JEE Main 2018 · Physics · Work, Energy and Power · hard

A uniform chain of length L and mass M lies on a smooth horizontal table with one quarter of its length hanging over the edge over the smooth edge.…

Single correct option

A uniform chain of length

L

and mass

M

lies on a smooth horizontal table with one quarter of its length hanging over the edge over the smooth edge. It is released from rest. The speed of the chain at the instant its last link leaves the table is

A. $\sqrt{gL}$
B. $\dfrac{1}{4}\sqrt{gL}$
C. $\sqrt{\dfrac{15gL}{32}}$
D. $\dfrac{1}{4}\sqrt{15gL}$

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Answer: (D) $\dfrac{1}{4}\sqrt{15gL}$

Take the table level as zero of potential energy. When a length

x

hangs, the hanging part has mass

Mx/L

with its centre of mass at depth

x/2

, so

U(x)=-\dfrac{Mgx^2}{2L}

. Initially

x=L/4

U_i=-\dfrac{MgL}{32}

. Finally

x=L

U_f=-\dfrac{MgL}{2}

. Energy conservation:

\tfrac12Mv^2=MgL\left(\tfrac12-\tfrac1{32}\right)=\tfrac{15}{32}MgL

, hence

v^2=\dfrac{15gL}{16}

and

v=\dfrac14\sqrt{15gL}

. Forgetting the factor

\tfrac12

in kinetic energy gives the second option.

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