A uniform chain of mass $m$ and length $l$ is on a smooth horizontal table with…

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
  1. $\sqrt{g l\left(1-\frac{1}{n^2}\right)}$
  2. $\sqrt{2 g l\left(1+\frac{1}{n^2}\right)}$
  3. $\sqrt{2 g l\left(1-\frac{1}{n^2}\right)}$
  4. $\sqrt{2 g l}$

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} $

Asked in: AP EAMCET 2018 (22 Apr Shift 2)

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