question_answer

A thin uniform rod of mass and length is hinged at the lower end to a level floor and strands vertically. It is now allowed to fall, then its upper end will strike the floor with the velocity :

A)  \[\sqrt{2gl}\]                                    

B) \[\sqrt{5gl}\]

C) \[\sqrt{3gl}\]                                     

D) \[\sqrt{mgl}\]

Correct Answer: C

Solution :

                The kinetic energy at the point Q is given by \[=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\frac{m{{l}^{2}}}{3}\frac{{{\upsilon }^{2}}}{{{l}^{2}}}\] \[=\frac{1}{2}\times \frac{1}{3}m{{\upsilon }^{2}}\]         ??(i) The potential energy at G \[=\frac{1}{2}mgl\]                          .....(ii) From equations (i) and (ii), we get                 \[\frac{1}{2}\frac{m{{\upsilon }^{2}}}{3}=\frac{1}{2}mgl\]                 \[\upsilon =\sqrt{3gl}\]