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. $\sqrt{2gl\left(1+\frac{1}{n^2}\right)}$
• C. $\sqrt{2gl\left(1-\frac{1}{n^2}\right)}$
• 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
$=\frac{-Mgl}{2n^2}$
$\bullet$ Potential energy of chain or it leaves table
$=\frac{-Mgl}{2}$
Kinetic energy = Loss of potential energy
$\Rightarrow \frac{1}{2}M{v}^2=\frac{Mgl}{2}\left(1-\frac{1}{n^2}\right)$
$\Rightarrow v=\sqrt{gl\left(1-\frac{1}{n^2}\right)}$