Mass of the part hanging is nM. Centre of mass of this part lies 2nL distance from the surface.
∴ Work done in lifting the hanging part to surface is
W=(nM)⋅g(2nL)
=2n2MgL
JEE Main 2019 — Physics Mechanics
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:
Held on 9 Apr 2019 · Verified 6 Jul 2026.
2n2MgL
n2MgL
nMgL
n22MgL
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