Question 26 of 183

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

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Answer: (A) $\dfrac{MgL}{2{n}^{2}}$

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

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