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 = Potential energy =- 2KE for escape, out, total mechanical energy of particle should become zero.
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