Question

A hole is drilled in a copper sheet. The diameter of hole is $4.24cm$ at ${27.0^0}C$. Diameter of the hole when it is heated to ${35^0}C$ is:

Answer 1

Hint: The initial diameter is given; we can find the radius and thus the area of the copper sheet. Assume coefficient of linear expansion for copper to be some variable. Then for the expansion in area, use this linear expansion coefficient to find the coefficient of superficial expansion. Then use the formula for thermal expansion. Also, as the diameter is length and not an area therefore, we can calculate the change in diameter using formula for linear thermal expansion.

Complete step by step solution:
When some object is heated, there is change in the shape of the object. For objects such as metal ruler or rod this change is along their length only. For objects such as sheets, plates which have an area, the change is throughout the area. Also, there is change in the volume of objects due to thermal heating.

We are given with a copper sheet and we need to find the change in its diameter. The diameter being length only will have linear expansion due to thermal heating.

The formula of linear thermal expansion for the diameter change for copper sheet will be given as
$\Delta D = {D_0}\alpha \Delta T$ ---- equation 1
Here, $\Delta D$ is the change in length
${D_0}$ is the initial length
$\alpha $ is coefficient for linear expansion of copper
$\Delta T$ is the change in temperature.
The given values are ${D_0} = 4.24cm$
$\Delta T = {35^0}C - {27^0}C = {8^0}C$
Substituting the given values in equation 1 , we get
$\Delta D = 4.24 \times \alpha \times 8$
$ \Rightarrow \Delta D = 17.92\alpha \,cm$

Therefore, the change in diameter will be $17.92\alpha \,cm$ when the sheet is heated to ${35^0}C$.

Note: Over small temperature ranges, the fractional thermal expansion of uniform linear objects is proportional to the change in temperature. This fact is used to construct thermometers based on the expansion of a thin tube of mercury or alcohol. If the value of coefficient for linear expansion of copper was given, we just needed to multiply that value to the final answer.