A) \[\sqrt{2gl}\]
B) \[\sqrt{5gl}\]
C) \[\sqrt{3gl}\]
D) \[\sqrt{mgl}\]
Correct Answer: C
Solution :
From law of conservation of energy Potential energy of fall = rotational kinetic energy Potential energy of fall \[=\frac{1}{2}mgl\] ... (i) Kinetic energy at \[\theta \] is given by \[=\frac{1}{2}\,I{{\omega }^{2}}\] where J is moment of inertia, \[\omega \] the angular velocity. For a rod \[I=\frac{m{{l}^{2}}}{3}\] and \[\omega =\frac{v}{r}\]\[\omega =\frac{v}{r}\] \[\therefore \] \[KE=\frac{1}{2}\frac{m{{l}^{2}}}{3}.\,\frac{{{v}^{2}}}{{{l}^{2}}}\] ?.. (ii) From Eqs. (i) and (ii), we get \[\frac{1}{2}.\frac{m{{l}^{2}}}{3}.\,\frac{{{v}^{2}}}{{{l}^{2}}}=\frac{1}{2}\,mgl\] \[\Rightarrow \] \[v=\sqrt{3gl}\]You need to login to perform this action.
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