Question

A uniform chain of mass $M$ and length $L$ is lying on a frictionless table in such a way that its $1/3$ part is hanging vertically down. The work done in pulling the chain up the table is:

• A. $\frac{MgL}{20}$
• B. $\frac{MgL}{18}$
• C. $2$MgL
• D. None of these

Answer 1

The correct option is B $\frac{MgL}{18}$

unit length of rope is $\frac{M}{L}$

given $\frac{L}{3}$ part of rope is hanging

we have to solve using integration methods, lets say ${}^{\prime }{x}^{\prime }$ part of rope is consided with an element ${}^{\prime }d{x}^{\prime }$ which is pulled up with limits from 0 to $\frac{L}{3}$

Potential energy$=mgh$

$dE=$integral of $(\frac{Mx}{L}\times g\times dx)$

$E=$integral of $(\frac{Mx}{L}\times g\times dx)$limits from $0to\frac{L}{3}$

$P.E=\frac{MgL}{18}$