A uniform rope is hanging vertically from the ceiling such that its free end just touches the horizontal floor of a room. The upper end of the rope is then released. At any instant during the fall of the rope, the total force exerted by it on the floor is n times the weight of that part of the rope which is on the floor at that time. What is the value of n?

Let m be the mass per unit length of the rope. Let x be the part of the rope on the floor at time t. The weight of this part is

          $F_1=mgx$
Now, if a small part dx falls on the floor in time dt, the force exerted by it is
          $$\begin{gathered} F_2=rate of change of momentum \\ =\frac{\left(mdx\right)v}{dt} \end{gathered}$$
Now $\frac{dx}{dt}=v$, where v is the velocity of that part of the rope at that instant. But $v^2=2gx$. Hence $F_2=m{v}^2$ $=m\times \left(2gx\right)=2mgx$. Total force $F=F_1+F_2=mgx+2mgx=3mgx=3F_1$