A) 3.2 m/s
B) 5.4 m/s
C) 7.6 m/s
D) 9.2 m/s
Correct Answer: B
Solution :
In this process potential energy of the metre stick will be converted into rotational kinetic energy. \[\operatorname{P}.E. of meter stick=mg\left( \frac{l}{2} \right)\] Because its centre of gravity lies at the middle point of the rod. Rotational kinetic energy \[E=\frac{1}{2}\,I\,{{\omega }^{2}}\] \[\operatorname{I} = M.I.\] of metre stick about point \[A=\frac{m{{l}^{2}}}{3}\] \[\omega =\] Angular speed of the rod while striking the ground \[{{\nu }_{B}}=\] Velocity of end B of metre stick while striking the ground By the law of conservation of energy, \[mg\left( \frac{l}{2} \right)=\frac{1}{2}\,I\,{{\omega }^{2}}=\frac{1}{2}\,\frac{m{{l}^{2}}}{3}\,{{\left( \frac{{{\nu }_{B}}}{l} \right)}^{2}}\] By solving we get, \[{{\nu }_{B}}=\,\,\sqrt{3gl}=\,\,\sqrt{3\times 10\times 1}=\,\,5.4\,\,m/s\]You need to login to perform this action.
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