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A ball of mass m is attached to the end of a thread fastened to the top of a vertical rod which is fitted to a horizontally revolving round table as shown. If the thread forms an angle $α$ with the vertical, the angular velocity $ω$ of table is

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$\sqrt{\frac{r}{gtan α}}$

$\sqrt{\frac{gtan α}{r + lsin α}}$

$\sqrt{\frac{g}{1 \cos α}}$

$\sqrt{\frac{gtan α}{r}}$

answer is C.

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Detailed Solution

$\begin{array}{l} Tsin α = {mω}^{2} R \\ Tcos α = mg \\ \tan α = \frac{ω^{2} R}{g} \\ ω = \sqrt{\frac{gtanα}{R}} = \sqrt{\frac{gtanα}{r + l sinα}} \end{array}$

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