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AP EAMCET · PHYSICS · Work Power Energy

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. A \(\sqrt{g l\left(1-\frac{1}{n^2}\right)}\)
  2. B \(\sqrt{2 g l\left(1+\frac{1}{n^2}\right)}\)
  3. C \(\sqrt{2 g l\left(1-\frac{1}{n^2}\right)}\)
  4. D \(\sqrt{2 g l}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

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…