A hole is drilled in a copper sheet. The diameter of the hole is $4.24\; cm$ at $27.0\,^{\circ} C$ What is the change in the diameter of the hole when the sheet is heated to $227\,^{\circ} C ?$ Coefficient of linear expansion of copper $=1.70 \times 10^{-5}\; K ^{-1}$
A hole is drilled in a copper sheet. The diameter of the hole is $4.24\; cm$ at $27.0\,^{\circ} C$ What is the change in the diameter of the hole when the sheet is heated to $227\,^{\circ} C ?$ Coefficient of linear expansion of copper $=1.70 \times 10^{-5}\; K ^{-1}$
Initial temperature, $T_{1}=27.0^{\circ} C$
Diameter of the hole at $T_{1}, d_{1}=4.24 cm$
Final temperature, $T_{2}=227^{\circ} C$
Diameter of the hole at $T_{2}=d_{2}$
Co-efficient of linear expansion of copper, $\alpha cu =1.70 \times 10^{-5} K ^{-1}$
For co-efficient of superficial expansion $\beta$, and change in temperature $\Delta T$, we have the relation:
$\frac{\text { Change in area }(\Delta A)}{\text { Original area }(A)}=\beta \Delta T$
$\frac{\left(\pi \frac{d_{2}^{2}}{4}-\pi \frac{d_{1}^{2}}{4}\right)}{\left(\pi \frac{d_{1}^{2}}{4}\right)}=\frac{\Delta A}{A}$
$\therefore \frac{\Delta A}{A}=\frac{d_{2}^{2}-d_{1}^{2}}{d_{1}^{2}}$
But $\beta=2 \alpha$
$\therefore \frac{d_{2}^{2}-d_{1}^{2}}{d_{1}^{2}}=2 \alpha \Delta T$
$\frac{d_{2}^{2}}{d_{1}^{2}}-1=2 \alpha\left(T_{2}-T_{1}\right)$
$\frac{d_{2}^{2}}{(4.24)^{2}}=2 \times 1.7 \times 10^{-5}(227-27)+1$
$d_{2}^{2}=17.98 \times 1.0068=18.1$
$\therefore d_{2}=4.2544 cm$
Change in diameter $=d_{2}-d_{1}=4.2544-4.24=0.0144 cm$
Hence, the diameter increases by $1.44 \times 10^{-2}\; cm .$
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