A uniform flexible chain of mass M and length L is held vertically with its lower end just touching the pan of a weighing machine. It is released…

Let $x$ be the length of chain already on the pan. Since the free part falls rigidly, the top has fallen $x$; with $x=\dfrac{3L}{4}$. The links landing have speed $v=\sqrt{2gx}$. The pan force has two parts. First, the weight of the chain already lying on it: $\dfrac{Mx}{L}g$. Second, the rate at which momentum is destroyed by the landing links: the mass arriving per unit time is $\dfrac{M}{L}v$, each loses speed $v$, so the force is $\dfrac{M}{L}v^2=\dfrac{M}{L}\,2gx$. Total reading $N=\dfrac{3Mgx}{L}$. At $x=\dfrac{3L}{4}$ this is $\dfrac{9Mg}{4}$. The option $\dfrac{3Mg}{4}$ is the weight alone, $\dfrac{3Mg}{2}$ counts only the momentum term, and $3Mg$ is the value when the whole chain has landed ($x=L$).