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How much should the temperature of a brass rod be increased so as to increase its length by $1 \%$ ? (Given, $\alpha$ for brass is $0.00002^{\circ} \mathrm{C}^{-1}$ ).
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The correct answer is:
$500^{\circ} \mathrm{C}$
Here, $\Delta \mathrm{T}=$ ?, $\frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{1}{100}$
$\alpha=0.00002^{\circ} \mathrm{C}^{-1}$
As $\Delta \mathrm{L}=\alpha \mathrm{L} \Delta \mathrm{T}$
$\therefore \quad \alpha \Delta \mathrm{T}=\frac{\Delta \mathrm{L}}{\mathrm{L}}$
or $\Delta \mathrm{T}=\frac{\Delta \mathrm{L}}{\mathrm{L} \alpha}=\frac{1}{100 \times 0.00002}$
$\Delta \mathrm{T}=\frac{10^5}{2 \times 10^2}=500^{\circ} \mathrm{C}$
$\alpha=0.00002^{\circ} \mathrm{C}^{-1}$
As $\Delta \mathrm{L}=\alpha \mathrm{L} \Delta \mathrm{T}$
$\therefore \quad \alpha \Delta \mathrm{T}=\frac{\Delta \mathrm{L}}{\mathrm{L}}$
or $\Delta \mathrm{T}=\frac{\Delta \mathrm{L}}{\mathrm{L} \alpha}=\frac{1}{100 \times 0.00002}$
$\Delta \mathrm{T}=\frac{10^5}{2 \times 10^2}=500^{\circ} \mathrm{C}$
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