Question

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm 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 the linear expansion of copper $=1.70\times {10}^{-5}{K}^{-1}$

Answer 1

Initial temperature, $T_1={27.0}^{o}C$
Diameter of the hole at $T_1,d_1=4.24cm$

Final temperature, $T_2={227}^{o}C$
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 β,and change in temperature $△T$, we have the relation:
Change in area $\left(△\right)$ / Original area (A) = $\beta △T$

$\left[\right(\pi d_2^2/4)-(\pi d_2^2/4\left)\right]/\left(\pi d_1^1/4\right)=△A/A$

$\therefore △A/A=(d_2^2-d_1^2)/d_1^2$

But $\beta =2\alpha$

$\therefore (d_2^2-d_1^2)/d_1^2=2\alpha △T$

$\left(d_2^2/d_1^2\right)-1=2\alpha (T_2-T_1)$

$d_2^2/{4.24}^2=2\times 1.7\times {10}^{-5}(227-27)+1$

$d_2^2=17.98\times 1.0068=18.1$

$\therefore d_2=4.2544cm$
Change in diameter $=d_2-d_1=4.2544-4.24=0.0144cm$

Hence, the diameter increases by $1.44\times {10}^{-2}cm$