=\beta \text{V}\vartriangle \text{T} $
$ \beta $ =coefficient of volume expansion and $ \beta =3\alpha $ .

Note
To solve the numerical keep in mind that there are different formulas for change in area, change in volume and change in length. The coefficient of linear expansion has many applications in our daily life. It is used in fixing iron rims with the wooden wheel firmly and if we find it difficult to remove the stopper from a glass bottle, we can heat the neck of the bottle. Now the neck of the bottle expands and the stopper comes out easily.