Answer:
a)[tex]6.318\ \rm m[/tex]
b[tex]144.89^\circ[/tex]
Explanation:
Given :
Let [tex]\alpha[/tex] be the coefficient of linear thermal expansion of aluminium alloy
When the temperate of the rod is increased then its length will be changed accordingly. Let [tex]l_0[/tex] be its length at freezing point of water at T=[tex]0^\circ[/tex]
[tex]l_1=l(1+\alpha \Delta T)\\\\\dfrac{l_1}{l}-1=\alpha \Delta T[/tex]
According to question we have
a)
[tex]\dfrac{6.3243}{l}-1=\alpha (16-0)\\\dfrac{6.3568}{l}-1=\alpha (100-0)[/tex]
Solving above two equations we get
[tex]l=6.318\ \rm m[/tex]
b) Let [tex]t^\circ[/tex] be the temperature of the rod when its length is 6.3689 m then we have
[tex]\dfrac{6.3689}{6.2838}-1=\alpha(t-0)[/tex]
Also we have
[tex]\dfrac{6.3243}{6.318}-1=\alpha (16-0)[/tex]
Solving above two equations we have
[tex]t=144.89^\circ[/tex]