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MGIMS WARDHA MGIMS WARDHA Solved Paper-2013

  • question_answer
    A glass marble projected horizontally from the top of a table falls of 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 :

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


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