Question

A thin copper wire of $l$ length increases in length by $1\%$ when heated from $0^\circ {\text{C}}$ to $100^\circ {\text{C}}$, if a thin copper plate of area $2l \times l$ is heated from $0^\circ {\text{C}}$ to $100^\circ {\text{C}}$, the percentage increase in its area will be.
A) $1\%$
B) $2\%$
C) $3\%$
D) $4\%$

Answer 1

In this question, use the concept of the coefficient of the thermal expansion of the material that is when a solid material is heated or cooled, then the length of that object increases or decreases which depends on the coefficient of thermal expansion of the material.

Complete step by step answer:
As we know that in the Linear thermal expansion, when a solid material heated or cooled, then the change in length of solid material depends on the coefficient of thermal expansion of the material, the temperature difference, and the initial length of the object that is,
$\Delta L = \alpha L\Delta T$
Here, $\alpha$ is the thermal expansion of the solid material, $L$is the original length of the solid material, and $\Delta T$ is the change in temperature of the material.
Now we calculate the change in temperature of the copper wire when heated.
$\Delta T = 100\;^\circ {\text{C}} - 0\;^\circ {\text{C}} \\\ \Rightarrow \Delta T = 100\;^\circ {\text{C}} \\\$
So, the change in the temperature is $100\;^\circ {\text{C}}$.
Now, we substitute the value of the change in the length of the wire that is $\dfrac{{\Delta L}}{L} = \dfrac{1}{{100}}$ and the value of the change in the temperature in the thermal expansion expression to calculate the coefficient of thermal expansion of the material as,
$\Delta L = \alpha L\left( {100} \right)$
Rearrange the equation as,
$\Rightarrow \dfrac{{\Delta L}}{L} = \alpha 100$
we substitute the value of the change in the length of the wire that is $\dfrac{{\Delta L}}{L} = \dfrac{1}{{100}}$ as,
$\Rightarrow \dfrac{1}{{100}} = \alpha 100$
After calculation we get,
$\Rightarrow \alpha = {10^{ - 4}}\,{^\circ C}$
So, the value of the coefficient of thermal expansion is ${10^{ - 4}}\,{^\circ C}$.
In Area Thermal Expansion, when a solid material heated or cooled, then the change in the area of solid material is,
$\Delta A = \beta A\Delta T$
Here, $\beta$ is the thermal expansion of the solid material, $A$is the original area of the solid material, and $\Delta T$ is the change in temperature of the material.
As we know that the plate will expand in two directions when heated. So, $\beta = 2\alpha$
Now, we substitute the value of the thermal expansion in the above formula as,
$\Delta A = 2\alpha A\Delta T$
Now, we rearrange the above expression as,
$\Rightarrow \dfrac{{\Delta A}}{A} = 2\alpha \Delta T$
Now, we substitute ${10^{ - 4}}$ for $\alpha$ and $100$ for $\Delta T$ in the above equation to find the percentage increase in the area of the plate as,
$\Rightarrow \dfrac{{\Delta A}}{A} = 2\left( {{{10}^{ - 4}}} \right)100$
After simplification we get,
$\Rightarrow \dfrac{{\Delta A}}{A} = 0.02$
In the percentage form we get,
$\therefore \dfrac{{\Delta A}}{A} = 2\%$

$\therefore$ The percentage increases in the area of the plate is $2\%$. Hence, option (B) is correct.

Note:
When a solid material wire is heated or cooled, then the length of the wire increases or decreases which depends on the coefficient of thermal expansion of the material. The increase or decrease in the length of wire depends on the change in temperature and the coefficient of thermal expansion of the material of the wire.