Question

A thin copper wire of length L increases in length by 1% when heated from $0^0$ C to ${100}^0$ C. If then a copper plate of area 2L × L is heated from $0^0$ C to ${100}^0$ C, the percentage increase in its area will be

Answer 1

$\Delta T=100-0={100}^{0}C$
$\frac{\Delta L}{L}=0.01$
$\Delta L=\alpha L\Delta T$
$\frac{\Delta L}{L}=\alpha \times 100$
$\alpha =\frac{.01}{100}=1\times {10}^{-4}$
Given, area of copper plate = 2L × L = 2$L^2$
By thermal expansion,
$\Delta A=\beta A\Delta T$
$\frac{\Delta A}{A}=\beta \Delta T,but\beta =2\alpha$
$\frac{\Delta A}{A}=2\alpha \Delta T$
$=2\times 1\times {10}^{-4}\times 100=2\times {10}^{-2}$
$\frac{\Delta A}{A}\times 100=2\times {10}^{-2}\times 100=2$