Question

A uniform cable of mass ${}^{\prime }{M}^{\prime }$ and length ${}^{\prime }{L}^{\prime }$ is placed on a horizontal surface such that its ${\left(\frac{1}{n}\right)}^{th}$ 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. $nMgL$
• C. $\frac{MgL}{2n^2}$
• D. $\frac{2MgL}{n^2}$

Answer 1

The correct option is C $\frac{MgL}{2n^2}$

Length of hanging part $=\frac{L}{n}$
Mass of hanging part $=\frac{M}{n}$
Weight of hanging part $=\frac{Mg}{n}$
Let ${}^{\prime }{C}^{\prime }$ be the centre of mass of the hanging part.

The hanging part can be assued to be a particle of weight $\frac{Mg}{n}$ at a distance $\frac{L}{2n}$ below the table top.

The work done in lifting it to the table top is equal to increase in its potential energy.

$\therefore W=\left(\frac{Mg}{n}\right)\left(\frac{L}{2n}\right)$
$\therefore W=\frac{MgL}{2n^2}$
Hence option $\left(A\right)$ is correct.