A)
\[L\left( 1+\frac{1}{3}\frac{Mg}{\pi Y{{R}^{2}}} \right)\] B)
\[L\left( 1+\frac{2}{3}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]
C)
\[L\left( 1+\frac{1}{9}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]
D)
\[L\left( 1+\frac{2}{9}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]
Correct Answer:
A Solution :
\[\frac{r-R}{x}=\frac{3R-R}{L}\Rightarrow r=R\left( 1+\frac{2x}{L} \right);Y=\frac{Mg}{\pi {{R}^{2}}\frac{dL}{dx}}\] \[dL\frac{Mg}{\pi {{R}^{2}}};\frac{dx}{{{\left( 1+\frac{2x}{L} \right)}^{2}}}\] \[\Delta L=\frac{Mg}{Y\pi {{R}^{2}}}\int\limits_{0}^{L}{\frac{dx}{{{\left( 1+\frac{2x}{L} \right)}^{2}}}=\frac{MgL}{3\pi {{R}^{2}}Y};}\] \[L'=L+\Delta K=L\left( 1+\frac{1}{3}\frac{Mg}{\pi {{R}^{2}}Y} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec