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Uttarakhand PMT Uttarakhand PMT Solved Paper-2007

  • question_answer
    A mass m slides from rest down the surface of a frictionless hemispherical bowl of radius r from the highest point A. The velocity of mass when it reaches the bottom is

    A)  \[\sqrt{2gr}\]           

    B)  \[\sqrt{mgr}\]

    C)  2mgr         

    D)  \[gr\]

    Correct Answer: A

    Solution :

     From the condition \[mgr=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow \] \[2gr={{v}^{2}}\] \[\Rightarrow \] \[v=\sqrt{2gr}\]


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