The angular velocity of rotation of star (of mass M and radius R) at which the matter start to escape from its equator will be

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$\sqrt{\frac{2 GM}{R^{3}}}$

$\sqrt{\frac{2 {GM}^{2}}{R}}$

$\sqrt{\frac{2 GR}{M}}$

$\sqrt{\frac{2 GM}{g}}$

answer is C.

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Detailed Solution

Escape velocity $v = \sqrt{\frac{2 GM}{R}}$
If star rotates with angular velocity $ω$
then $ω = \frac{v}{R} = \frac{1}{R} \sqrt{\frac{2 GM}{R}} = \sqrt{\frac{2 GM}{R^{3}}}$

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