Question

A hole is drilled in a copper sheet. The diameter of the hole is $4.24 \,cm$ at $27^{\circ} C$. 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 $=1.70 \times 10^{-5 \circ} C ^{-1}$ )

• A. $1.44 \times 10^{-2} cm$
• B. $1.44 \times 10^{-3} cm$
• C. $1.44 \times 10^{-1} cm$
• D. 1.44 cm

Answer 1

Answer: (A)

In this problem superficial expansion of copper sheet will be involved on heating. Here, area of hole at $27^{\circ} C$,
$S_{1}=\frac{\pi D_{1}^{2}}{4}=\frac{\pi}{4} \times(4.24)^{2} cm ^{2}$
If $D_{2} cm$ is the diameter of the hole at $227^{\circ} C$, then area of the hole at $227^{\circ} C$,
$S_{2}=\frac{\pi D_{2}^{2}}{4} cm ^{2}$
Coefficient of superficial expansion of copper is
$\beta=2 \alpha=2 \times 1.70 \times 10^{-5}=3.4 \times 10^{-5}{ }^{\circ} C ^{-1}$
Increase in area $=S_{2}-S_{1}=\beta S_{1} \Delta T$
or $S_{2}=S_{1}+\beta S_{1} \Delta T=S_{1}(1+\beta \Delta T)$
or $\frac{\pi D_{2}^{2}}{4}=\frac{\pi}{4}(4.24)^{2}\left[1+3.4 \times 10^{-5} \times(227-27)\right]$
or $D_{2}^{2}=(4.24)^{2} \times 1.0068$ or $D_{2}=4.2544 \,cm$
Change in diameter $=D_{2}-D_{1}=4.2544-4.24$
$=0.0144 \,cm =1.44 \times 10^{-2} cm$