A chain of mass $m$ and length $l$ is placed on a table with one-sixth of it hanging freely from the table edge. The amount of work done to pull the chain on to the table is

\frac{m g l}{4}

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\frac{m g l}{6}

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\frac{m g l}{72}

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\frac{m g l}{36}

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Solution

The correct option is C $\frac{m g l}{72}$

Given, the total mass of chain $= m$

Length of chain $= l$

So, mass per unit length is $(λ = m / l)$

If we consider a small element $d x$ at a distance ' $x$ ' from the table, then the small mass element $d m$ is given as:

$d m = \frac{m}{l} d x$

So, work done against gravity in lifting the small mass element $d m$ through $x$ distance is :

$d W = d m g x = (m / l) g x d x$

Now, total work done is obtained by integrating $d W$ from limits $x = 0$ to $x = \frac{l}{6}$ as one-sixth of the chain hangs from table.

$\int_{0}^{W} d W = \int \frac{m g}{l} x (d x) = \frac{m g}{l} \int_{0}^{\frac{l}{6}} x d x$

$W = \frac{m g}{2 l} {(\frac{l}{6})}^{2} = \frac{m g l}{72}$

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