Question

A thin copper wire of length $L$ increase in length by $1\%$ when heated from temperature $T1$ to $T2$ .What is the percentage change in area when a thin copper plate having dimensions $2L \times L$ is heated from $T1$ to $T2$ $?$
$\left( A \right)\,\,1\%$
$\left( B \right)\,\,2\%$
$\left( C \right)\,\,3\%$
$\left( D \right)\,\,4\%$

Answer 1

Hint
From the question, the length of the wire is increased so we have to find the actual length of the wire, then find the difference of the change in length. Initial temperature of the wire is increased to another temperature. So we have to find the change in temperature of the wire.
The expression for finding the change in the area of wire is $\Delta A = \beta A\Delta T$
Where,
$\Delta A$ be the change in area of the wire,
$\Delta T$ be the change in temperature of the wire,
$A$be the original area of wire
$\beta$ is the volumetric coefficient of thermal expansion.

Complete step by step solution
We know that, change in temperature $\Delta T = T2 - T1$.
$\Delta T = 100 - 0$
$\Delta T = {100^ \circ }C$
Change in length $\dfrac{{\Delta l}}{l} = length = l = 1\% = \dfrac{1}{{100}}$
Hence $\dfrac{{\Delta l}}{l} = 0.01$
Change in length $\Delta l = \alpha L\Delta T$
From the above equation,
Change in length $\dfrac{{\Delta l}}{l} = \alpha \times 100$
Substitute the known values in the above equation,
$0.01 = \alpha \times 100$
Find $\alpha$from the above equation so,
$\alpha = \dfrac{{0.01}}{{100}}$
$\alpha = 1 \times {10^{ - 4}}$
From the question we know that,
Area of copper plate$= 2l \times l = 2{l^2}$
By the Thermal expansion theory,
Change in area =$\Delta A = \beta A\Delta T$
$\dfrac{{\Delta A}}{A} = \beta \Delta T$
But we know that the volumetric coefficient of thermal expansion is twice the linear coefficient of thermal expansion.
that is $\beta = 2\alpha$
where $\alpha$ is the linear coefficient of thermal expansion.
Substitute the $\beta$ value in the above equation
$\dfrac{{\Delta A}}{A} = 2\alpha \Delta T$
Substitute the known values in the above equation,
$\dfrac{{\Delta A}}{A} = 2 \times 1 \times {10^{ - 4}} \times 100$
$\dfrac{{\Delta A}}{A} = 2 \times {10^{ - 2}}$
$\dfrac{{\Delta A}}{A} \times 100 = 2 \times {10^{ - 2}} \times 100$
$\dfrac{{\Delta A}}{A} = 2\% \,\left( {\dfrac{2}{{100}}} \right)$
Hence from the above option, $2\%$ is the correct answer.
Thus, the option (B) is correct.

Note
In the question, we know that temperature increases so we have to find the percentage of change in temperature of the wire. We have to find the change in area by thermal expansion theory because the area of the plate becomes changed.