JEE Main 2019PhysicsWork Power EnergyHardMCQ

JEE Main 2019Work Power Energy Question with Solution

JEE Main 2019 (09 Apr Shift 1)

Question

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

Choose an option

Show full solutionCorrect option: A
Correct answer
AMgL2n2

Step-by-step explanation

Mass of the part hanging is Mn. Centre of mass of this part lies L2n distance from the surface.
solution
Work done in lifting the hanging part to surface is
W=Mn·gL2n
=MgL2n2

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About this question

This is a previous-year question from JEE Main 2019, covering the Work Power Energy chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.