A) \[\sqrt{g/2}\]
B) \[\sqrt{g}\]
C) \[\sqrt{2g}\]
D) \[2\sqrt{g}\]
Correct Answer: D
Solution :
[c] \[N\,\,cos \theta = mg and N\,\,sin \theta = m{{\omega }^{2}}r\]\[\therefore \tan \theta =\frac{{{\omega }^{2}}r}{g}\] Given \[y={{x}^{2}}\] \[\therefore \frac{dy}{dx}=2x\] or \[tan\,\theta =2\times 1=2\] ...(ii) From above equations, we get \[\omega = \sqrt{2g}\,\,~\left( r=1m \right)\]You need to login to perform this action.
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