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VMMC VMMC Medical Solved Paper-2010

  • question_answer
    If pendulum bob on a 2m string is displaced \[60{}^\circ \] from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path?

    A) \[\sqrt{2}m/s\]                

    B) \[\sqrt{2\times 9.8}m/s\]            

    C) 4.43 m/s             

    D) \[1/\sqrt{2}m/s\]

    Correct Answer: B

    Solution :

    Velocity of bob \[v=\sqrt{2gl(1-cos\theta )}\] \[=\sqrt{2\times 9.8l(1-\cos {{60}^{o}})}\] \[v=\sqrt{2\times 9.8}\,m/s\]


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