JEE Main & Advanced Sample Paper JEE Main Sample Paper-36

  • question_answer
    A simple pendulum with a bob of mass 'm' 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)  \[2\,gH\]

    C)  \[mgH\]                       

    D)  \[\sqrt{2\,g\,H}\]

    Correct Answer: D

    Solution :

     At B, the velocity is maximum. Taking vertical downward motion of body from A to B, we have, \[u=0,a=g,\,s=H\]and v=? As \[{{v}^{2}}={{u}^{2}}+2\]as; so \[{{v}^{2}}=0+2g\,H\] or \[v=\sqrt{2gH}\]


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