Question

A chain of length $L$ and mass $m$ is allowed to fall on a table such that the part falling on the table comes to rest simultaneously. The force acting on the table when $l$ part of it has lied on the table is:

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

Answer 1

The correct option is A $\frac{3mlg}{L}$

Force acting $=$ force due to $x$ length $+$ force due to impulse by table

force due to $x$ length is $\frac{mgx}{L}$

force due to impulse is $\frac{mV}{L}\times \frac{dx}{dt}=\frac{m{V}^2}{L}$
Also, $v^2=2xg$

So, force due to impulse is $\frac{2mgx}{L}$

So, total force is $\frac{3mgx}{L}$