A chain is held on a frictionless table with ${(\frac{1}{n})}^{t h}$ of its length hanging over the edge. If the chain has length L and mass M, how much work is required to pull the hanging part back on the table?

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Solution

The correct option is A
The chain is pulled very slowly. Let $λ$ is the mass per unit length of the chain. Consider a differential element of the chain of length dy at a depth y from the surface of the table. The force acting on the element due to gravity is equal to $λ d y g$ .
So the work done in pulling this element of the chain on the table is dW = F(y)

mass of this element is $d m = λ y$
$d W = λ g y d y = \frac{M}{L} g y d y W = \frac{M g}{L} \int_{0}^{L / n} y d y = \frac{M g L}{2 n^{2}}$

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