A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its ((1/n))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 :

Question

A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its ((1/n))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) (MgL/n2) (B) (MgL/2n2) (C) (2MgL/n2) (D) nMgL

Tardigrade Admin · Accepted Answer

Mass of the hanging part = (M/n) hCOM = (L/2n) work done W = mghCOM = ((M/n) ) g ((L/2n)) = (MgL/2n2) Option (2)

Q. A uniform cable of mass $^{'} M^{'}$ and length $^{'} L^{'}$ is placed on a horizontal surface such that its $(\frac{1}{n})^{t h}$ 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 :

$\frac{M gL}{n ^{2}}$

$\frac{M gL}{2 n ^{2}}$

$\frac{2 M gL}{n ^{2}}$

$n M gL$

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

Q. A uniform cable of mass ′M′ and length ′L′ is placed on a horizontal surface such that its (n1​)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 :

n2MgL​

2n2MgL​

n22MgL​

nMgL