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  4. To increase the length of brass rod by 2 % its temperature should increase by (α=0.00002 °C-1)

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Q. To increase the length of brass rod by $2\%$ its temperature should increase by $\left(\alpha=0.00002\,{}^{\circ}C^{-1}\right)$

Thermal Properties of Matter

Solution:

$\frac{\Delta L}{L}=2\%=\frac{2}{100}$
$\alpha=0.00002\,{}^{\circ}C^{-1}$
As $\Delta L=\alpha\,L\,\Delta T$
or $\Delta T=\frac{\Delta L}{\alpha L}=\frac{2}{100 \times 0.00002}=\frac{1}{0.001}$
$=10^{3}=1000\,{}^{\circ}C$