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JEE Main & Advanced Physics Two Dimensional Motion Question Bank Uniform Circular Motion

  • question_answer
    A cyclist goes round a circular path of circumference 34.3 m in \[\sqrt{22}\] sec. the angle made by him, with the vertical, will be                                                 [MH CET 2000]

    A)             \[{{45}^{o}}\]

    B)             \[{{40}^{o}}\]

    C)             \[{{42}^{o}}\]         

    D)             \[{{48}^{o}}\]

    Correct Answer: A

    Solution :

                    \[2\pi r=34.3\] Þ \[r=\frac{34.3}{2\pi }\] and \[v=\frac{2\pi r}{T}=\frac{2\pi r}{\sqrt{22}}\]     Angle of binding \[\theta ={{\tan }^{-1}}\left( \frac{{{v}^{2}}}{rg} \right)=45{}^\circ \]       


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