A thin copper wire of length L increase in le | vedclass

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Question

$\Delta T=100-0=100^{\circ} \mathrm{C}$

$\Delta L / L=$ length $=L=1 \%=1 / 100=0.01$

$\Delta L=\alpha L \Delta T$

$\Delta L / L=\alpha x 100

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\Delta T=100-0=100^{\circ} \mathrm{C}$

$\Delta L / L=$ length $=L=1 \%=1 / 100=0.01$

$\Delta L=\alpha L \Delta T$

$\Delta L / L=\alpha x 100$

$0.01=\alpha x 100$

$\alpha=0.01 / 100=1 	imes 10^{-4}$

Given Area of copper plate $=2 L x L=2 L^{2}$

By thermal expansion theory,

$\Delta A=\beta A \Delta T$

$\Delta A / A=\beta \Delta T$

but,

$\beta=2 \alpha$

$\Delta A / A=2 \alpha \Delta T$

$=2 	imes 1 	imes 10^{-4} 	imes 100$

$=2 	imes 10^{-} 2$

$\Delta A / A 	imes 100=2 	imes 10^{-2} 	imes 100=2 \%$

A thin copper wire of length $L$ increase in length by $1\%$ when heated from temperature $T_1$ to $T_2$. What is the percentage change in area when a thin copper plate having dimensions $2L\times L$ is heated from $T_1$ to $T_2$ ? ....... $\%$

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