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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    A cyclist riding the bicycle at a speed of \[14\sqrt{3}\text{ }m{{s}^{-1}}\]takes a turn around a circular road of radius\[20\sqrt{3}\]m without skidding. Given, \[g=9.8\text{ }m{{s}^{-2}},\]what is his inclination to the vertical?

    A)  \[30{}^\circ \]                                  

    B)  \[90{}^\circ \]                  

    C)  \[45{}^\circ \]                                  

    D)  \[60{}^\circ \]

    E)  \[0{}^\circ \]

    Correct Answer: D

    Solution :

    \[\tan \theta =\frac{{{v}^{2}}}{rg}=\frac{{{(14\sqrt{3})}^{2}}}{20\sqrt{3}\times 9.8}=\sqrt{3}\] \[\Rightarrow \]               \[\theta =60{}^\circ \]


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