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 strickes the table and comes to rest. What force the chain will exert on the table at the moment 'y' part of length falls on the table ? .

Explanation

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