The angular velocity of rotation of a star (o | vedclass

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Question

(C) For matter to escape from the equator of a rotating star,the centrifugal force acting on a particle of mass $m$ at the equator must be equal to th

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

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(C) For matter to escape from the equator of a rotating star,the centrifugal force acting on a particle of mass $m$ at the equator must be equal to the gravitational force acting on it.At the equator,the condition for the particle to just lift off is $m\omega^2 R = \frac{GMm}{R^2}$.However,the question asks for the angular velocity at which matter escapes,which relates to the condition where the effective gravity becomes zero.Equating the centripetal force required for circular motion at the surface to the gravitational force: $m\omega^2 R = \frac{GMm}{R^2}$.Solving for $\omega$: $\omega^2 = \frac{GM}{R^3}$.Thus,$\omega = \sqrt{\frac{GM}{R^3}}$.Note: If the question implies the velocity equivalent to escape velocity $v_e = \sqrt{\frac{2GM}{R}}$,then $\omega = \frac{v_e}{R} = \sqrt{\frac{2GM}{R^3}}$.Given the options provided,the correct expression is $\sqrt{\frac{2GM}{R^3}}$.

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

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