A very fleible uniform chain of mass M and length L is suspended vertically such that its lower end just touches the surface of a table. When the upper end of the chain is released . it falls with each link coming to test the instant it strikes the table.

Explanation

Since chain is uniform, the mass of y part of chain will be `((M)/(L)y)`. When this part reaches the table, its total force exerted must be equal to
weight of y part resting on table + Force due to momentum imparted
`= (M)/(L)yg +(((M)/(L)dy)sqrt(2gy))/(dt)=(Mg)/(L)y+(M)/(L)y+(M)/(L)v sqrt(2gy)`
`(as(dy)/(dt)=v)=(Mg)/(L)y +(M)/(L)sqrt(2gy).sqrt(2gy)=3(My)/(L)g`