A chain of mass M and length L is initially held in vertical position as shown in the figure. It is released from rest, then which of the following graph(s) is/are correct. Here N represents normal force by the ground on the chain and v represents velocity of free end

$Consider an elemental part of length dx at height x from floor. Just before hiting the floor, its velocity is v=\sqrt{2gx}\text{\hspace{0.17em}\hspace{0.17em}and }$

$\begin{array}{l}mass element is dm=\frac{M}{L}dx. \\ After hiting floor this element comes to rest.\\ \text{}\therefore F_{impulsive}=\frac{dP}{dt}=\frac{\left|0-\left(dm\right)\left(\sqrt{2gx}\right)\right|}{dt}\end{array}$

$\begin{array}{l}At this moment the length of heap of chain on floor is x.\\ \text{}\therefore F_{heap}=\left(\frac{M}{L}x\right)g\text{}\\ Thus, N=F_{heap}+F_{impulsive}=\frac{Mgx}{L}+\frac{2Mgx}{L}=\frac{3Mgx}{L}\end{array}$

$=\sqrt{2gx}\frac{dm}{dt}=\sqrt{2gx}\frac{M}{L}\frac{dx}{dt}=\sqrt{2gx}\frac{M}{L}\sqrt{2gx}=\frac{2Mgx}{L}$

$\begin{array}{l}Case 1 : 0<x<L\text{}\\ N=\frac{3Mgx}{L} \\ Straight line pas\sin g through origin.\end{array}$

$\begin{array}{l}Case 2 : x>L\text{}\\ N=Mg\\ \text{}Whole chain is on the floor and N becomes cons\tan t.\end{array}$