A very flexible chain of length L and mass M is vertically suspended with its lower end just touching the table. If it is released so that each link strikes the table and comes to rest, what force the chain will exert on table at the moment 'y 'part of length falls on table ?

Explanation

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