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

Options

A. `sqrt(gl(1-(1)/(n^2)))`
B. `sqrt(2gl(1+(1)/(n^2)))`
C. `sqrt(2gl(1-(1)/(n^2)))`
D. `sqrt(2gl)`

Answer

A

Explanation

Taking surface of table zero level of potential energy,
Potential energy of chain with `1/n` part hanging
`=(-Mgl)/(2n^2)`
Kinetic energy= Loss of potential energy
`rArr 1/2 Mv^2 = (Mgl)/(2) (1-(1)/(n^2))`
`rArr v sqrt(gl (1-(1)/(n^2)))`