Question

A thin copper wire of length $l$ metre increases in length by $2\% $ when heated through ${10^ \circ }C$. What is the percentage increase in area when a square copper sheet of length $l$ metre is heated through ${10^ \circ }C$?
(A) $4\% $
(B) $8\% $
(C) $16\% $
(D) None of these

Answer 1

Hint
To solve this question, we have to calculate the coefficient of linear expansion using the formula of the thermal expansion for the copper rod. From there we have to obtain the value of coefficient of superficial expansion with which the final answer would be evaluated.

Formula Used: The formulae used in solving this question are given by
$\Rightarrow \Delta l = \alpha l\Delta T$
$\Rightarrow \Delta A = \beta A\Delta T$
Here $\alpha $ is the coefficient of linear expansion, $\beta $ is the coefficient of superficial expansion, and the rest of the symbols have their usual meanings.

Complete step by step answer
Let the coefficient of linear expansion of copper be $\alpha $.
We know that the change in length of a wire in terms of the change in temperature is given by
$\Rightarrow \Delta l = \alpha l\Delta T$ …...(1)
Dividing both sides by $l$ we get
$\Rightarrow \dfrac{{\Delta l}}{l} = \alpha \Delta T$
For the percentage change, we multiply both sides by $100$
$\Rightarrow \dfrac{{\Delta l}}{l} \times 100 = 100\alpha \Delta T$
According to the question, we have
$\Rightarrow \dfrac{{\Delta l}}{l} \times 100 = 2\% $, and
$\Rightarrow \Delta T = {10^ \circ }C$
Substituting these in (1)
$\Rightarrow 2 = 100\alpha \left( {10} \right)$
$\Rightarrow \alpha = 2 \times {10^{ - 3}}{/^ \circ }C$ …..(2)
Now, we know that the coefficient of superficial expansion, $\beta $ of a material is related to the coefficient of linear expansion by
$\Rightarrow \beta = 2\alpha $ ……..(3)
Let the increase in area of the copper sheet be $\Delta A$.
We know that the change in area is related to the change in temperature by
 $\Rightarrow \dfrac{{\Delta A}}{A} = \beta \Delta T$
From (3)
$\Rightarrow \dfrac{{\Delta A}}{A} = 2\alpha \Delta T$
For the percentage change, we multiply both sides by $100$
$\Rightarrow \dfrac{{\Delta A}}{A} \times 100 = 200\alpha \Delta T$
According to the question, we have$\Delta T = {10^ \circ }C$. Also from (2) we have $\alpha = 2 \times {10^{ - 3}}{/^ \circ }C$. Substituting these above we get
$\dfrac{{\Delta A}}{A} \times 100 = 200\left( {2 \times {{10}^{ - 3}}} \right)\left( {10} \right)$
$\dfrac{{\Delta A}}{A} \times 100 = 4$
So the percentage increase in the area of the copper sheet comes out to be equal to $4\% $.
Hence the correct answer is option (A).

Note
In both the cases the change in temperature is the same. So using the relation between the coefficients of the linear expansion and that of the superficial expansion we can directly predict the percentage increase in the area as twice that of the percentage change in the length. Also we should note that the length of the copper sheet is just extra information which is not needed at all.