A uniform chain of mass m and length l is on a smooth horizontal table with ( 1 n )^ th part of its length is hanging from one end of the table. The velocity of the chain, when it completely slips off the table is

Step-by-step solution

Taking surface of table as zero level of potential energy, - Potential energy of chain with 1 n th part hanging = -M g l 2 n^2 - Potential energy of chain or it leaves table = -M g l 2 Kinetic energy = Loss of potential energy array rlrl & 1 2 M v^2 & = M g l 2 (1- 1 n^2 ) v & = g l (1- 1 n^2 ) array