Question

When the temperature of a thin copper coin is raised by ${80}^{\circ }\text{C}$, its diameter increases by $0.2\text{\%}$. The coefficient of linear expansion$\left(\alpha \right)$ and percentage rise in the area of coin will be :

• A. $\alpha =0.80\times {10}^{-4}{/}^{\circ }\text{C}$, percentage rise in the area of coin is $0.6\text{\%}$
• B. $\alpha =0.25\times {10}^{-4}{/}^{\circ }\text{C}$, percentage rise in the area of coin is $0.4\text{\%}$
• C. $\alpha =1.25\times {10}^{-4}{/}^{\circ }\text{C}$, percentage rise in the area of coin is $0.2\text{\%}$
• D. $\alpha =0.50\times {10}^{-4}{/}^{\circ }\text{C}$, percentage rise in the area of coin is $0.80\text{\%}$

Answer 1

The correct option is B $\alpha =0.25\times {10}^{-4}{/}^{\circ }\text{C}$, percentage rise in the area of coin is $0.4\text{\%}$

If the diameter increases by $0.2\text{\%}$ radius will also increase by $0.2\text{\%}$.

$\Rightarrow \frac{\Delta r}{r}\times 100=0.2$

Since area of a face $A=\pi r^2$

$\Rightarrow \frac{\Delta A}{A}\times 100=2\times \frac{\Delta r}{r}\times 100$

$\text{\%}$ increase in area is

$\frac{\Delta A}{A}\times 100=2\times 0.2=0.4\%$
Increase in diameter due to temperature rise is analogous to linear expansion. The coin will tend to expand radially outwards.
$\Rightarrow \Delta r=r\alpha \Delta T$

$\%$ increase in radius can be given as,

$\frac{\Delta r}{r}\times 100=\left(\alpha \Delta T\right)\times 100$

$\Rightarrow 0.2=\alpha \times 80\times 100$

$\therefore \alpha =\frac{2}{8}\times {10}^{-4}=0.25\times {10}^{-4}{/}^{\circ }\text{C}$

Hence coefficient of linear expansion is $0.25\times {10}^{-4}{/}^{\circ }\text{C}$

So option $\left(b\right)$ is correct.