A chain is held on a frictionless table with one third of its length hanging over the edge. The total length of the chain is l and the mass is m. The work required to pull the hanging part back to the table is

2 m g l

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\frac{m g l}{5}

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\frac{m g l}{9}

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\frac{m g l}{18}

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Solution

The correct option is D $\frac{m g l}{18}$
Length of hanging chaing is $l / 3$ , and mass of hanging chain is $M = m / 3$ .

The center of mass of hanging chain is at $h = \frac{l / 3}{2} = \frac{l}{6}$ from the table top.

So work done , $W = M g h = \frac{m g l}{3 \times 6} = \frac{m g l}{18}$

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A chain is held on a frictionless table with one third of its length hanging over the edge. The total length of the chain is l and the mass is m. The work required to pull the hanging part back to the table is

The correct option is D mgl18Length of hanging chaing is l/3 , and mass of hanging chain is M=m/3 .The center of mass of hanging chain is at h=l/32=l6 from the table top.So work done ,W=Mgh=mgl3×6=mgl18

The correct option is D $\frac{m g l}{18}$
Length of hanging chaing is $l / 3$ , and mass of hanging chain is $M = m / 3$ .

The center of mass of hanging chain is at $h = \frac{l / 3}{2} = \frac{l}{6}$ from the table top.

So work done , $W = M g h = \frac{m g l}{3 \times 6} = \frac{m g l}{18}$