JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Question Bank Gravitation Potential

  • question_answer
    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                          [MH CET 1999]

    A)             \[\sqrt{\frac{2G{{M}^{2}}}{R}}\]

    B)               \[\sqrt{\frac{2GM}{g}}\]

    C)             \[\sqrt{\frac{2GM}{{{R}^{3}}}}\]

    D)               \[\sqrt{\frac{2GR}{M}}\]

    Correct Answer: C

    Solution :

                    Escape velocity \[v=\sqrt{\frac{2GM}{R}}\] If star rotates with angular velocity w then \[\omega =\frac{v}{R}=\frac{1}{R}\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2GM}{{{R}^{3}}}}\]


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