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Uttarakhand PMT Uttarakhand PMT Solved Paper-2011

  • question_answer
    A glass marble projected horizontally from the top of a table falls at a distance \[x\] from the edge of the table. If h is the height of the table, then the velocity of projection is

    A)  \[h\sqrt{\frac{g}{2x}}\]

    B)  \[x\sqrt{\frac{g}{2h}}\]

    C)  \[gxh\]

    D)  \[gx+h\]

    Correct Answer: B

    Solution :

     \[t=\sqrt{\frac{2h}{g}}\] \[\therefore \] \[x=v\sqrt{\frac{2h}{g}}\] Or \[v=s\sqrt{\frac{g}{2h}}\]


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