A) 1 sec
B) 10 sec
C) 8 sec
D) 4 sec
Correct Answer: D
Solution :
Speed at the highest point must be \[v>\sqrt{(rg)}\] Now, \[v=r\omega =r\frac{2\pi }{T}\] \[\therefore \] \[\frac{2\pi }{T}>\sqrt{rg}\] \[\Rightarrow \] \[T<\frac{2\pi r}{\sqrt{rg}}<2\pi \sqrt{\frac{r}{g}}\] \[\therefore \] \[T=2\pi \sqrt{\frac{4}{9.8}}=\mathbf{4}\,\,\mathbf{sec}\]
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