A) 1 sec
B) 2 sec
C) 3 sec
D) 4 sec
Correct Answer: C
Solution :
Minimum angular velocity \[{{\omega }_{\min }}=\sqrt{g/R}\] \[\therefore \,\,{{T}_{\max }}=\frac{2\pi }{{{\omega }_{\min }}}=2\pi \sqrt{\frac{R}{g}}\]=\[=2\pi \sqrt{\frac{2}{10}}=2\sqrt{2}\cong 3s\]
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