question_answer

A rod of length; is held vertically stationary with its lower end located at a position P on the horizontal plane. When the rod is released to topple about P. the velocity of the upper end of the rod with which it hits the ground is

A)  \[\sqrt{\frac{g}{l}}\]                                      

B)  \[\sqrt{3gl}\]

C)  \[3\sqrt{\frac{g}{l}}\]                                   

D)  \[\sqrt{\frac{3g}{l}}\]

Correct Answer: B

Solution :

  When rod fall on horizontal surface then rod will. rotate about point P then its potential energy change into rotational kinetic energy. From law of conservation of energy, \[M.f.\frac{l}{2}=\frac{l}{2}I.{{\omega }^{2}}\] Since, moment of inertia of rod about point is \[I=\left( \frac{M{{l}^{2}}}{3} \right)\] \[\therefore \]                  \[Mg.\frac{l}{2}=\frac{1}{2}.\left( \frac{M{{l}^{2}}}{3} \right){{\omega }^{2}}\] \[gl=\frac{{{l}^{2}}}{3}\left( \frac{{{v}^{2}}}{{{l}^{2}}} \right)\] \[\Rightarrow \]               \[v=\sqrt{3gl}\]