A) \[5\,\,rad/s\]
B) \[\sqrt{30}\,rad/s\]
C) \[\sqrt{60}\,rad/s\]
D) \[10\,rad/s\]
Correct Answer: A
Solution :
\[{{T}_{\max }}=m\omega _{^{\max }}^{2}r+mg\] Þ \[\frac{{{T}_{\max }}}{m}={{\omega }^{2}}r+g\] Þ\[\frac{30}{0.5}-10={{\omega }^{2}}_{\max }r\]Þ\[{{\omega }_{\max }}=\sqrt{\frac{50}{r}}=\sqrt{\frac{50}{2}}=5\,rad/s\]
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