Question

A uniform cable of mass $'M'$ and length $'L'$ is placed on a horizontal surface such that its $\bigg(\frac{1}{n}\bigg)^{th}$ part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be :

• A. $\frac{MgL}{n^2}$
• B. $\frac{MgL}{2n^2}$
• C. $\frac{2MgL}{n^2}$
• D. $nMgL$

Answer 1

Answer: (B)

$\frac{MgL}{2n^2}$

Answer 2

Mass of the hanging part = $\frac{M}{n}$
$h_{COM} \, = \, \frac{L}{2n}$
work done W = mgh$_{COM} = \bigg(\frac{M}{n} \bigg) g \bigg(\frac{L}{2n}\bigg) \, = \, \frac{MgL}{2n^2}$
Option (2)