A) \[\sqrt{\cos \alpha }\]
B) \[\cos \alpha \]
C) \[\sin \alpha \]
D) \[\sqrt{\sin \alpha }\]
Correct Answer: A
Solution :
When the rod makes an angle of \[\alpha \] displacement of centre of mass \[=\frac{l}{2}\cos \alpha \] \[mg\frac{l}{2}\cos \alpha =\frac{1}{2}l{{\omega }^{2}}\] \[mg\frac{l}{2}\cos \alpha =\frac{m{{l}^{2}}}{6}{{\omega }^{2}}\] \[\omega =\sqrt{\frac{3g\cos \alpha }{l}}\] speed of end \[=\omega \times l=\sqrt{3g\cos \alpha l}\] hence \[\omega \] is proportional to\[\sqrt{\cos \alpha }\]
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