A) \[{{\left( \frac{g}{L} \right)}^{1/2}}\]
B) \[{{\left( \frac{3g}{4L} \right)}^{1/2}}\]
C) \[{{\left( \frac{3\sqrt{3}g}{2L} \right)}^{1/2}}\]
D) \[{{\left( \frac{3g}{2L} \right)}^{1/2}}\]
Correct Answer: D
Solution :
[d]: The fall of centre of gravity h is given by \[\frac{\left( \frac{L}{2}-h \right)}{\left( \frac{L}{2} \right)}=\cos {{60}^{0}}\]or\[h=\frac{L}{2}(1-cos{{60}^{o}})\] \[\therefore \]Decrease in potential energy \[=Mgh=Mg\frac{L}{2}(1-cos{{60}^{o}})\] Kinetic energy of rotation\[=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\times \frac{M{{L}^{2}}}{3}{{\omega }^{2}}\] [\[I=\frac{M{{L}^{2}}}{3}\](because rod is rotating about an axis passing through its one end)] According to law of conservation of energy, \[Mg\frac{L}{2}(1-cos{{60}^{o}})=\frac{M{{L}^{2}}}{6}{{\omega }^{2}}\] \[\therefore \]\[\omega =\sqrt{\frac{6g}{L}}\sin {{30}^{o}}=\sqrt{\frac{6g}{L}}\left( \frac{1}{2} \right)=\sqrt{\frac{3g}{2L}}\]You need to login to perform this action.
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