 # A thin copper wire of length L increases by 1% when heated from temperature T1 to T2. What is the percentage change in area when a thin copper plate having dimensions 2l×l is heated from T1 to T2?

The coefficient of linear expansion (α) is half the coefficient of superficial expansion( β).  Mathematically the equation can be expressed as, β=2α

Given:

$$\Delta T = 100 – 0 = 100^{\circ}C$$ $$\Delta L/L = Length = L = 1% = 1/100 = 0.01$$ $$\Delta L = \alpha x 100$$ $$0.01 =\alpha x 100$$ $$\alpha = 0.01 / 100 = 1 * 10^{-4}$$

Given Area of copper plate =$$2L * L = 2L^{2}$$

Accoring to thermal expansion theory,

$$\Delta A = \beta A\Delta T$$ $$\Delta A / A = \beta\Delta T$$

Here$$\beta = 2\alpha$$ $$\Delta A / A = 2\alpha\Delta T$$

=$$2 * 1 * 10^{-4} * 100$$

= $$2 * 10^{-2}$$ $$\Delta A / A * 100$$

= $$2 * 10^{-2}$$

= 2%

Therefore, the percentage change in the area when a thin copper plate having dimensions 2L*L is heated from T1 to T2 is 2 percentage or 2%

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