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Question:

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

MgL/n2

nMgL

MgL2/n2

2MgL/n2

Solution:

Correct option is B. MgL/2n2
Mass of the hanging part = M/n
COM = L/2n
Work done W = mghCOM = (M/n)g(L/2n) = MgL/2n2