Question

A uniform chain has mass $M$ and length $L$. It is lying on a smooth horizontal table with half of its length hanging vertically downward. The work done in pulling the chain up the table is:

• A. $MgL/2$
• B. $MgL/4$
• C. $MgL/8$
• D. $MgL/16$

Answer 1

Answer: (C)

$MgL/8$

The per unit length of the chain $=\left(\frac{M}{L}\right)$

Let us consider a finite small length of the chain $dx$ at a distance $x$ from the bottom. To move $dx$ to the top, a force equal to the weight of chain $x$ will have to be applied in upward direction.

Weight of chain of length $x=\left(\frac{M}{L}x\right)g$

Small amount of work done in moving $dx$ to the top

$dW=Fdx=\left(\frac{M}{L}x\right)g\times dx$

The total amount of work done in moving the half length of the hanging chain on the table will be:

$W={\int }_0^{L/2}\frac{Mg}{L}xdx=\frac{Mg}{L}{\int }_0^{L/2}xdxW=\frac{Mg}{L}{\left[\frac{x^2}{2}\right]}_0^{L/2}=\frac{MgL}{8}$

Option C is correct.