A) \[\sqrt{\frac{g}{I}}\]
B) \[\sqrt{3gI}\]
C) \[3\sqrt{\frac{g}{I}}\]
D) \[\sqrt{\frac{3g}{I}}\]
Correct Answer: B
Solution :
In this process potential energy of the meter stick will be converted into rotational kinetic energy. PE of metre stick \[=\frac{mgl}{2}\] Because its centre of gravity lies at the middle of the rod. Rotational kinetic energy \[E=\frac{1}{2}I{{\omega }^{2}}\] \[I\]= moment of inertia of metre stick about point \[A=\frac{m{{l}^{2}}}{3}\]. By the law of conservation of energy \[mg\left( \frac{l}{2} \right)=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\frac{m{{l}^{2}}}{3}{{\left( \frac{{{v}_{B}}}{l} \right)}^{2}}\] By solving, we get \[{{v}_{B}}=\sqrt{3gl}\]You need to login to perform this action.
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