Question

A thin metal square plate has length l. When it is heated from $0^{\circ }C$ to ${100}^{\circ }C$, its length increases by $1\mathrm{\%}$. What is the percentage increase in the area of the plate?
A. 2.00%
B. 2.01%
C. 2.02%
D. 2.03%

Answer 1

Hint: Let the initial length and area be l and A respectively. Now, given that the length increases by 1% determine the new length. This increase in length results in an increase in the area of the square plate. Find the new area by taking the square of the new length. To determine the percentage increase, take the ratio of the difference between the initial and final areas to the initial area and multiply the resulting quantity by 100 to obtain the appropriate result.

Formula used:
Percentage increase in area: ${\displaystyle \frac{A_f-A_i}{A_i}}\times 100\mathrm{\%}$

Complete answer:
We are given that the initial length of the sides of the square plate is l.
Let the area of the square plate be $A=l\times l=l^2$.
Now, when the square plate is heating, its length increases by $1\mathrm{\%}$. Assuming a uniform increase in length, let the new length of the sides of the square plate be $l^{\prime }$.
$l^{\prime }=l+1\mathrm{\%}.l=l+{\displaystyle \frac{1}{100}}l=l+0.01l=1.01l$
Therefore, the new area of the square plate becomes:
$A^{\prime }=l^{\prime }\times l^{\prime }=1.01l^2=1.0201l^2$
Therefore, the percentage increase in the area of the plate can be given as:
${\displaystyle \frac{A^{\prime }-A}{A}}\times 100={\displaystyle \frac{1.0201l^2-l^2}{l^2}}\times 100={\displaystyle \frac{0.0201l^2}{l^2}}\times 100=0.0201\times 100=2.01\mathrm{\%}$

So, the correct answer is “Option B”.

Note:
Usually, the change in length with a change in temperature is characterized by the coefficient of linear thermal expansion. It is important to remember that the coefficient of thermal expansion in general, measures the fractional change in size per degree change in temperature but at a constant pressure. In the above problem, since we were already given the numerical quantity by which the length changes, we did not have to worry about the coefficient of thermal expansion. However, most problems call for the calculation of change in size with changing temperature when only the coefficient in thermal expansion is given. In such a case, assuming linear expansion:
${\displaystyle \frac{\mathrm{\Delta }L}{L}}=\alpha \mathrm{\Delta }T$, where ${\displaystyle \frac{\mathrm{\Delta }L}{L}}$ is the fractional change in length, $\alpha$ is the linear coefficient of thermal expansion, and $\mathrm{\Delta }T$ is the change in temperature.
However, we can also directly find the fractional change in area by bringing in the coefficient of superficial expansion, given as:
${\displaystyle \frac{\mathrm{\Delta }A}{A}}=\beta \mathrm{\Delta }T$, where ${\displaystyle \frac{\mathrm{\Delta }A}{A}}$ is the fractional change in area, $\beta$ is the superficial coefficient of thermal expansion, and $\mathrm{\Delta }T$ is the change in temperature.