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  3. RAJASTHAN ­ PET

RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • 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)  \[mgH\]            

    B)  \[\sqrt{2gH}\]

    C)  \[2gH\]             

    D)  zero

    Correct Answer: B

    Solution :

     The potential energy of pendulum at point C will be equal to kinetic energy at point B \[\frac{1}{2}mV_{B}^{2}=mgH\] \[{{v}_{B}}=\sqrt{2gH}\]


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