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Question: A thin metal square plate has side of length l. When it is heated from \(0^\circ C\) to \(100^\circ ...

A thin metal square plate has side of length l. When it is heated from 0∘C0^\circ C to 100∘C100^\circ C ,its length increases by 1%. What is the percentage increase in the area of the plate?
A) 2.00%
B) 2.02%
C) 2.03%
D) 2.01%

Explanation

Solution

Let the initial length and area be I 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.

Complete step by step answer: The length of the square plate is given to be l. We assume that the area of the plate is A. Then,
A=l×l=l2A = l \times l = {l^2}
Now, when the square plate is heating, its length is increasing by 1%. We assume that the increase in length is uniform, so let the new length of sides of square plate be l’. Mathematically,
l′=l+1%.l ⇒l′=l+1100l ⇒l′=1.01l  l' = l + 1\% .l \\\ \Rightarrow l' = l + \dfrac{1}{{100}}l \\\ \Rightarrow l' = 1.01l \\\
Let the new area of square plate be A’. The new area would become,
A′=l′×l′=(1.01l)2=1.0201l2A' = l' \times l' = {\left( {1.01l} \right)^2} = 1.0201{l^2}
Therefore, we can write the percentage increase in the area of plate as:
Δ%=A′−AA×100=1.0201l2−l2l2×100=0.0201l2l2×100 =0.0201×100=2.01%  \Delta \% = \dfrac{{A' - A}}{A} \times 100 = \dfrac{{1.0201{l^2} - {l^2}}}{{{l^2}}} \times 100 = \dfrac{{0.0201{l^2}}}{{{l^2}}} \times 100 \\\ = 0.0201 \times 100 = 2.01\% \\\
So, the correct answer is D)2.01%

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 P 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.