A) 4 m
B) 3 m
C) 2.5 m
D) 2 m
Correct Answer: A
Solution :
The velocity of body at bottom of inclined plane \[v=\sqrt{2gh}=\sqrt{10g}\] Velocity at minimum point for revolves on circular path will be\[\sqrt{5gR}\] \[\therefore \] \[\sqrt{5gR}=\sqrt{10g}\] \[\Rightarrow \] \[R=2\text{ }m\]You need to login to perform this action.
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