The temperature of a metal coin is increased | vedclass

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Question

(b)

Percentage increase in area

$=\frac {Final \,area - Initial \,area}{Initial \,area} 	imes 100$

$=\frac{\pi\left(r_2^2-r_1^2ight)}{\pi

Vedclass · Accepted Answer

$0.30 \%$

Vedclass · Answer

(b)

Percentage increase in area

$=\frac {Final \,area - Initial \,area}{Initial \,area} 	imes 100$

$=\frac{\pi\left(r_2^2-r_1^2ight)}{\pi r_1^2} 	imes 100$

$=\frac{\left(r_1+\Delta right)^2-r_1^2}{r_1^2} 	imes 100$

where, $\Delta r=$ increase in radius.

$=\left(\frac{r_1^2+2 r_1 \Delta r+\Delta r^2-r_1^2}{r_1^2}ight) 	imes 100$

$=\frac{2 r_1 \Delta r+\Delta r^2}{r_1^2} 	imes 100$

As $\Delta r$ is small, we can neglect $\Delta r^2$.

$\approx \frac{2 r_1 \Delta r}{r_1^2}=100$

$=2(\frac{\Delta r}{r_1} 	imes 100)$

$=2 	imes 0.15 \%=0.30 \%$

The temperature of a metal coin is increased by $100^{\circ} C$ and its diameter increases by $0.15 \%$. Its area increases by nearly

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