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BHU PMT BHU PMT (Screening) Solved Paper-2011

  • question_answer
    The particle is released from a height h. At a certain height, its KE is two times that of potential energy. Height and speed of the particle at that instant are

    A)  \[\frac{h}{3},\sqrt{\frac{2gh}{3}}\]                         

    B)  \[\frac{h}{3},2\sqrt{\frac{gh}{3}}\]

    C)  \[\frac{2h}{3},\sqrt{\frac{2gh}{3}}\]                      

    D)  \[\frac{h}{3},\sqrt{2gh}\]

    Correct Answer: B

    Solution :

                     Total mechanical energy\[=mgh\] \[\frac{KE}{PE}=\frac{2}{1}\] \[KE=\frac{2}{3}mgh\]and \[PE=\frac{1}{3}mgh\] Height from the ground at this instant\[h'=\frac{h}{3'}\] and speed of particle at this instant \[v=\sqrt{2g(h-h')}=\sqrt{2g\left( \frac{2h}{3} \right)}\] \[=2\sqrt{\frac{gh}{3}}\]


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