A) \[\frac{3}{2}\,r\]
B) \[\frac{2}{3}\,r\]
C) \[\frac{1}{2}\,g{{t}^{2}}\]
D) \[\frac{{{v}^{2}}}{2g}\]
Correct Answer: B
Solution :
The body will loose contact when centripetal acceleration becomes equal to the component of acceleration due to gravity along the radius. Velocity at \[v=\,\,\sqrt{2g(r-h)}\,(\because \,\,{{v}^{2}}-{{u}^{2}}=2gx)\] Centripetal acceleration will be \[\frac{{{v}^{2}}}{r}\]. It should be equal to the component of g along PO. Hence, \[\frac{{{v}^{2}}}{r}=\,\,g\,\cos \theta \] or \[\frac{2g\,\,(r-h)}{r}=g\,\,\times \,\,\frac{h}{r}\] Solving, we get \[h=\frac{2r}{3}\]You need to login to perform this action.
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