MGIMS WARDHA MGIMS WARDHA Solved Paper-2015

  • question_answer
    The angle to which a cyclist bends when he covers a circular path of 34.3 m in\[\sqrt{22}s\]is

    A)  \[20{}^\circ \]                  

    B)  \[30{}^\circ \]   

    C)  \[60{}^\circ \]                                  

    D)  \[45{}^\circ \]

    Correct Answer: D

    Solution :

                    Here, \[l=2\pi r\] \[\Rightarrow \]               \[r=\frac{l}{2\pi }=\frac{34.3}{3.14\times 2}\] Also        \[v=\frac{l}{t}=\frac{34.3}{\sqrt{22}}\] The minimum velocity of a cyclist on a circular path is                 \[v=\sqrt{rg\tan \theta }\] \[\Rightarrow \]               \[\tan \theta =\frac{{{v}^{2}}}{rg}=\frac{34.3\times 34.3}{22}\times \frac{3.14\times 2}{34.3\times 9.8}\] \[\Rightarrow \]               \[\tan \theta =1=\tan 45{}^\circ \] So               \[\theta =45{}^\circ \]


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