Question

A uniform chain of mass $m$ and length $l$ is on a smooth horizontal table with $\left( \frac{1}{n}\right)^{th}$ part of its length is hanging from one end of the table. The velocity of the chain when it completely slips off the table is

• A. $\sqrt{gl \left(1- \frac{1}{n^{2}}\right)} $
• B. $\sqrt{2 gl \left(1 + \frac{1}{n^{2}}\right)} $
• C. $\sqrt{ 2 gl \left(1- \frac{1}{n^{2}}\right)} $
• D. $\sqrt{2gl } $

Answer 1

Answer: (A)

Taking surface of table as zero level of potential energy,
$\bullet$ Potential energy of chain with $\frac{1}{n}$ th part hanging
$=\frac{-M g l}{2 n^{2}}$
$\bullet$ Potential energy of chain or it leaves table
$=\frac{-M g l}{2}$
Kinetic energy = Loss of potential energy
$\Rightarrow \frac{1}{2} M v^{2}=\frac{M g l}{2}\left(1-\frac{1}{n^{2}}\right)$
$\Rightarrow v=\sqrt{g l\left(1-\frac{1}{n^{2}}\right)}$