Question

A hole is drilled in a copper sheet. The diameter of the hole is $4.24cm$ at ${27.0}^{o}C$. What is the change in the diameter of the hole when the sheet is heated to ${227}^{o}C$? Coefficient of linear expansion of copper is $1.70\times {10}^{-5}{K}^{-1}$.

• A. $1.44\times {10}^{-2}cm$
• B. $2.44\times {10}^{-3}cm$
• C. $1.44\times {10}^{-2}mm$
• D. $2.44\times {10}^{-3}mm$

Answer 1

The correct option is A $1.44\times {10}^{-2}cm$

$T_2={27}^{o}C$

$D_1=4.24cm$

$T_2={227}^{o}C$

$D_2=?$

We know that '$\beta$' area thermal expansion is given as

$\beta =\left(\frac{\Delta A}{A}\right)\times \frac{1}{\Delta T}$ ...(1)

When $\Delta A$ is change is area and $\Delta T$ is the charge is temperature

$\Delta T=227-27$

$\Delta T={200}^{o}C$

Putting values is (1)

$A=\frac{(\pi D_2^2/4-\pi D_1^2/4)}{\pi D_1^2/4}\times \frac{1}{{200}^{o}C}$

Now $\beta =2\alpha$ where $\alpha$ is linear thermal coefficient, and given in the question

$\beta =2\times 1.7\times {10}^{-5}$

$2\times 1.7\times {10}^{-5}=\frac{(x{D}_2^2/4-x{D}_1^2/4)}{x{D}_1^2/4}\times \frac{1}{200}$

$\frac{D_2^2-D_1^2}{D_1^2}=200\times 2\times 1.7\times {10}^{-5}$

$D_2^2=\left(4.24{)}^2\right(6.0\times {10}^{-3}+1)$

$D_2=4.24144$

So change is Diameter $D_2-D_1=4.24144-4.24$

$=0.00144$

$=1.44\times {10}^{-2}mm$