A uniform cable of mass M and length L is placed on a horizontal surface such that its 1 n t h part is…

1. $\frac{MgL}{2n^2}$
2. $\frac{MgL}{n^2}$
3. $nMgL$
4. $\frac{2MgL}{n^2}$

Solution

Mass of the part hanging is $\frac{M}{n}$. Centre of mass of this part lies $\frac{L}{2n}$ distance from the surface.

$\therefore$ Work done in lifting the hanging part to surface is
$W=\left(\frac{M}{n}\right)·g\left(\frac{L}{2n}\right)$
$=\frac{MgL}{2n^2}$

Asked in: JEE Main 2019 (09 Apr Shift 1)

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