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A thin copper wire of length L increases by 1% when heated from 0°C to 100°C. If a thin copper plate of area 2L × L is heated from °C to 100°C. The percentage increase in the area of the plate will be

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answer is D.

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Detailed Solution

$Delta L = L propto Delta theta$

$Rightarrow frac{{Delta L}}{L} times 100% = alpha Delta theta times 100% = 1%$

Now, A = L × 2L = 2L²

$Rightarrow frac{{Delta A}}{A} = 2frac{{Delta L}}{L}$

$therefore frac{{Delta A}}{A}{kern 1pt} times 100% = 2left( {frac{{Delta L}}{L} times 100% } right) = 2 times 1% = 2%$

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