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A rod of length l is held vertically stationary with its lower and located at a point 'P', on the horizontal plane. When the rod is released to topple about 'P', the velocity of the upper end of the rod with which it hits the ground is :

\[\begin {array} {1 1} (1)\;\sqrt{\frac{g}{l}} & \quad (2)\;\sqrt {3gl} \\ (3)\;3 \sqrt{\frac {g}{l}} & \quad (4)\;\sqrt{\frac{3g}{l}} \end {array}\]

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$(2)\;\sqrt {3gl}$
answered Nov 7, 2013 by pady_1

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