A uniform chain of mass $m$ and length $l$ is on a smooth horizontal table with $\left(\frac{1}{n}\right)^{\text {th }}$ 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

• Correct answer
  A

  $\sqrt{g l\left(1-\frac{1}{n^2}\right)}$
• B

  $\sqrt{2 g l\left(1+\frac{1}{n^2}\right)}$
• C

  $\sqrt{2 g l\left(1-\frac{1}{n^2}\right)}$
• D

  $\sqrt{2 g l}$

Solution

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

Hence, the correct answer is option: 'A'.