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

Explanation

With respect to the top of the table, the initial potential energy of the chain,
`U_(1)` = PE of the chain lying on the table + PE of the hanging part of the chain
`=L(1-(1)/(n))mg(0)+(m)/(n)g""((-L)/(2n))=-(mgL)/(2n^(2))`
P.E of the chain, when it just slips off the table,
`U_(2)= mg((-L)/(2))=-(mgL)/(2)`
From jaw of conservation of energy
`Delta K = - Delta U K_(f) - K_(i) = - (U_(f) - U _(i))`
`thereforeK_(i)=0,K_(f)=-[-(mgL)/(2)-(-(mgL)/(2n^(2)))]`
`K_(f)= (mgL)/(2)[1-(1)/(n^(2))]`
It .V. is the velocity of the chain, then, `(1)/(2)mv^(2)=(mgL)/(2)[1-(1)/(n^(2))]`
`thereforev=sqrt(gL[1-(1)/(n^(2))])`