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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 38

  • question_answer
    There are three point objects A, B and C of mass M, m and m respectively placed on a smooth surface on an x-y plane. Among them A and C are fixed while B can move. There is a massless target F that is placed at distance r from the object B, and the line joining B and T makes an angle of  \[30{}^\circ \]withx-axis. Then the value of M (in terms of m) so that the point object B can hit the target  \[T(a>>r)\] is (assuming only mutual gravitation force of attraction between the point objects)

    A) \[m\]                            

    B) \[\sqrt{2}m\]

    C) \[\sqrt{3}m\]     

    D)        \[8m\]

    Correct Answer: C

    Solution :

    [c] \[\frac{{{F}_{y}}}{{{F}_{x}}}=\tan 30{}^\circ \Rightarrow \,\frac{\frac{G{{m}^{2}}}{{{a}^{2}}}}{\frac{GMm}{{{a}^{2}}}}=\frac{1}{\sqrt{3}}\,\,\,\,\therefore \,\,\,M=\sqrt{3}m\]


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