A) \[m{{v}^{2}}\]
B) \[\frac{1}{2}m{{v}^{2}}\]
C) \[\frac{3}{2}m{{v}^{2}}\]
D) \[2\,m{{v}^{2}}\]
Correct Answer: A
Solution :
KE of revolving particle = \[\frac{1}{2}\,m{{v}^{2}}\] Potential energy =- 2KE \[=\,\,-m{{v}^{2}}\] for escape, out, total mechanical energy of particle should become zero. \[KE+PE=0\] \[KE-m{{v}^{2}}=0\] \[KE=m{{v}^{2}}\]You need to login to perform this action.
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