Question

A uniform chain of mass $m$ and length $l$ hangs on a thread and touches the surface of the table by its lower end. The thread breaks suddenly, find the force exerted by the table on the chain when half of its length has fallen on the table. The fallen part does not form a heap.

• A. $\frac{mg}{2}$
• B. $mg$
• C. $\frac{3mg}{2}$
• D. $2mg$

Answer 1

The correct option is C $\frac{3mg}{2}$


Mass per unit length, $\lambda =\frac{m}{l}$

Let $dx$ be the length of chain reaching the surface in time $dt$.

At any instant of time, let velocity of part of chain which is just coming in contact with the surface be $v$.

Momentum carried by $dm$ mass, $dp=\left(dm\right)v$

$\therefore$ Rate of change of momentum of chain

$\frac{dp}{dt}=\left(\frac{dm}{dt}\right)v$

$\frac{dp}{dt}=\left(\frac{\lambda dx}{dt}\right)v=\left(\lambda v\right)v=\lambda v^2$

Since the chain has fallen through a height of $\frac{l}{2}$

$v=\sqrt{2gh}=\sqrt{2g\left(\frac{l}{2}\right)}=\sqrt{gl}$

$\therefore F_i=\frac{dp}{dt}=\lambda v^2=\frac{m}{l}\left(gl\right)=mg$

Force due to weight of the chain lying on the table

$F_2=\frac{m}{2}g$

$\therefore \text{Total force},F=F_1+F_2=\frac{3mg}{2}$

Hence, option $\left(\text{C}\right)$ is correct.

Why this question? Concept: Force acting on the chain is due to weight of the tying chain and change in momentum of the chain coming in contact with the table surface.