Manipal Medical Manipal Medical Solved Paper-2013

  • question_answer
    If a cycle wheel of radius 4m completes one revolution in two seconds. Then acceleration of the cycle is

    A)  \[4{{\pi }^{2}}m/{{s}^{2}}\]

    B)  \[2{{\pi }^{2}}\,m/{{s}^{2}}\]

    C)  \[{{\pi }^{2}}\,m/{{s}^{2}}\]

    D)  \[4\,m/{{s}^{2}}\]

    Correct Answer: A

    Solution :

     Here, radius of cycle wheel \[r=4m\] Frequency of revolution\[f=\frac{1}{2}rad/s\] The acceleration \[a=r{{\omega }^{2}}=r{{(2\pi f)}^{2}}\] \[=4\times {{\left( 2\pi \times \frac{1}{2} \right)}^{2}}\] \[=4{{\pi }^{2}}m/{{s}^{2}}\]


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