The natural length of the spring will be "8 cm".
According to Hooke's law,
Now,
→ [tex]6 = \int_{10}^{12} K(x-L)dx[/tex]
[tex]= K[\frac{x^2}{2}-Lx ]_{10}^{12}[/tex]
[tex]= K(22-2L)[/tex]...(eq. 1)
and,
→ [tex]10= \int_{12}^{14}K(x-L)dx[/tex]
[tex]= L[\frac{x^2}{2} -Lx]_{12}^{14}[/tex]
[tex]= K(26-2L)[/tex]...(eq. 2)
By dividing "eq 2" by "eq. 1", we get
→ [tex]\frac{10}{3} = \frac{K(26-2L)}{K(22-2L)}[/tex]
[tex]\frac{5}{3} = \frac{26-2L}{22-2L}[/tex]
[tex]110-10L=78-6L[/tex]
[tex]110-78 = 10L-6L[/tex]
[tex]L = \frac{32}{4}[/tex]
[tex]= 8 \ cm[/tex]
Thus the above approach is correct.
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