A) \[L\left( 1+\frac{2}{9}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]
B) \[L\left( 1+\frac{1}{9}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]
C) \[L\left( 1+\frac{1}{3}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]
D) \[L\left( 1+\frac{2}{3}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]
Correct Answer: C
Solution :
[c] Consider a small element dx of radius r, \[r=\frac{2R}{L}x+R\] At equilibrium change in length of the wire \[\int\limits_{0}^{1}{dL}=\int{\frac{Mgdx}{\pi {{\left[ \frac{2R}{L}x+R \right]}^{2}}y}}\] Taking limit from 0 to L \[\Delta L=\frac{Mg}{\pi y}\left[ -\frac{1}{\left[ \frac{2Rx}{L}+R \right]_{0}^{L}}\times \frac{L}{2R} \right]=\frac{MgL}{3\pi {{R}^{2}}y}\] The equilibrium extended length of wire\[=L+\Delta L\] \[=L+\frac{MgL}{3\pi {{R}^{2}}Y}=L\left( 1+\frac{1}{3}\frac{Mg}{\pi Y{{R}^{2}}} \right)\]You need to login to perform this action.
You will be redirected in
3 sec