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Answer:
The correct answer is "299.88 mm".
Explanation:
The given values are:
Length,
[tex]l_o=300 \ mm[/tex]
Load,
[tex]P=50 \ kN[/tex]
i.e.,
[tex]=50\times 10^3 \ N[/tex]
[tex]E=200 \ GN/m^2[/tex]
[tex]=2\times 10^5 \ M/mm^2[/tex]
Now,
Stress,
⇒ [tex]\sigma=\frac{P}{A}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{50\times 10^3}{25\times 25}[/tex]
⇒ [tex]=80 \ N/mm^2[/tex]
The change in length will be:
⇒ [tex]\Delta l=\frac{Pl}{AE}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{50\times 10^3\times 300}{25\times 25\times 2\times 10^5}[/tex]
⇒ [tex]=\frac{15000\times 10^3}{1250\times 10^5}[/tex]
⇒ [tex]=0.12 \ mm[/tex]
hence,
⇒ [tex]L=l_o-\Delta l[/tex]
[tex]=300-0.12[/tex]
[tex]=299.88 \ mm[/tex]