Question

A hole is drilled in a copper sheet. The diameter of the hole is $4.24cm$ 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 $27^{\circ }C$,
$S_1=\frac{\pi D_1^2}{4}=\frac{\pi }{4}\times (4.24{)}^{2}c{m}^2$
If $D_{2}cm$ is the diameter of the hole at $227^{\circ }C$, then area of the hole at $227^{\circ }C$,
$S_2=\frac{\pi D_2^2}{4}c{m}^2$
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 $D_2=4.2544cm$
Change in diameter $=D_2-D_1=4.2544-4.24$
$=0.0144cm=1.44\times 10^{-2}cm$