Q 11-05-186JEE MainJEE Main 2019 (9 Apr, Shift 1)Easy
A uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $\left(\frac1n\right)^{\text{th}}$ 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
Answer: (A) $\dfrac{MgL}{2n^2}$
The hanging part has mass $M/n$ and length $L/n$; its centre of mass is $\dfrac{L}{2n}$ below the edge. Lifting it raises this centre of mass by $\dfrac{L}{2n}$:
$$W = \frac Mn\,g\,\frac{L}{2n} = \frac{MgL}{2n^2}$$
Solution by Sreeraj P, M.Sc Physics