A thin copper wire of length L increases in length by 1% when heated from temperature ${{\text{T}}_{\text{1}}}\text{to}{{\text{T}}_{\text{2}}}$ what is the percentage change in area when a thin copper plate is heated having dimensions $\text{2L }\!\!\times\!\!\text{ L}$ is heated from ${{\text{T}}_{\text{1}}}\text{to}{{\text{T}}_{\text{2}}}$
a) 1%
b) 2%
c) 3%
d) 4%

- Hint: The material of the wire as well as the sheet is the same i.e. copper. Hence we can write the coefficient of linear expansion ($\alpha $) is half the coefficient of superficial expansion( $\beta $), mathematically the relation can be written as, $\beta =2\alpha ...(1)$. The percentage increase is basically the ratio of increase in dimension to the original dimension of the body. Hence we can relate the expansion of the two above bodies using the relation i.e. $\beta =2\alpha $ determine the percentage change in the area of the sheet.

Complete step-by-step solution
To begin with, let us write the equation of linear expansion and superficial expansion and express them in terms of the percentage increase in each of them.
The equation of linear expansion in a rod of length L is given by,
$L=l(1+\alpha \Delta T)$ where L is the length of the rod or the wire at the final change in temperature, $l$ is the length of the rod or the wire at some initial temperature, $\alpha $ is the coefficient of linear expansion and $\Delta T$ is the change in temperature from the initial length of the rod to the final length of the rod. The above equation can also be written as,
$\begin{align}
  & L=l(1+\alpha \Delta T) \\
 & \alpha =\dfrac{L-l}{l\Delta T}...(2) \\
\end{align}$
The term $\dfrac{L-l}{l}$ is analogous to the percentage increase in the length of the wire.
Now let us write the equation of superficial expansion.
$A=a(1+\beta \Delta T)$ where A is the area of the sheet at final change in temperature, a is the area of the at some initial temperature, $\beta $is the coefficient of superficial expansion and $\Delta T$ is the change in temperature from the initial area of the sheet to the final area of the sheet. The above equation can also be written as,
$\begin{align}
  & A=a(1+\beta \Delta T) \\
 & \beta =\dfrac{A-a}{a\Delta T}....\text{(3) } \\
\end{align}$
The term $\dfrac{A-a}{a}$ is analogous to the percentage increase in the area of the sheet.
The expansions of both sheet and the wire is taking place at the same change in temperature i.e. from ${{\text{T}}_{\text{1}}}\text{to}{{\text{T}}_{\text{2}}}$. Now let us write equation 1 and substitute equation 2 and 3 in that.
$\begin{align}
  & \beta =2\alpha \\
 & \dfrac{A-a}{a\Delta T}=2\left( \dfrac{L-l}{l\Delta T} \right) \\
\end{align}$
$\text{ }\!\!\Delta\!\!\text{ T}$ is same for both the wire as well as the sheet, hence
$\begin{align}
  & \dfrac{A-a}{a}=2\left( \dfrac{L-l}{l} \right) \\
 & \dfrac{A-a}{a}=2\left( 1\% \right) \\
 & \dfrac{A-a}{a}=2\% \\
\end{align}$
Hence the percentage increase in the area of the sheet is 2%.

Note: We could use the equation 1 as the material of the sheet and the wire were the same i.e. copper. The term $\dfrac{A-a}{a}$ is taken as the percentage increase in the area for simplicity. The percentage increase is actually given by $\dfrac{A-a}{a}\times 100$. We could just use the term $\dfrac{A-a}{a}$ as the increase in the area we equated this to the percentage increase in the length of the wire i.e. $\dfrac{L-l}{l}$.