A) \[\sqrt{gr}\]
B) \[g/r\]
C) \[{{\left( \frac{g}{r} \right)}^{3/2}}\]
D) \[gr\]
Correct Answer: A
Solution :
[a] Let velocity at A \[={{v}_{A}}\] and velocity at B\[={{v}_{B}}\] Applying conservation of energy at A & B \[\frac{1}{2}mv{}_{A}^{2}+2gmr=\frac{1}{2}mv{}_{B}^{2}\] \[v{}_{B}^{2}=v{}_{A}^{2}+4gr\].........(i) Now as it is moving in circular path it has centripetal force. At point A \[\Rightarrow \]\[T+mg=\frac{mv{}_{A}^{2}}{r}\] for minimum velocity \[T\ge 0\] or \[\frac{mv{}_{A}^{2}}{r}\ge mg\]\[\Rightarrow \]\[v_{A}^{1}\ge gr\]\[\Rightarrow \]\[{{v}_{A}}\ge \sqrt{gr}\]You need to login to perform this action.
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