Solveeit Logo

Question

Question: When the temperature of a copper coin is raised by \(80{}^\circ C\), its diameter increases by 0.2%....

When the temperature of a copper coin is raised by 80C80{}^\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 0.25×104C10.25\times {{10}^{-4}}{}^\circ {{C}^{-1}}

Explanation

Solution

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 80C80{}^\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,
ΔT=80C\Delta T=80{}^\circ C
We know that the change in diameter is given by the equation,
ΔD=D(1+αΔT)\Delta D=D\left( 1+\alpha \Delta T \right), where ‘D’ is the original diameter, ‘α\alpha ’ is the coefficient of expansion and ‘ΔT\Delta T’ is the change in temperature.
From this equation we can write,
ΔDD=αΔT\Rightarrow \dfrac{\Delta D}{D}=\alpha \Delta T
We are given the change in diameter by 0.2%. Therefore,
ΔDD=αΔT=0.2\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,
ΔA=A(1+βΔT)\Delta A=A\left( 1+\beta \Delta T \right), where ‘A’ is the original area and ‘β\beta ’ is the coefficient of areal expansion.
Therefore we can write,
ΔAA=βΔT\Rightarrow \dfrac{\Delta A}{A}=\beta \Delta T
We know that,
β=2α\beta =2\alpha
Therefore we get,
ΔAA=2αΔT\Rightarrow \dfrac{\Delta A}{A}=2\alpha \Delta T
From earlier calculations we got, αΔT=0.2\alpha \Delta T=0.2%
Therefore,
ΔAA=2×0.2\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,
ΔV=V(1+γΔT)\Delta V=V\left( 1+\gamma \Delta T \right), where ‘V’ is the original volume and ‘γ\gamma ’ is the coefficient of expansion of volume.
From this,
ΔVV=γΔT\Rightarrow \dfrac{\Delta V}{V}=\gamma \Delta T
We know that, γ=3α\gamma =3\alpha . Thus,
ΔVV=3αΔT\Rightarrow \dfrac{\Delta V}{V}=3\alpha \Delta T
ΔVV=3×0.2\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,
ΔLL=αΔT\dfrac{\Delta L}{L}=\alpha \Delta T, where ‘α\alpha ’ is the coefficient of linear expansion.
We know that,
αΔT=0.2\alpha \Delta T=0.2%=\dfrac{0.2}{100} and ΔT=80\Delta T=80
Therefore,
0.2100=α×80\Rightarrow \dfrac{0.2}{100}=\alpha \times 80
From this we get.
α=0.280×100\Rightarrow \alpha =\dfrac{0.2}{80\times 100}
α=0.25×104C1\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,
ΔX=X(1+αΔT)\Delta X=X\left( 1+\alpha \Delta T \right), where ‘X’ is the original thickness.
Therefore we get,
ΔXX=αΔT\Rightarrow \dfrac{\Delta X}{X}=\alpha \Delta T
We are given that, αΔT=0.2\alpha \Delta T=0.2%
Thus we get,
ΔXX=αΔT=0.2\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.