A) \[\omega =\sqrt{\frac{3g\sin \alpha }{2L}}\]
B) \[\omega =\sqrt{\frac{2L}{3g\,\sin \alpha }}\]
C) \[\omega =\sqrt{\frac{6g\sin \alpha }{L}}\]
D) \[\omega =\sqrt{\frac{L}{g\sin \alpha }}\]
Correct Answer: A
Solution :
By the law of conservation of energy, P.E. of rod = rotational K.E.You need to login to perform this action.
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