Question

A circular hole in an aluminium plate is 1 cm in diameter at $0^{\circ }$. What is the diameter when the temperature of the plate is raised to ${100}^{\circ }$. (coefficient of linear expansion of aluminium $\alpha =2.55\times {10}^{-5}{/}^{\circ }$)

• A. $(1.1051{)}^{1/2}cm$
• B. $(10.0051{)}^{1/2}cm$
• C. $(1.51{)}^{1/2}cm$
• D. $(1.0051{)}^{1/2}cm$

Answer 1

The correct option is D $(1.0051{)}^{1/2}cm$

Given that $D_1=1cm,{\theta }_1=0^{\circ }C,D_2=$?
${\theta }_2={100}^{\circ }Cand\alpha =2.55\times {10}^{-5}{/}^{\circ }C\Delta T={100}^{\circ }C$
Coefficient of superficial expansion of aluminium $=\beta =2\alpha =5.10\times {10}^{-5}{/}^{\circ }C$
Surface area of the hole at \(0^{\circ}C\)
$A_1=\frac{\pi D_1^2}{4}=\frac{\pi }{4}({10}^{-2}{)}^2$
The surface area of hole at \(100^{\circ}C\) is given by
$A_2=A_1(1+\beta \Delta T)=\frac{\pi }{4}\left({10}^{-2}{)}^2\right[1+5.1\times {10}^{-5}\times 100]$
If $D_2$ is the diameter of the hole at \(100^{\circ}C\), then $A_2=\frac{\pi }{4}$
Hence $\frac{\pi D_2^2}{4}=\frac{\pi }{4}\times {10}^{-4}[1+5.1\times {10}^{-3}]or,D_2^2={10}^{-4}\times 1.0051or,D_2={10}^{-2}\times (1.0051{)}^{1/2}m=(1.0051{)}^{1/2}cm$