question_answer

A thin uniform rod of mass m 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 \[\frac{1}{2}\frac{m{{\upsilon }^{2}}}{3}=\frac{1}{2}mgl\] \[\upsilon =\sqrt{3gl}\]