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JIPMER Jipmer Medical Solved Paper-2015

  • question_answer
    A simple pendulum with a bob of mass in oscillates from A to C and back to A such that PB is H. If the acceleration due to gravity is g, then the velocity of the bob as it passes through B is:

    A)  zero                     

    B)         \[2gH\]

    C)  \[mgH\]             

    D)         \[\sqrt{2gH}\]

    Correct Answer: D

    Solution :

    We know that potential energy at A (or C) = Kinetic energy at B. Thus\[\frac{1}{2}mv_{B}^{2}\]or\[{{v}_{B}}=\sqrt{2gH}.\]


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