Question

The temperature of a metal coin is increased by ${100^ \circ }C$ and its diameter increases by $0.15\%$. Its area increases by nearly
(A) 0.15%
(B) 0.60%
(C) 0.30%
(D) 0.0225%

Answer 1

In the given question we have to find the percentage change in area of the coin. We have been given the percentage change in the diameter of the coin. For this we will use the formula of the area of a circle, $A = \,\pi \,{r^2}$. The change in the area of the coin is dependent on the radius and the $radius\, = \,\dfrac{d}{2}$.

Complete step by step answer:
Let us consider the original diameter of the coin to be $d$
Therefore, the initial Area of the coin, ${A_1}\, = \,\pi {\left( {\dfrac{d}{2}} \right)^2}\, = \,\pi \dfrac{{{d^2}}}{4}$
After the $0.15\%$ increase in the diameter, the diameter will be,
$\begin{array}{l} d'\, = \,d\, + \,\dfrac{{0.15}}{{100}}d\\\ \Rightarrow d'\, = \,d(1 + 0.0015)\\\ \Rightarrow d'\, = \,1.0015d \end{array}$
The area of the coin after the change in diameter:
${A_2}\, = \,\pi {\left( {\dfrac{{d'}}{2}} \right)^2}\, = \,\pi \left( {\dfrac{{{{(1.0015)}^2}{d^2}}}{4}} \right)\, = \,\pi \dfrac{{{d^2}}}{4}{\left( {1.0015} \right)^2}$
${A_2}\, = \,{A_1}{(1.0015)^2}$
Hence, the percentage change in area of coin:
$\begin{array}{l} \Delta A'\, = \,\dfrac{{{A_2} - {A_1}}}{{{A_1}}} \times \,100\\\ \Rightarrow \Delta A'\, = \,\dfrac{{{A_1}{{(1.0015)}^2} - {A_1}}}{{{A_1}}} \times \,100\\\ \Rightarrow \Delta A'\, = \,\dfrac{{(1.00300225){A_1} - {A_1}}}{{{A_1}}} \times \,100\\\ \Rightarrow \Delta A'\, = \,\dfrac{{{A_1}(1.00300225 - 1)}}{{{A_1}}} \times \,100\\\ \Rightarrow \Delta A'\, = \,0.0030\,\, \times \,100\\\ \Rightarrow \Delta A'\, = \,0.30\% \end{array}$
Therefore, the area increases by 0.30% and the option (C) is correct.

Note: It should be noted that while calculating the final diameter of the coin, proper decimals positions are considered and while taking the square of the radius the value should not be rounded off until the last step.