Respuesta :
Answer:
[tex]\delta L\%=22.5\%[/tex]
Explanation:
Assuming that the thermal expansion of brass cube occurs isotropically (i.e. equal in all the directions).
Given:
- length of the cube, [tex]l=10\ cm=0.1\ m[/tex]
- change in temperature of the cube, [tex]\Deta T=200\^{\circ}C[/tex]
- coefficient of volume expansion, [tex]\beta=57\times 10^{-6}\ ^{\circ}C^{-1}[/tex]
Hence volume of the cube:
[tex]V=10^{-3}\ m^3[/tex]
Now the volume of the cube after expansion:
[tex]\delta V=V.\beta.\Delta T[/tex]
[tex]\delta V=10^{-3}\times 57\times 10^{-6}\times 200[/tex]
[tex]\delta V=1.14\times 10^{-5}\ m^3[/tex]
Therefore,
[tex]\delta L=0.0225\ m[/tex]
Now the percentage change in the edges of the cube:
[tex]\delta L\%=\frac{\delta L}{L} \times 100[/tex]
[tex]\delta L\%=\frac{0.0225}{0.1} \times 100[/tex]
[tex]\delta L\%=22.5\%[/tex]
