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Question

Physics Question on Work-energy theorem

A uniform chain of mass mm and length ll is on a smooth horizontal table with (1n)th\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

gl(1−1n2)\sqrt{gl \left(1- \frac{1}{n^{2}}\right)}

B

2gl(1+1n2)\sqrt{2 gl \left(1 + \frac{1}{n^{2}}\right)}

C

2gl(1−1n2)\sqrt{ 2 gl \left(1- \frac{1}{n^{2}}\right)}

D

2gl\sqrt{2gl }

Answer

gl(1−1n2)\sqrt{gl \left(1- \frac{1}{n^{2}}\right)}

Explanation

Solution

Taking surface of table as zero level of potential energy, ∙\bullet Potential energy of chain with 1n\frac{1}{n} th part hanging =−Mgl2n2=\frac{-M g l}{2 n^{2}} ∙\bullet Potential energy of chain or it leaves table =−Mgl2=\frac{-M g l}{2} Kinetic energy = Loss of potential energy ⇒12Mv2=Mgl2(1−1n2)\Rightarrow \frac{1}{2} M v^{2}=\frac{M g l}{2}\left(1-\frac{1}{n^{2}}\right) ⇒v=gl(1−1n2)\Rightarrow v=\sqrt{g l\left(1-\frac{1}{n^{2}}\right)}