Question

When the temperature of a copper coin is raised by $80{}^\circ C$, its diameter increases by 0.2%.
A. Percentage rise in the area of face is 0.4%
B. Percentage rise in the thickness is 0.4%
C. percentage rise in the volume is 0.6%
D. Coefficient of linear expansion of copper is $0.25\times {{10}^{-4}}{}^\circ {{C}^{-1}}$

Answer 1

Hint: We are given the change in temperature and the change in diameter of a coin. To solve the question we need to find the expansion in area, thickness, volume and also the coefficient of linear expansion of the coin using the known formula. By comparing the results with the given options we will get the solution.

Complete answer:
In the question it is said that the temperature of a coin is increased by $80{}^\circ C$, due to this there is an increase in the diameter of the coin by 0.2%.
Therefore we have the change in temperature,
$\Delta T=80{}^\circ C$
We know that the change in diameter is given by the equation,
$\Delta D=D\left( 1+\alpha \Delta T \right)$, where ‘D’ is the original diameter, ‘$\alpha $’ is the coefficient of expansion and ‘$\Delta T$’ is the change in temperature.
From this equation we can write,
$\Rightarrow \dfrac{\Delta D}{D}=\alpha \Delta T$
We are given the change in diameter by 0.2%. Therefore,
$\Rightarrow \dfrac{\Delta D}{D}=\alpha \Delta T=0.2%$
Now let us calculate the change in area of the face of the coin.
We have the equation for change in area as,
$\Delta A=A\left( 1+\beta \Delta T \right)$, where ‘A’ is the original area and ‘$\beta $’ is the coefficient of areal expansion.
Therefore we can write,
$\Rightarrow \dfrac{\Delta A}{A}=\beta \Delta T$
We know that,
$\beta =2\alpha $
Therefore we get,
$\Rightarrow \dfrac{\Delta A}{A}=2\alpha \Delta T$
From earlier calculations we got, $\alpha \Delta T=0.2%$
Therefore,
$\Rightarrow \dfrac{\Delta A}{A}=2\times 0.2%=0.4%$
Therefore the percentage rise in the area of the face is 0.4%.
Hence the first statement is correct.
Now let us calculate the change in volume. The equation is given as,
$\Delta V=V\left( 1+\gamma \Delta T \right)$, where ‘V’ is the original volume and ‘$\gamma $’ is the coefficient of expansion of volume.
From this,
$\Rightarrow \dfrac{\Delta V}{V}=\gamma \Delta T$
We know that, $\gamma =3\alpha $. Thus,
 $\Rightarrow \dfrac{\Delta V}{V}=3\alpha \Delta T$
$\Rightarrow \dfrac{\Delta V}{V}=3\times 0.2%=0.6%$
Therefore the percentage rise in volume is 0.6%.
Hence the third statement is also correct.
Now we can calculate the coefficient of linear expansion of copper.
The equation for linear expansion is given as,
$\dfrac{\Delta L}{L}=\alpha \Delta T$, where ‘$\alpha $’ is the coefficient of linear expansion.
We know that,
$\alpha \Delta T=0.2%=\dfrac{0.2}{100}$ and $\Delta T=80$
Therefore,
$\Rightarrow \dfrac{0.2}{100}=\alpha \times 80$
From this we get.
$\Rightarrow \alpha =\dfrac{0.2}{80\times 100}$
$\Rightarrow \alpha =0.25\times {{10}^{-4}}{}^\circ {{C}^{-1}}$
Thus we prove the third statement also as correct.

So, the correct answer is “Option A,C and D”.

Note:
We know that the equation for expansion in volume is given as,
$\Delta X=X\left( 1+\alpha \Delta T \right)$, where ‘X’ is the original thickness.
Therefore we get,
$\Rightarrow \dfrac{\Delta X}{X}=\alpha \Delta T$
We are given that, $\alpha \Delta T=0.2%$
Thus we get,
$\Rightarrow \dfrac{\Delta X}{X}=\alpha \Delta T=0.2%$
Hence the percentage rise in thickness is 0.2% and not 0.6% as given in option B.