Answer :
Answer:
Therefore it is not possible because aluminium will melt first at 939.82 kelvin.
Explanation:
Given:
- mass of steel sphere, [tex]m_s=1.58\ kg[/tex]
- mass of aluminium plate, [tex]m_a=0.752\ kg[/tex]
radius of sphere is 10% larger than the radius of the hole in aluminium plate.
We have:
- density of steel, [tex]\rho_s=7750\ kg.m^{-3}[/tex]
- coefficient of linear thermal expansion of aluminium, [tex]\alpha_a=23\times 10^{-6}\ m.m^{-1}.^{\circ}C^{-1}[/tex]
- coefficient of volumetric thermal expansion of steel, [tex]\alpha_s=17\times 10^{-6}\ m^3.m^{-3}.^{\circ}C^{-1}[/tex]
Now, we find the volume of the given sphere:
[tex]V=\frac{m}{\rho_s}[/tex]
[tex]V=\frac{1.58}{7750}[/tex]
[tex]V=2.0387\times 10^{-4}\ m^3[/tex]
We know that the volume of sphere is given as:
[tex]V=\frac{4}{3}\pi.r^3[/tex]
[tex]r=0.0365\ m[/tex]
Now the perimeter of the of the 2-D projection of the sphere:
[tex]P=2\pi.r[/tex]
[tex]P=2\pi\times 0.0365[/tex]
[tex]P=0.2293\ m[/tex]
Therefore radius of hole in the aluminium plate:
[tex]r=1.10\times r_h[/tex]
[tex]r_h=\frac{r}{1.10}[/tex]
[tex]r_h=\frac{0.0365}{1.10}[/tex]
[tex]r_h=0.0332\ m[/tex]
Now experiment of the hole:
[tex]P_h=2\pi.r_h[/tex]
[tex]P_h=2\pi\times 0.0332[/tex]
[tex]P_h=0.2085\ m[/tex]
We know the equation of linear thermal expansion is given as:
[tex]\delta l=l.\alpha.\Delta T[/tex]
where:
[tex]\delta l=[/tex] change in length due to change in temperature
[tex]l=[/tex] initial original length
[tex]\alpha =[/tex]coefficient of linear thermal expansion
[tex]\Delta T=[/tex] change in temperature
Now, using the above concept:
[tex]P+\delta P=P_h+\delta P_h[/tex]
[tex]0.2293+P\times \alpha_s\times \Delta T=0.2085+P_h\times \alpha_a\times \Delta T[/tex]
[tex]0.2293+0.2293\times (17\times 10^{-6})\times \Delta T=0.2085+0.2085\times \(23\times 10^{-6})\times \Delta T[/tex]
[tex]0.8974\times 10^{-6}\times \Delta T=0.0208[/tex]
[tex]\Delta T=23178.06\ K[/tex]
Therefore it is not possible because aluminium will melt first at 939.82 kelvin.